# Exercise 4.5 Solutions

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Exercise 4.5 Solutions
Find adjoint of each of the matrices in Exercises 1 and 2.
1. {\begin{bmatrix} 1 & 2 \\[5pt] 3 & 4 \end{bmatrix}}
The adjoint of a 2 × 2 matrix can be found by using any of the following two methods.
Method 1: In this method, we find the adjoint of the given 2 x 2 matrix by performing the following steps:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{1}^{\textbf{\large{Interchange}}}} & & 2 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ 3 & & \textcolor{green}{\underbrace{4}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} 4 & 2 \\[5pt] 3 & 1 \end{bmatrix}}
● Change the sign of the elements in the secondary diagonal {\begin{bmatrix} 4 & & \textcolor{red}{\overbrace{2}^{\textbf{\large{Change Sign}}}} \\[10pt] & \textcolor{red}{\Huge⤢} & \\ \textcolor{red}{\underbrace{3}_{\textbf{\large{Change Sign}}}} & & 1 \end{bmatrix} = \begin{bmatrix} 4 & -2 \\[5pt] -3 & 1 \end{bmatrix}}
The resulting matrix will be the adjoint of the given matrix.
∴ The adjoint of the matrix {\begin{bmatrix} 1 & 2 \\[5pt] 3 & 4 \end{bmatrix}} is {\begin{bmatrix} 4 & -2 \\[5pt] -3 & 1 \end{bmatrix}.}
Method 2: In this method, we find the adjoint of the given 2 × 2 matrix by performing the following steps.
i.
Find the Minor of each element {a_{ij}}.
ii.
Find the Cofactor of each element {a_{ij}} which will be {A_{ij}}
iii.
Find the Transpose the Cofactor matrix to get the adjoint of the given matrix.
Let’s consider the given matrix as
{\text{A} = \begin{bmatrix} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} 1 & 2 \\[5pt] 3 & 4 \end{bmatrix}}
The Minors, Cofactors are calculated as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
Cofactor
{\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij}}
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{4}\end{vmatrix}}
4
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1} × 4} = 4
{a_{12} = 2}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{4}}\end{vmatrix}}
3
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2} × 3} = -3
{a_{21} = 3}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{4}}\end{vmatrix}}
2
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1} × 2} = -2
{a_{22} = 4}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{4}}\end{vmatrix}}
1
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2} × 1} = 1
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} \\[5pt] \text{A}_{12} & \text{A}_{22} \end{bmatrix}}
{= \begin{bmatrix} 4 & -2 \\[5pt] -3 & 1 \end{bmatrix}}
∴ The adjoint of the matrix {\begin{bmatrix} 1 & 2 \\[5pt] 3 & 4 \end{bmatrix}} is {\begin{bmatrix} 4 & -2 \\[5pt] -3 & 1 \end{bmatrix}.}
Find adjoint of each of the matrices in Exercises 1 and 2.
2. {\begin{bmatrix} 1 & -1 & 2 \\[5pt] 3 & 0 & -2 \\[5pt] 1 & 0 & 3 \end{bmatrix}}
we find the adjoint of the given 2 × 2 matrix by performing the following steps.
i.
Find the Minor of each element {a_{ij}} of the given matrix.
ii.
Find the Cofactor of each element {a_{ij}} which will be {A_{ij}.}
iii.
Find the Transpose the Cofactor matrix to get the adjoint of the given matrix.
Let’s consider the given matrix as
{\text{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\[5pt] a_{21} & a_{22} & a_{23} \\[5pt] a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} 1 & -1 & 2 \\[5pt] 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}}
The Minors are calculated as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{3} & \textcolor{green}{5} \\[5pt] \textcolor{red}{\cancel{-2}} & \textcolor{green}{0} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 3 & 5 \\[5pt] 0 & 1 \end{vmatrix} = 3(1) - 5(0) = 3 - 0 = 3}
{a_{12} = 0}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{3}} & \textcolor{green}{5} \\[5pt] \textcolor{green}{-2} & \textcolor{red}{\cancel{0}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 2 & 5 \\[5pt] -2 & 1 \end{vmatrix} = 2(1) - 5(-2) = 2 + 10 = 12}
{a_{13} = 0}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{2} & \textcolor{green}{3} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{green}{-2} & \textcolor{green}{0} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 3 \\[5pt] -2 & 0 \end{vmatrix} = 2(0) - 3(-2) = 0 + 6 = 6}
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{red}{\cancel{-2}} & \textcolor{green}{0} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} -1 & 2 \\[5pt] 0 & 1 \end{vmatrix} = (-1)1 - 2(0) = -1 - 0 = -1}
{a_{22} = 1}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{green}{-2} & \textcolor{red}{\cancel{0}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] -2 & 1 \end{vmatrix} = 1(1) - 2(-2) = 1 + 4 = 5}
{a_{23} = 0}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{green}{-2} & \textcolor{green}{0} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] -2 & 0 \end{vmatrix} = 1(0) - (-1)(-2) = 0 - 2 = -2}
{a_{31} = 0}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{3} & \textcolor{green}{5} \\[5pt] \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} -1 & 2 \\[5pt] 3 & 5 \end{vmatrix} = (-1)5 - 2(3) = -5 - 6 = -11}
{a_{32} = 0}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{2} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{3}} & \textcolor{green}{5} \\[5pt] \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] 2 & 5 \end{vmatrix} = 1(5) - 2(2) = 5 - 4 = 1}
{a_{33} = 1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{2} & \textcolor{green}{3} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 2 & 3 \end{vmatrix} = 1(3) - (-1)2 = 3 + 2 = 5}
Now, the Cofactors are calculated as follows:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 3}
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = 1(3) = 3}
{a_{12} = -1}
{\text{M}_{12} = 12}
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)12 = -12}
{a_{13} = 2}
{\text{M}_{13} = 6}
{\text{A}_{13} = (-1)^{1 + 3}\text{M}_{13} = 1(6) = 6}
{a_{21} = 2}
{\text{M}_{21} = -3}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)(-1) = 1}
{a_{22} = 3}
{\text{M}_{22} = 5}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(5) = 5}
{a_{23} = 5}
{\text{M}_{23} = -2}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)(-2) = 2}
{a_{31} = -2}
{\text{M}_{31} = -11}
{\text{A}_{31} = (-1)^{3 + 1}\text{M}_{31} = 1(-11) = -11}
{a_{32} = 0}
{\text{M}_{32} = 1}
{\text{A}_{32} = (-1)^{3 + 2}\text{M}_{32} = (-1)1 = -1}
{a_{33} = 1}
{\text{M}_{33} = 5}
{\text{A}_{33} = (-1)^{3 + 3}\text{M}_{33} = 1(5) = 5}
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
{= \begin{bmatrix} 3 & 1 & -11 \\[5pt] -12 & 5 & -1 \\[5pt] 6 & 2 & 5 \end{bmatrix}}
∴ The adjoint of the matrix {\begin{bmatrix} 1 & -1 & 2 \\[5pt] 3 & 0 & -2 \\[5pt] 1 & 0 & 3 \end{bmatrix}} is {\begin{bmatrix} 3 & 1 & -11 \\[5pt] -12 & 5 & -1 \\[5pt] 6 & 2 & 5 \end{bmatrix}.}
Verify {\text{A} (adj \text{ A}) = (adj \text{ A) A = |A| I}} in Exercises 3 and 4
3. {\begin{bmatrix} 2 & 3 \\[5pt] -4 & -6 \end{bmatrix}}
To solve this problem, we need to find.
I.
II.
|A|
I. To find {adj \text{ A}}
The adjoint of the given matrix can be found by using two methods.
Method 1: In this method, we find the adjoint of the given 2 x 2 matrix by performing the following steps:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{2}^{\textbf{\large{Interchange}}}} & & 3 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ -4 & & \textcolor{green}{\underbrace{-6}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} -6 & 3 \\[5pt] -4 & 2 \end{bmatrix}}
● Change the sign of the elements in the secondary diagonal {\begin{bmatrix} -6 & & \textcolor{red}{\overbrace{3}^{\textbf{\large{Change Sign}}}} \\[10pt] & \textcolor{red}{\Huge⤢} & \\ \textcolor{red}{\underbrace{-4}_{\textbf{\large{Change Sign}}}} & & 2 \end{bmatrix} = \begin{bmatrix} -6 & -3 \\[5pt] 4 & 2 \end{bmatrix}}
The resulting matrix will be the adjoint of the given matrix.
∴ The adjoint of the matrix {\begin{bmatrix} 2 & 3 \\[5pt] -4 & -6 \end{bmatrix}} is {\begin{bmatrix} -6 & -3 \\[5pt] 4 & 2 \end{bmatrix}} ———-❶
Method 2: In this method, we find the adjoint of the given 2 × 2 matrix by performing the following steps.
i.
Find the Minor of each element {a_{ij}}.
ii.
Find the Cofactor of each element {a_{ij}} which will be {A_{ij}}
iii.
Find the Transpose the Cofactor matrix to get the adjoint of the given matrix.
Let’s consider the given matrix as
{\text{A} = \begin{bmatrix} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} 2 & 3 \\[5pt] -4 & -6 \end{bmatrix}}
The Minors, Cofactors are calculated as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
Cofactor
{\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij}}
{a_{11} = 2}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{red}{\cancel{-4}} & \textcolor{green}{-6}\end{vmatrix}}
-6
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1}(-6)} = -6
{a_{12} = 3}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{green}{-4} & \textcolor{red}{\cancel{-6}}\end{vmatrix}}
-4
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2}(-4)} = 4
{a_{21} = -4}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{3} \\[5pt] \textcolor{red}{\cancel{-4}} & \textcolor{red}{\cancel{-6}}\end{vmatrix}}
3
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1}(3)} = -3
{a_{22} = -6}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{red}{\cancel{-4}} & \textcolor{red}{\cancel{-6}}\end{vmatrix}}
2
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2}(2)} = 2
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} \\[5pt] \text{A}_{12} & \text{A}_{22} \end{bmatrix}}
{= \begin{bmatrix} -6 & -3 \\[5pt] 4 & 2 \end{bmatrix}} ———-❶
II. To find |A|
|A|
{= \begin{bmatrix} 2 & 3 \\[5pt] -4 & -6 \end{bmatrix}}
= 2(-6) – 3(-4)
= -12 + 12
= 0 ———-❷
Now,
{\text{A} (adj \text{ A})}
{= \begin{bmatrix} 2 & 3 \\[5pt] -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\[5pt] 4 & 2 \end{bmatrix}} (from ❶)
{= \begin{bmatrix} 2(-6) + 3(4) & 2(-3) + 3(2) \\[5pt] (-4)(-6) + (-6)4 & (-4)(-3) + (-6)2 \end{bmatrix}}
{= \begin{bmatrix} -12 + 12 & -6 + 6 \\[5pt] 24 - 24 & 12 + -12 \end{bmatrix}}
{= \begin{bmatrix} 0 & 0 \\[5pt] 0 & 0 \end{bmatrix}}
= O
Similarly,
{(adj \text{ A})} A
{= \begin{bmatrix} -6 & -3 \\[5pt] 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\[5pt] -4 & -6 \end{bmatrix}} (from ❶)
{= \begin{bmatrix} (-6)2 + (-3)(-4) & (-6)3 + (-3)(-6) \\[5pt] 4(2) + 2(-4) & 4(3) + 2(-6) \end{bmatrix}}
{= \begin{bmatrix} -12 + 12 & -18 + 18 \\[5pt] 8 - 8 & 12 + 12 \end{bmatrix}}
{= \begin{bmatrix} 0 & 0 \\[5pt] 0 & 0 \end{bmatrix}}
= O
Now,
|A|I
{= 0 × \begin{bmatrix} 1 & 0 \\[5pt] 0 & 1 \end{bmatrix}} (from ❷)
{= \begin{bmatrix} 0 & 0 \\[5pt] 0 & 0 \end{bmatrix}}
= O
∴ It is verified that {\text{A} (adj \text{ A}) = (adj \text{ A}) \text{A} = \text{|A|I}}
Verify {\text{A} (adj \text{ A}) = (adj \text{ A) A = |A| I}} in Exercises 3 and 4
4. {\begin{bmatrix} 1 & -1 & 2 \\[5pt] 3 & 0 & -2 \\[5pt] 1 & 0 & 3 \end{bmatrix}}
To solve this problem, we need to find.
I.
II.
|A|
I. To find {adj \text{ A}}
To find {adj \text{ A},} we have to find the Cofactors of the elements from A. However, for finding the Cofactors, we need to find the Minors of the elements of A.
The Minors of the elements of A can be found as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{0} & \textcolor{green}{-2} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 0 & -2 \\[5pt] 0 & 3 \end{vmatrix} = 0(-3) - (-2)0\big) = 0 - 0 = 0}
{a_{12} = -1}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{0}} & \textcolor{green}{-2} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 3 & -2 \\[5pt] 1 & 3 \end{vmatrix} = 3(3) - (-2)1\big) = 9 + 2 = 11}
{a_{13} = 2}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{0} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 3 & 0 \\[5pt] 1 & 0 \end{vmatrix} = 3(0) - 0(1) = 0 - 0 = 0}
{a_{21} = 3}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} -1 & 2 \\[5pt] 0 & 3 \end{vmatrix} = (-1)3 - 2(0) = -3 - 0 = -3}
{a_{22} = 0}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] 1 & 3 \end{vmatrix} = 1(3) - 2(1) = 3 - 2 = 1}
{a_{23} = -2}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 1 & 0 \end{vmatrix} = 1(0) - (-1)1 = 0 + 1 = 1}
{a_{31} = 1}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{0} & \textcolor{green}{-2} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} -1 & 2 \\[5pt] 0 & -2 \end{vmatrix} = (-1)(-2) - 2(0) = 2 - 0 = 2}
{a_{32} = 0}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{2} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{0}} & \textcolor{green}{-2} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] 3 & -2 \end{vmatrix} = 1(-2) - 2(3) = -2 - 6 = -8}
{a_{33} = 3}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{0} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 3 & 0 \end{vmatrix} = 1(0) - (-1)3\big) = 0 + 3 = 3}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 0}
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = 1(0) = 0}
{a_{12} = -1}
{\text{M}_{12} = 11}
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)11 = -11}
{a_{13} = 2}
{\text{M}_{13} = 0}
{\text{A}_{13} = (-1)^{1 + 3}\text{M}_{13} = 1(0) = 0}
{a_{21} = 3}
{\text{M}_{21} = -3}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)(-3) = 3}
{a_{22} = 0}
{\text{M}_{22} = 1}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(1) = 1}
{a_{23} = -2}
{\text{M}_{23} = 1}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)1 = -1}
{a_{31} = 1}
{\text{M}_{31} = 2}
{\text{A}_{31} = (-1)^{3 + 1}\text{M}_{31} = 1(2) = 2}
{a_{32} = 0}
{\text{M}_{32} = -8}
{\text{A}_{32} = (-1)^{3 + 2}\text{M}_{32} = (-1)(-8) = 8}
{a_{33} = 3}
{\text{M}_{33} = 3}
{\text{A}_{33} = (-1)^{3 + 3}\text{M}_{33} = 1(3) = 3}
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
{= \begin{bmatrix} 0 & 3 & 2 \\[5pt] -11 & 1 & 8 \\[5pt] 0 & -1 & 3 \end{bmatrix}} ———-❶
II. To find |A|
|A|
{= \begin{vmatrix} a_{11} & a_{12} & a_{13} \\[5pt] a_{21} & a_{22} & a_{23} \\[5pt] a_{31} & a_{32} & a_{33} \end{vmatrix}}
{= a_{11}\text{A}_{11} + a_{12}\text{A}_{12} + a_{13}\text{A}_{13}}
= 1(0) + (-1)(-11) + 2(0)
= 0 + 11 + 0
= 11 ———-❷
Now
{= \begin{bmatrix} 1 & -1 & 2 \\[5pt] 3 & 0 & -2 \\[5pt] 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\[5pt] -11 & 1 & 8 \\[5pt] 0 & -1 & 3 \end{bmatrix}} (from ❶)
{= \begin{bmatrix} 1(0) + (-1)(-11) + 2(0) & 1(3) + (-1)1 + 2(-1) & 1(2) + (-1)8 + 2(3) \\[5pt] 3(0) + 0(-11) + (-1)0 & 3(3) + 0(1) + (-2)(-1) & 3(2) + 0(8) + (-2)3 \\[5pt] 1(0) + 0(-11) + 3(0) & 1(3) + 0(1) + 3(-1) & 1(2) + 0(8) + 3(3) \end{bmatrix}}
{= \begin{bmatrix} 0 + 11 + 0 & 3 - 1 - 2 & 2 - 8 + 6 \\[5pt] 0 + 0 + 0 & 9 + 0 + 2 & 6 + 0 - 6 \\[5pt] 0 + 0 + 0 & 3 + 0 + -3 & 2 + 0 + 9 \end{bmatrix}}
{= \begin{bmatrix} 11 & 0 & 0 \\[5pt] 0 & 11 & 0 \\[5pt] 0 & 0 & 11 \end{bmatrix}}
{= 11 \begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{bmatrix}}
= |A|I (from ❷) ———-❸
Similarly,
{= \begin{bmatrix} 0 & 3 & 2 \\[5pt] -11 & 1 & 8 \\[5pt] 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\[5pt] 3 & 0 & -2 \\[5pt] 1 & 0 & 3 \end{bmatrix}} (from ❶)
{= \begin{bmatrix} 0(1) + 3(3) + 2(1) & 0(-1) + 0(0) + 2(0) & 0(2) + 3(-2) + 2(3) \\[5pt] (-11)1 + 1(3) + 8(1) & (-11)(-1) + 1(0) + 8(0) & (-11)2 + 1(-2) + 8(3) \\[5pt] 0(1) + (-1)3 + 3(1) & 0(-1) + (-1)0 + 3(0) & 0(2) + (-1)(-2) + 3(3) \end{bmatrix}}
{= \begin{bmatrix} 0 + 9 + 2 & 0 + 0 + 0 & 0 - 6 + 6 \\[5pt] -11 + 3 + 8 & 11 + 0 + 0 & -22 - 2 + 24 \\[5pt] 0 - 3 + 3 & 0 + 0 + 0 & 0 + 2 + 9 \end{bmatrix}}
{= \begin{bmatrix} 11 & 0 & 0 \\[5pt] 0 & 11 & 0 \\[5pt] 0 & 0 & 11 \end{bmatrix}}
{= 11 × \begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{bmatrix}}
= |A|I (from ❷) ———-❹
∴ From ❸ and ❹, it is verified that {\text{A} (adj \text{ A}) = (adj \text{ A}) \text{A} = \text{|A|I}}
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
5. {\begin{bmatrix} 2 & -2 \\[5pt] 4 & 3 \end{bmatrix}}
We know that a square matrix is invertible if and only if it is a nonsingular matrix. So, let’s first check whether the given matrix is nonsingular or not. Now,
|A|
{= \begin{bmatrix} 2 & -2 \\[5pt] 4 & 3 \end{bmatrix}}
= 2(3) – (-2)4
= 6 + 8
= 14 ———-❶
As |A| ≠ 0, A is nonsingular and hence A is invertible.
As we know, {\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A}),} let’s first find {adj \text{ A}}
The adjoint of the given 2 × 2 matrix can be found by using two methods.
Method 1: In this method, we find the adjoint of the given 2 x 2 matrix by performing the following steps:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{2}^{\textbf{\large{Interchange}}}} & & -2 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ 4 & & \textcolor{green}{\underbrace{3}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} 3 & -2 \\[5pt] 4 & 2 \end{bmatrix}}
● Change the sign of the elements in the secondary diagonal {\begin{bmatrix} 3 & & \textcolor{red}{\overbrace{-2}^{\textbf{\large{Change Sign}}}} \\[10pt] & \textcolor{red}{\Huge⤢} & \\ \textcolor{red}{\underbrace{4}_{\textbf{\large{Change Sign}}}} & & 2 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\[5pt] -4 & 2 \end{bmatrix}}
The resulting matrix will be the adjoint of the given matrix.
∴ The adjoint of the matrix {\begin{bmatrix} 2 & -2 \\[5pt] 4 & 3 \end{bmatrix}} is {\begin{bmatrix} 3 & 2 \\[5pt] -4 & 2 \end{bmatrix}} ———-❷
Method 2: In this method, we find the adjoint of the given 2 × 2 matrix by performing the following steps.
i.
Find the Minor of each element {a_{ij}}.
ii.
Find the Cofactor of each element {a_{ij}} which will be {A_{ij}}
iii.
Find the Transpose the Cofactor matrix to get the adjoint of the given matrix.
Let’s consider the given matrix as
{\text{A} = \begin{bmatrix} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} 2 & -2 \\[5pt] 4 & 3 \end{bmatrix}}
The Minors, Cofactors are calculated as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
Cofactor
{\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij}}
{a_{11} = 2}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{green}{3}\end{vmatrix}}
3
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1}(3)} = 3
{a_{12} = -2}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{green}{4} & \textcolor{red}{\cancel{3}}\end{vmatrix}}
4
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2}(4)} = -4
{a_{21} = 4}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{-2} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{3}}\end{vmatrix}}
-2
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1}(-2)} = 2
{a_{22} = 3}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{3}}\end{vmatrix}}
2
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2}(2)} = 2
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} \\[5pt] \text{A}_{12} & \text{A}_{22} \end{bmatrix}}
{= \begin{bmatrix} 3 & 2 \\[5pt] -4 & 2 \end{bmatrix}} ———-❷
Now
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{14}\begin{bmatrix} 3 & 2 \\[5pt] -4 & 2 \end{bmatrix}} (from ❶ and ❷)
∴ The inverse of the matrix {\begin{bmatrix} 2 & -2 \\[5pt] 4 & 3 \end{bmatrix}} is {= \dfrac{1}{14}\begin{bmatrix} 3 & 2 \\[5pt] -4 & 2 \end{bmatrix}}
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
6. {\begin{bmatrix} -1 & 5 \\[5pt] -3 & 2 \end{bmatrix}}
We know that a square matrix is invertible if and only if it is a nonsingular matrix. So, let’s first check whether the given matrix is nonsingular or not. Now,
|A|
{= \begin{bmatrix} -1 & 5 \\[5pt] -3 & 2 \end{bmatrix}}
= (-1)2 – 5(-3)
= -2 + 15
= 13 ———-❶
As |A| ≠ 0, A is nonsingular and hence A is invertible.
As we know, {\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A}),} let’s first find {adj \text{ A}}
The adjoint of the given 2 × 2 matrix can be found by using two methods.
Method 1: In this method, we find the adjoint of the given 2 x 2 matrix by performing the following steps:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{-1}^{\textbf{\large{Interchange}}}} & & 5 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ -3 & & \textcolor{green}{\underbrace{2}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} 2 & 5 \\[5pt] -3 & -1 \end{bmatrix}}
● Change the sign of the elements in the secondary diagonal {\begin{bmatrix} 2 & & \textcolor{red}{\overbrace{5}^{\textbf{\large{Change Sign}}}} \\[10pt] & \textcolor{red}{\Huge⤢} & \\ \textcolor{red}{\underbrace{-3}_{\textbf{\large{Change Sign}}}} & & -1 \end{bmatrix} = \begin{bmatrix} 2 & -5 \\[5pt] 3 & -1 \end{bmatrix}}
The resulting matrix will be the adjoint of the given matrix.
∴ The adjoint of the matrix {\begin{bmatrix} -1 & 5 \\[5pt] -3 & 2 \end{bmatrix}} is {\begin{bmatrix} 2 & -5 \\[5pt] 3 & -1 \end{bmatrix}} ———-❷
Method 2: In this method, we find the adjoint of the given 2 × 2 matrix by performing the following steps.
i.
Find the Minor of each element {a_{ij}}.
ii.
Find the Cofactor of each element {a_{ij}} which will be {A_{ij}}
iii.
Find the Transpose the Cofactor matrix to get the adjoint of the given matrix.
Let’s consider the given matrix as
{\text{A} = \begin{bmatrix} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} 2 & -2 \\[5pt] 4 & 3 \end{bmatrix}}
The Minors, Cofactors are calculated as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
Cofactor
{\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij}}
{a_{11} = -1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{red}{\cancel{-3}} & \textcolor{green}{2} \end{vmatrix}}
2
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1}(2)} = 2
{a_{12} = 5}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{green}{-3} & \textcolor{red}{\cancel{2}} \end{vmatrix}}
-3
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2}(-3)} = 3
{a_{21} = -3}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{-1}} & \textcolor{green}{5} \\[5pt] \textcolor{red}{\cancel{-3}} & \textcolor{red}{\cancel{2}} \end{vmatrix}}
5
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1}(5)} = -5
{a_{22} = 2}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{-1} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{red}{\cancel{-3}} & \textcolor{red}{\cancel{2}} \end{vmatrix}}
-1
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2}(-1)} = -1
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} \\[5pt] \text{A}_{12} & \text{A}_{22} \end{bmatrix}}
{= \begin{bmatrix} 2 & -5 \\[5pt] 3 & -1 \end{bmatrix}} ———-❷
Now
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{13}\begin{bmatrix} 2 & -5 \\[5pt] 3 & -1 \end{bmatrix}} (from ❶ and ❷)
∴ The inverse of the matrix {\begin{bmatrix} -1 & 5 \\[5pt] -3 & 2 \end{bmatrix}} is {= \dfrac{1}{13}\begin{bmatrix} 2 & -5 \\[5pt] 3 & -1 \end{bmatrix}}
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
7. {\begin{bmatrix} 1 & 2 & 3 \\[5pt] 0 & 2 & 4 \\[5pt] 0 & 0 & 5 \end{bmatrix}}
To solve this problem, we need to find.
I.
II.
|A|
I. To find {adj \text{ A}}
To find {adj \text{ A},} we have to find the Cofactors of the elements from A. However, for finding the Cofactors, we need to find the Minors of the elements of A.
The Minors of the elements of A can be found as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{2} & \textcolor{green}{4} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{0} & \textcolor{green}{5} \end{vmatrix}}
{= \begin{vmatrix} 2 & 4 \\[5pt] 0 & 5 \end{vmatrix} = 2(5) - 4(0) = 10 - 0 = 10}
{a_{12} = 2}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{2}} & \textcolor{green}{4} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{0}} & \textcolor{green}{5} \end{vmatrix}}
{= \begin{vmatrix} 0 & 4 \\[5pt] 0 & 5 \end{vmatrix} = 0(5) - 4(0) = 0 - 0 = 0}
{a_{13} = 3}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{2} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{0} & \textcolor{red}{\cancel{5}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 2 \\[5pt] 0 & 0 \end{vmatrix} = 0(0) - 2(0) = 0 - 0 = 0}
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{2} & \textcolor{green}{3} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{0} & \textcolor{green}{5} \end{vmatrix}}
{= \begin{vmatrix} 2 & 3 \\[5pt] 0 & 5 \end{vmatrix} = 2(5) - 3(0) = 10 - 0 = 10}
{a_{22} = 2}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{2}} & \textcolor{green}{3} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{0}} & \textcolor{green}{5} \end{vmatrix}}
{= \begin{vmatrix} 1 & 3 \\[5pt] 0 & 5 \end{vmatrix} = 1(5) - 3(0) = 5 - 0 = 5}
{a_{23} = 4}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{2} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{0} & \textcolor{red}{\cancel{5}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] 0 & 0 \end{vmatrix} = 1(0) - 2(0) = 0 - 0 = 0}
{a_{31} = 0}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{2} & \textcolor{green}{3} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{2} & \textcolor{green}{4} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{5}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 3 \\[5pt] 2 & 4 \end{vmatrix} = 2(4) - 3(2) = 8 - 6 = 2}
{a_{32} = 0}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{2}} & \textcolor{green}{3} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{2}} & \textcolor{green}{4} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{5}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 3 \\[5pt] 0 & 4 \end{vmatrix} = 1(4) - 3(0) = 4 - 0 = 4}
{a_{33} = 5}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{2} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{2} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{5}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] 0 & 2 \end{vmatrix} = 1(2) - 2(0) = 2 - 0 = 2}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 10}
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = 1(10) = 10}
{a_{12} = 2}
{\text{M}_{12} = 0}
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)0 = 0}
{a_{13} = 3}
{\text{M}_{13} = 0}
{\text{A}_{13} = (-1)^{1 + 3}\text{M}_{13} = 1(0) = 0}
{a_{21} = 0}
{\text{M}_{21} = 10}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)10 = -10}
{a_{22} = 2}
{\text{M}_{22} = 5}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(5) = 5}
{a_{23} = 4}
{\text{M}_{23} = 0}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)0 = 0}
{a_{31} = 0}
{\text{M}_{31} = 2}
{\text{A}_{31} = (-1)^{3 + 1}\text{M}_{31} = 1(2) = 2}
{a_{32} = 0}
{\text{M}_{32} = 4}
{\text{A}_{32} = (-1)^{3 + 2}\text{M}_{32} = (-1)4 = -4}
{a_{33} = 5}
{\text{M}_{33} = 2}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1(2) = 2}
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
{= \begin{bmatrix} 10 & -10 & 2 \\[5pt] 0 & 5 & -4 \\[5pt] 0 & 0 & 2 \end{bmatrix}} ———-❶
II. To find |A|
Let’s find the determinant by expanding {\text{C}_1} (as it has more zeroes)
|A|
{= \begin{vmatrix} a_{11} & a_{12} & a_{13} \\[5pt] a_{21} & a_{22} & a_{23} \\[5pt] a_{31} & a_{32} & a_{33} \end{vmatrix}}
{= a_{11}\text{A}_{11} + a_{21}\text{A}_{21} + a_{31}\text{A}_{31}}
= 1(10) + 0(-10) + 0(2)
= 10 + 0 + 0
= 10 ———-❷
As |A| ≠ 0, A is nonsingular and hence A is invertible.
Now
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{10}\begin{bmatrix} 10 & -10 & 2 \\[5pt] 0 & 5 & -4 \\[5pt] 0 & 0 & 2 \end{bmatrix}} (from ❶ and ❷)
∴ The inverse of the matrix {\begin{bmatrix} 1 & 2 & 3 \\[5pt] 0 & 2 & 4 \\[5pt] 0 & 0 & 5 \end{bmatrix}} is {= \dfrac{1}{10}\begin{bmatrix} 10 & -10 & 2 \\[5pt] 0 & 5 & -4 \\[5pt] 0 & 0 & 2 \end{bmatrix}}
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
8. {\begin{bmatrix} 1 & 0 & 0 \\[5pt] 3 & 3 & 0 \\[5pt] 5 & 2 & -1 \end{bmatrix}}
To solve this problem, we need to find.
I.
II.
|A|
I. To find {adj \text{ A}}
To find {adj \text{ A},} we have to find the Cofactors of the elements from A. However, for finding the Cofactors, we need to find the Minors of the elements of A.
The Minors of the elements of A can be found as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{3} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{5}} & \textcolor{green}{2} & \textcolor{green}{-1} \end{vmatrix}}
{= \begin{vmatrix} 3 & 0 \\[5pt] 2 & -1 \end{vmatrix} = 3(-1) - 0(2) = -3 - 0 = -3}
{a_{12} = 0}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{3}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{5} & \textcolor{red}{\cancel{2}} & \textcolor{green}{-1} \end{vmatrix}}
{= \begin{vmatrix} 3 & 0 \\[5pt] 5 & -1 \end{vmatrix} = 3(-1) - 0(5) = -3 - 0 = -3}
{a_{13} = 0}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{3} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{5} & \textcolor{green}{2} & \textcolor{red}{\cancel{-1}} \end{vmatrix}}
{= \begin{vmatrix} 3 & 3 \\[5pt] 5 & 2 \end{vmatrix} = 3(2) - 3(5) = 6 - 15 = -9}
{a_{21} = 3}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{5}} & \textcolor{green}{2} & \textcolor{green}{-1} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] 2 & -1 \end{vmatrix} = 0(-1) - 0(2) = 0 - 0 = 0}
{a_{22} = 3}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{5} & \textcolor{red}{\cancel{2}} & \textcolor{green}{-1} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 5 & -1 \end{vmatrix} = 1(-1) - 0(5) = -1 - 0 = -1}
{a_{23} = 0}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{5} & \textcolor{green}{2} & \textcolor{red}{\cancel{-1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 5 & 2 \end{vmatrix} = 1(2) - 0(5) = 2 - 0 = 2}
{a_{31} = 5}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{3} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{5}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-1}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] 3 & 0 \end{vmatrix} = 0(0) - 0(3) = 0 - 0 = 0}
{a_{32} = 2}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{3}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{5}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 3 & 0 \end{vmatrix} = 1(0) - 0(3) = 0 - 0 = 0}
{a_{33} = -1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{3} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{5}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 3 & 3 \end{vmatrix} = 1(3) - 0(3) = 3 - 0 = 3}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = -3}
{\text{A}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × -3 = -3}
{a_{12} = 0}
{\text{M}_{12} = -3}
{\text{A}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × -3 = 3}
{a_{13} = 0}
{\text{M}_{13} = -9}
{\text{A}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × -9 = -9}
{a_{21} = 3}
{\text{M}_{21} = 0}
{\text{A}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × 0 = 0}
{a_{22} = 3}
{\text{M}_{22} = -1}
{\text{A}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × -1 = -1}
{a_{23} = 0}
{\text{M}_{23} = 2}
{\text{A}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × 2 = -2}
{a_{31} = 5}
{\text{M}_{31} = 0}
{\text{A}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × 0 = 0}
{a_{32} = 2}
{\text{M}_{32} = 0}
{\text{A}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × 0 = 0}
{a_{33} = -1}
{\text{M}_{33} = 3}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × 3 = 3}
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
{= \begin{bmatrix} -3 & 0 & 0 \\[5pt] 3 & -1 & 0 \\[5pt] -9 & -2 & 3 \end{bmatrix}} ———-❶
II. To find |A|
Let’s find the determinant by expanding {\text{R}_1} (as it has more zeroes)
|A|
{= \begin{vmatrix} a_{11} & a_{12} & a_{13} \\[5pt] a_{21} & a_{22} & a_{23} \\[5pt] a_{31} & a_{32} & a_{33} \end{vmatrix}}
{= a_{11}\text{A}_{11} + a_{12}\text{A}_{12} + a_{31}\text{A}_{13}}
= 1(-3) + 0(0) + 0(0)
= -3 + 0 + 0
= -3 ———-❷
As |A| ≠ 0, A is nonsingular and hence A is invertible.
Now
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{-3}\begin{bmatrix} -3 & 0 & 0 \\[5pt] 3 & -1 & 0 \\[5pt] -9 & -2 & 3 \end{bmatrix}} (from ❶ and ❷)
{= \dfrac{-1}{3}\begin{bmatrix} -3 & 0 & 0 \\[5pt] 3 & -1 & 0 \\[5pt] -9 & -2 & 3 \end{bmatrix}}
∴ The inverse of the matrix {\begin{bmatrix} 1 & 0 & 0 \\[5pt] 3 & 3 & 0 \\[5pt] 5 & 2 & -1 \end{bmatrix}} is {= \dfrac{-1}{3}\begin{bmatrix} -3 & 0 & 0 \\[5pt] 3 & -1 & 0 \\[5pt] -9 & -2 & 3 \end{bmatrix}}
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
9. {\begin{bmatrix} 2 & 1 & 3 \\[5pt] 4 & -1 & 0 \\[5pt] -7 & 2 & 1 \end{bmatrix}}
To solve this problem, we need to find.
I.
II.
|A|
I. To find {adj \text{ A}}
To find {adj \text{ A},} we have to find the Cofactors of the elements from A. However, for finding the Cofactors, we need to find the Minors of the elements of A.
The Minors of the elements of A can be found as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{a_{11} = 2}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{green}{-1} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{-7}} & \textcolor{green}{2} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} -1 & 0 \\[5pt] 2 & 1 \end{vmatrix} = (-1)1 - 0(2) = -1 - 0 = -1}
{a_{12} = 1}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{green}{4} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{-7} & \textcolor{red}{\cancel{2}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 4 & 0 \\[5pt] -7 & 1 \end{vmatrix} = 4(1) - 0(-7) = 4 - 0 = 4}
{a_{13} = 3}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{green}{4} & \textcolor{green}{-1} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{-7} & \textcolor{green}{2} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 4 & -1 \\[5pt] -7 & 2 \end{vmatrix} = 4(2) - (-1)(-7) = 8 - 7 = 1}
{a_{21} = 4}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{1} & \textcolor{green}{3} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{-7}} & \textcolor{green}{2} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & 3 \\[5pt] 2 & 1 \end{vmatrix} = 1(1) - 3(2) = 1 - 6 = -5}
{a_{22} = -1}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{1}} & \textcolor{green}{3} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{-7} & \textcolor{red}{\cancel{2}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 2 & 3 \\[5pt] -7 & 1 \end{vmatrix} = 2(1) - 3(-7) = 2 + 21 = 23}
{a_{23} = 0}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{green}{1} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{-7} & \textcolor{green}{2} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 1 \\[5pt] -7 & 2 \end{vmatrix} = 2(2) - 1(-7) = 4 + 7 = 11}
{a_{31} = -7}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{1} & \textcolor{green}{3} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{green}{-1} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{-7}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 3 \\[5pt] -1 & 0 \end{vmatrix} = 1(0) - 3(-1)\big) = 0 + 3 = 3}
{a_{32} = 2}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{1}} & \textcolor{green}{3} \\[5pt] \textcolor{green}{4} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{-7}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 3 \\[5pt] 4 & 0 \end{vmatrix} = 2(0) - 3(4) = 0 - 12 = -12}
{a_{33} = 1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{green}{1} & \textcolor{red}{\cancel{3}} \\[5pt] \textcolor{green}{4} & \textcolor{green}{-1} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{-7}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 1 \\[5pt] 4 & -1 \end{vmatrix} = 2(-1) - 1(4) = -2 - 4 = -6}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 2}
{\text{M}_{11} = -1}
{\text{A}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × -1 = -1}
{a_{12} = 1}
{\text{M}_{12} = 4}
{\text{A}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × 4 = -4}
{a_{13} = 3}
{\text{M}_{13} = 1}
{\text{A}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × 1 = 1}
{a_{21} = 4}
{\text{M}_{21} = -5}
{\text{A}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × -5 = 5}
{a_{22} = -1}
{\text{M}_{22} = 23}
{\text{A}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × 23 = 23}
{a_{23} = 0}
{\text{M}_{23} = 11}
{\text{A}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × 11 = -11}
{a_{31} = -7}
{\text{M}_{31} = 3}
{\text{A}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × 3 = 3}
{a_{32} = 2}
{\text{M}_{32} = -12}
{\text{A}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × -12 = 12}
{a_{33} = 1}
{\text{M}_{33} = -6}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × -6 = -6}
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
{= \begin{bmatrix} -1 & 5 & 3 \\[5pt] -4 & 23 & 12 \\[5pt] 1 & -11 & -6 \end{bmatrix}} ———-❶
II. To find |A|
Let’s find the determinant by expanding {\text{C}_3}
|A|
{= \begin{vmatrix} a_{11} & a_{12} & a_{13} \\[5pt] a_{21} & a_{22} & a_{23} \\[5pt] a_{31} & a_{32} & a_{33} \end{vmatrix}}
{= a_{13}\text{A}_{13} + a_{23}\text{A}_{23} + a_{33}\text{A}_{33}}
= 3(1) + 0(-11) + 1(-6)
= 3 + 0 – 6
= -3 ———-❷
As |A| ≠ 0, A is nonsingular and hence A is invertible.
Now
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{-3}\begin{bmatrix} -1 & 5 & 3 \\[5pt] -4 & 23 & 12 \\[5pt] 1 & -11 & -6 \end{bmatrix}} (from ❶ and ❷)
{= \dfrac{-1}{3}\begin{bmatrix} -1 & 5 & 3 \\[5pt] -4 & 23 & 12 \\[5pt] 1 & -11 & -6 \end{bmatrix}}
∴ The inverse of the matrix {\begin{bmatrix} 2 & 1 & 3 \\[5pt] 4 & -1 & 0 \\[5pt] -7 & 2 & 1 \end{bmatrix}} is {= \dfrac{-1}{3}\begin{bmatrix} -1 & 5 & 3 \\[5pt] -4 & 23 & 12 \\[5pt] 1 & -11 & -6 \end{bmatrix}}
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
10. {\begin{bmatrix} 1 & -1 & 2 \\[5pt] 0 & 2 & -3 \\[5pt] 3 & -2 & 4 \end{bmatrix}}
To solve this problem, we need to find.
I.
II.
|A|
I. To find {adj \text{ A}}
To find {adj \text{ A},} we have to find the Cofactors of the elements from A. However, for finding the Cofactors, we need to find the Minors of the elements of A.
The Minors of the elements of A can be found as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{2} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{-2} & \textcolor{green}{4} \end{vmatrix}}
{= \begin{vmatrix} 2 & -3 \\[5pt] -2 & 4 \end{vmatrix} = 2(4) - (-3)(-2) = 8 - 6 = 2}
{a_{12} = -1}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{2}} & \textcolor{green}{-3} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{-2}} & \textcolor{green}{4} \end{vmatrix}}
{= \begin{vmatrix} 0 & -3 \\[5pt] 3 & 4 \end{vmatrix} = 0(4) - (-3)3 = 0 + 9 = 9}
{a_{13} = 2}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{2} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{-2} & \textcolor{red}{\cancel{4}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 2 \\[5pt] 3 & -2 \end{vmatrix} = 0(-2) - 2(3) = 0 - 6 = -6}
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{-2} & \textcolor{green}{4} \end{vmatrix}}
{= \begin{vmatrix} -1 & 2 \\[5pt] -2 & 4 \end{vmatrix} = (-1)4 - 2(-2) = -4 + 4 = 0}
{a_{22} = 2}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{-2}} & \textcolor{green}{4} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] 3 & 4 \end{vmatrix} = 1(4) - 2(3) = 4 - 6 = -2}
{a_{23} = -3}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{-2} & \textcolor{red}{\cancel{4}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 3 & -2 \end{vmatrix} = 1(-2) - (-1)3 = -2 + 3 = 1}
{a_{31} = 3}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{2} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{4}} \end{vmatrix}}
{= \begin{vmatrix} -1 & 2 \\[5pt] 2 & -3 \end{vmatrix} = (-1)(-3) - 2(2) = 3 - 4 = -1}
{a_{32} = -2}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{2} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{2}} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{4}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] 0 & -3 \end{vmatrix} = 1(-3) - 2(0) = -3 - 0 = -3}
{a_{33} = 4}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{2} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{4}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 0 & 2 \end{vmatrix} = 1(2) - (-1)0 = 2 - 0 = 2}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 2}
{\text{A}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × 2 = 2}
{a_{12} = -1}
{\text{M}_{12} = 9}
{\text{A}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × 9 = -9}
{a_{13} = 2}
{\text{M}_{13} = -6}
{\text{A}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × -6 = -6}
{a_{21} = 0}
{\text{M}_{21} = 0}
{\text{A}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × 0 = 0}
{a_{22} = 2}
{\text{M}_{22} = -2}
{\text{A}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × -2 = -2}
{a_{23} = -3}
{\text{M}_{23} = 1}
{\text{A}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × 1 = -1}
{a_{31} = 3}
{\text{M}_{31} = -1}
{\text{A}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × -1 = -1}
{a_{32} = -2}
{\text{M}_{32} = -3}
{\text{A}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × -3 = 3}
{a_{33} = 4}
{\text{M}_{33} = 2}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × 2 = 2}
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
{= \begin{bmatrix} 2 & 0 & -1 \\[5pt] -9 & -2 & 3 \\[5pt] -6 & -1 & 2 \end{bmatrix}} ———-❶
II. To find |A|
Let’s find the determinant by expanding {\text{C}_1,} we have,
|A|
{= \begin{vmatrix} a_{11} & a_{12} & a_{13} \\[5pt] a_{21} & a_{22} & a_{23} \\[5pt] a_{31} & a_{32} & a_{33} \end{vmatrix}}
{= a_{11}\text{A}_{11} + a_{21}\text{A}_{21} + a_{31}\text{A}_{31}}
= 1(2) + 0(0) + 3(-1)
= 2 + 0 – 3
= -1 ———-❷
As |A| ≠ 0, A is nonsingular and hence A is invertible.
Now
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{-1}\begin{bmatrix} 2 & 0 & -1 \\[5pt] -9 & -2 & 3 \\[5pt] -6 & -1 & 2 \end{bmatrix}} (from ❶ and ❷)
{= -\begin{bmatrix} 2 & 0 & -1 \\[5pt] -9 & -2 & 3 \\[5pt] -6 & -1 & 2 \end{bmatrix}}
{= \begin{bmatrix} -2 & 0 & 1 \\[5pt] 9 & 2 & -3 \\[5pt] 6 & 1 & -2 \end{bmatrix}}
∴ The inverse of the matrix {\begin{bmatrix} 1 & -1 & 2 \\[5pt] 0 & 2 & -3 \\[5pt] 3 & -2 & 4 \end{bmatrix}} is {= \begin{bmatrix} -2 & 0 & 1 \\[5pt] 9 & 2 & -3 \\[5pt] 6 & 1 & -2 \end{bmatrix}}
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
11. {\begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & \cos \text{α} & \sin \text{α} \\[5pt] 0 & \sin \text{α} & -\cos \text{α} \end{bmatrix}}
To solve this problem, we need to find.
I.
II.
|A|
I. To find {adj \text{ A}}
To find {adj \text{ A},} we have to find the Cofactors of the elements from A. However, for finding the Cofactors, we need to find the Minors of the elements of A.
The Minors of the elements of A can be found as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{\cos \text{α}} & \textcolor{green}{\sin \text{α}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{\sin \text{α}} & \textcolor{green}{-\cos \text{α}} \end{vmatrix}}
{= \begin{vmatrix} \cos \text{α} & \sin \text{α} \\[5pt] \sin \text{α} & -\cos \text{α} \end{vmatrix}}
{ = (\cos \text{α})(-\cos \text{α}) - (\sin \text{α})(\sin \text{α})}
{= -\cos^2 \text{α} - \sin^2 \text{α}}
{= -(\cos ^2 \text{α} + \sin ^2 \text{α})}
= -1
{a_{12} = 0}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{\cos \text{α}}} & \textcolor{green}{\sin \text{α}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{\sin \text{α}}} & \textcolor{green}{-\cos \text{α}} \end{vmatrix}}
{= \begin{vmatrix} 0 & \sin \text{α} \\[5pt] 0 & -\cos \text{α} \end{vmatrix}}
{= 0(-\cos \text{α}) - (\sin \text{α})0}
= 0 – 0
= 0
{a_{13} = 0}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{\cos \text{α}} & \textcolor{red}{\cancel{\sin \text{α}}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{\sin \text{α}} & \textcolor{red}{\cancel{-\cos \text{α}}} \end{vmatrix}}
{= \begin{vmatrix} 0 & \cos \text{α} \\[5pt] 0 & \sin \text{α} \end{vmatrix}}
{= 0(\sin \text{α}) - (\cos \text{α})0}
= 0 – 0
= 0
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{\cos \text{α}}} & \textcolor{red}{\cancel{\sin \text{α}}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{\sin \text{α}} & \textcolor{green}{-\cos \text{α}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] \sin \text{α} & -\cos \text{α} \end{vmatrix}}
{= 0(-\cos \text{α}) - 0(\sin \text{α})}
= 0 – 0
= 0
{a_{22} = \cos \text{α}}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{\cos \text{α}}} & \textcolor{red}{\cancel{\sin \text{α}}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{\sin \text{α}}} & \textcolor{green}{-\cos \text{α}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & -\cos \text{α} \end{vmatrix}}
{= 1(-\cos \text{α}) - 0(0)}
{= -\cos \text{α} - 0}
{= -\cos \text{α}}
{a_{23} = \sin \text{α}}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{\cos \text{α}}} & \textcolor{red}{\cancel{\sin \text{α}}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{\sin \text{α}} & \textcolor{red}{\cancel{-\cos \text{α}}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & \sin \text{α} \end{vmatrix}}
{= 1(\sin \text{α}) - 0(0)}
{= \sin \text{α} - 0}
{= \sin \text{α}}
{a_{31} = 0}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{\cos \text{α}} & \textcolor{green}{\sin \text{α}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{\sin \text{α}}} & \textcolor{red}{\cancel{-\cos \text{α}}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] \cos \text{α} & \sin \text{α} \end{vmatrix}}
{= 0(\sin \text{α}) - 0(\cos \text{α})}
= 0 – 0
= 0
{a_{32} = \sin \text{α}}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{\cos \text{α}}} & \textcolor{green}{\sin \text{α}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{\sin \text{α}}} & \textcolor{red}{\cancel{-\cos \text{α}}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & \sin \text{α} \end{vmatrix}}
{= 1(\sin \text{α}) - 0(0)}
{= \sin \text{α} - 0}
{= \sin \text{α}}
{a_{33} = -\cos \text{α}}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{\cos \text{α}} & \textcolor{red}{\cancel{\sin \text{α}}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{\sin \text{α}}} & \textcolor{red}{\cancel{-\cos \text{α}}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & \cos \text{α} \end{vmatrix}}
{= 1(\cos \text{α}) - 0(0)}
{= \cos \text{α} - 0}
{= \cos \text{α}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = -1}
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = 1(-1) = -1}
{a_{12} = 0}
{\text{M}_{12} = 0}
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)0 = 0}
{a_{13} = 0}
{\text{M}_{13} = 0}
{\text{A}_{13} = (-1)^{1 + 3}\text{M}_{13} = 1(0) = 0}
{a_{21} = 0}
{\text{M}_{21} = 0}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)0 = 0}
{a_{22} = \cos \text{α}}
{\text{M}_{22} = -\cos \text{α}}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(-\cos \text{α}) = -\cos \text{α}}
{a_{23} = \sin \text{α}}
{\text{M}_{23} = \sin \text{α}}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)(\sin \text{α}) = -\sin \text{α}}
{a_{31} = 0}
{\text{M}_{31} = 0}
{\text{A}_{31} = (-1)^{3 + 1}\text{M}_{31} = 1(0) = 0}
{a_{32} = \sin \text{α}}
{\text{M}_{32} = \sin \text{α}}
{\text{A}_{32} = (-1)^{3 + 2}\text{M}_{32} = (-1)(\sin \text{α}) = -\sin \text{α}}
{a_{33} = -\cos \text{α}}
{\text{M}_{33} = \cos \text{α}}
{\text{A}_{33} = (-1)^{3 + 3}\text{M}_{33} = 1(\cos \text{α}) = \cos \text{α}}
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
{= \begin{bmatrix} -1 & 0 & 0 \\[5pt] 0 & -\cos \text{α} & -\sin \text{α} \\[5pt] 0 & -\sin \text{α} & \cos \text{α} \end{bmatrix}} ———-❶
II. To find |A|
Let’s find the determinant by expanding {\text{R}_1,} we have,
|A|
{= \begin{vmatrix} a_{11} & a_{12} & a_{13} \\[5pt] a_{21} & a_{22} & a_{23} \\[5pt] a_{31} & a_{32} & a_{33} \end{vmatrix}}
{= a_{11}\text{A}_{11} + a_{12}\text{A}_{12} + a_{13}\text{A}_{13}}
= 1(-1) + 0(0) + 0(0)
= -1 + 0 + 0
= -1 ———-❷
As |A| ≠ 0, A is nonsingular and hence A is invertible.
Now
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{-1}\begin{bmatrix} -1 & 0 & 0 \\[5pt] 0 & -\cos \text{α} & -\sin \text{α} \\[5pt] 0 & -\sin \text{α} & \cos \text{α} \end{bmatrix}} (from ❶ and ❷)
{= -\begin{bmatrix} -1 & 0 & 0 \\[5pt] 0 & -\cos \text{α} & -\sin \text{α} \\[5pt] 0 & -\sin \text{α} & \cos \text{α} \end{bmatrix}}
{= \begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & \cos \text{α} & \sin \text{α} \\[5pt] 0 & \sin \text{α} & -\cos \text{α} \end{bmatrix}}
∴ The inverse of the matrix {\begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & \cos \text{α} & \sin \text{α} \\[5pt] 0 & \sin \text{α} & -\cos \text{α} \end{bmatrix}} is {= \begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & \cos \text{α} & \sin \text{α} \\[5pt] 0 & \sin \text{α} & -\cos \text{α} \end{bmatrix}}
📝Note: In this case, it can be noted that the given matrix is inverse of itself i.e. {\text{A}^{-1} = \text{A}}
12. Let {\text{A} = \begin{bmatrix} 3 & 7 \\[5pt] 2 & 5 \end{bmatrix}} and {\text{B} = \begin{bmatrix} 6 & 8 \\[5pt] 7 & 9 \end{bmatrix},} Verify that {\text{(AB)}^{-1} = \text{B}^{-1}\text{A}^{-1.}}
Given that
{\text{A} = \begin{bmatrix} 3 & 7 \\[5pt] 2 & 5 \end{bmatrix}} and {\text{B} = \begin{bmatrix} 6 & 8 \\[5pt] 7 & 9 \end{bmatrix}}
|A|
{= \begin{vmatrix} 3 & 7 \\[5pt] 2 & 5 \end{vmatrix}}
= 3(5) – 7(2)
= 15 – 14
= 1 ———-❶
|B|
{= \begin{vmatrix} 6 & 8 \\[5pt] 7 & 9 \end{vmatrix}}
= 6(9) – 8(7)
= 54 – 56
= -2 ———-❷
As |A| ≠ 0 and |B| ≠ 0, both A and B are nonsingular and hence invertible i.e. {\text{A}^{-1}} and {\text{B}^{-1}} exist.
Now,
AB
{= \begin{bmatrix} 2 & 7 \\[5pt] 2 & 5 \end{bmatrix} \begin{bmatrix} 6 & 8 \\[5pt] 7 & 9 \end{bmatrix}}
{= \begin{bmatrix} 3(6) + 7(7) & 3(8) + 7(9) \\[5pt] 2(6) + 5(7) & 2(8) + 5(9) \end{bmatrix}}
{= \begin{bmatrix} 18 + 49 & 24 + 63 \\[5pt] 12 + 35 & 16 + 45 \end{bmatrix}}
{= \begin{bmatrix} 67 & 87 \\[5pt] 47 & 61 \end{bmatrix}}
Now,
|AB|
{= \begin{vmatrix} 67 & 87 \\[5pt] 47 & 61 \end{vmatrix}}
= 67(61) – 87(47)
= 4087 – 4089
= -2 ———-❸
As |AB| ≠ 0, AB is nonsingular and hence AB is invertible i.e. {\text{(AB)}^{-1}} exists.
Now, we find the adjoints of A, B and AB so that we can proceed with finding their inverses.
To find {adj \text{ A}}
Method I:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{3}^{\textbf{\large{Interchange}}}} & & 7 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ 2 & & \textcolor{green}{\underbrace{5}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} 5 & 7 \\[5pt] 2 & 3 \end{bmatrix}}
● Change the sign of the elements in the secondary diagonal {\begin{bmatrix} 5 & & \textcolor{red}{\overbrace{7}^{\textbf{\large{Change Sign}}}} \\[10pt] & \textcolor{red}{\Huge⤢} & \\ \textcolor{red}{\underbrace{2}_{\textbf{\large{Change Sign}}}} & & 3 \end{bmatrix} = \begin{bmatrix} 5 & -7 \\[5pt] -2 & 3 \end{bmatrix}}
The resulting matrix will be the adjoint of the given matrix.
∴ The adjoint of the matrix {\begin{bmatrix} 3 & 7 \\[5pt] 2 & 5 \end{bmatrix}} is {\begin{bmatrix} 5 & -7 \\[5pt] -2 & 3 \end{bmatrix}.} ———-❹
Method 2: In this method, we find the adjoint of the given 2 × 2 matrix by performing the following steps.
i.
Find the Minor of each element {a_{ij}}.
ii.
Find the Cofactor of each element {a_{ij}} which will be {\text{A}_{ij}}
iii.
Find the Transpose the Cofactor matrix to get the adjoint of the given matrix.
Let’s consider the given matrix as
{\text{A} = \begin{bmatrix} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} 3 & 7 \\[5pt] 2 & 5 \end{bmatrix}}
The Minors, Cofactors are calculated as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
Cofactor
{\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij}}
{a_{11} = 3}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{7}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{5}\end{vmatrix}}
5
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1}(4)} = 5
{a_{12} = 7}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{7}} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{5}}\end{vmatrix}}
2
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2}(3)} = -2
{a_{21} = 2}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{3}} & \textcolor{green}{7} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{5}}\end{vmatrix}}
7
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1}(2)} = -7
{a_{22} = 5}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{3} & \textcolor{red}{\cancel{7}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{5}}\end{vmatrix}}
3
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2}(3)} = 1
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} \\[5pt] \text{A}_{12} & \text{A}_{22} \end{bmatrix}}
{= \begin{bmatrix} 5 & -7 \\[5pt] -2 & 3 \end{bmatrix}}
∴ The adjoint of the matrix {\begin{bmatrix} 3 & 7 \\[5pt] 2 & 5 \end{bmatrix}} is {\begin{bmatrix} 5 & -7 \\[5pt] -2 & 3 \end{bmatrix}.} ———-❹
To find {adj \text{ B}}
Method I:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{6}^{\textbf{\large{Interchange}}}} & & 8 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ 7 & & \textcolor{green}{\underbrace{9}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} 9 & 8 \\[5pt] 7 & 6 \end{bmatrix}}
● Change the sign of the elements in the secondary diagonal {\begin{bmatrix} 9 & & \textcolor{red}{\overbrace{8}^{\textbf{\large{Change Sign}}}} \\[10pt] & \textcolor{red}{\Huge⤢} & \\ \textcolor{red}{\underbrace{7}_{\textbf{\large{Change Sign}}}} & & 6 \end{bmatrix} = \begin{bmatrix} 9 & -8 \\[5pt] -7 & 6 \end{bmatrix}}
The resulting matrix will be the adjoint of the given matrix.
∴ The adjoint of the matrix {\begin{bmatrix} 6 & 8 \\[5pt] 7 & 9 \end{bmatrix}} is {\begin{bmatrix} 9 & -8 \\[5pt] -7 & 6 \end{bmatrix}.} ———-❺
Method 2: In this method, we find the adjoint of the given 2 × 2 matrix by performing the following steps.
i.
Find the Minor of each element {b_{ij}}.
ii.
Find the Cofactor of each element {b_{ij}} which will be {\text{B}_{ij}}
iii.
Find the Transpose the Cofactor matrix to get the adjoint of the given matrix.
Let’s consider the given matrix as
{\text{B} = \begin{bmatrix} b_{11} & b_{12} \\[5pt] b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 6 & 8 \\[5pt] 7 & 9 \end{bmatrix}}
The Minors, Cofactors are calculated as below:
Element
{b_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
Cofactor
{\text{B}_{ij} = (-1)^{i + j}\text{M}_{ij}}
{b_{11} = 6}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{6}} & \textcolor{red}{\cancel{8}} \\[5pt] \textcolor{red}{\cancel{7}} & \textcolor{green}{9}\end{vmatrix}}
9
{\text{B}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1}(9)} = 9
{b_{12} = 8}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{6}} & \textcolor{red}{\cancel{8}} \\[5pt] \textcolor{green}{7} & \textcolor{red}{\cancel{9}}\end{vmatrix}}
7
{\text{B}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2}(7)} = -7
{b_{21} = 7}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{6}} & \textcolor{green}{8} \\[5pt] \textcolor{red}{\cancel{7}} & \textcolor{red}{\cancel{9}}\end{vmatrix}}
8
{\text{B}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1}(8)} = -8
{b_{22} = 9}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{6} & \textcolor{red}{\cancel{8}} \\[5pt] \textcolor{red}{\cancel{7}} & \textcolor{red}{\cancel{9}}\end{vmatrix}}
6
{\text{B}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2}(6)} = 6
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{B}_{11} & \text{A}_{21} \\[5pt] \text{A}_{12} & \text{A}_{22} \end{bmatrix}}
{= \begin{bmatrix} 9 & -8 \\[5pt] -7 & 6 \end{bmatrix}}
∴ The adjoint of the matrix {\begin{bmatrix} 6 & 8 \\[5pt] 7 & 9 \end{bmatrix}} is {\begin{bmatrix} 9 & -8 \\[5pt] -7 & 6 \end{bmatrix}.} ———-❺
To find {adj \text{ AB}}
Method I:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{67}^{\textbf{\large{Interchange}}}} & & 87 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ 47 & & \textcolor{green}{\underbrace{61}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} 61 & 87 \\[5pt] 47 & 67 \end{bmatrix}}
● Change the sign of the elements in the secondary diagonal {\begin{bmatrix} 61 & & \textcolor{red}{\overbrace{87}^{\textbf{\large{Change Sign}}}} \\[10pt] & \textcolor{red}{\Huge⤢} & \\ \textcolor{red}{\underbrace{47}_{\textbf{\large{Change Sign}}}} & & 67 \end{bmatrix} = \begin{bmatrix} 61 & -87 \\[5pt] -47 & 67 \end{bmatrix}}
The resulting matrix will be the adjoint of the given matrix.
∴ The adjoint of the matrix {\begin{bmatrix} 67 & 87 \\[5pt] 47 & 61 \end{bmatrix}} is {\begin{bmatrix} 61 & -87 \\[5pt] -47 & 67 \end{bmatrix}.} ———-❻
Method 2: In this method, we find the adjoint of the given 2 × 2 matrix by performing the following steps.
i.
Find the Minor of each element {c_{ij}}.
ii.
Find the Cofactor of each element {c_{ij}} which will be {\text{AB}_{ij}}
iii.
Find the Transpose the Cofactor matrix to get the adjoint of the given matrix.
Let’s consider the given matrix as
{\text{AB} = \begin{bmatrix} c_{11} & c_{12} \\[5pt] c_{21} & c_{22} \end{bmatrix} = \begin{bmatrix} 67 & 87 \\[5pt] 47 & 61 \end{bmatrix}}
The Minors, Cofactors are calculated as below:
Element
{c_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
Cofactor
{\text{AB}_{ij} = (-1)^{i + j}\text{M}_{ij}}
{c_{11} = 67}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{67}} & \textcolor{red}{\cancel{87}} \\[5pt] \textcolor{red}{\cancel{47}} & \textcolor{green}{61}\end{vmatrix}}
9
{\text{AB}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1}(9)} = 9
{c_{12} = 87}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{67}} & \textcolor{red}{\cancel{87}} \\[5pt] \textcolor{green}{47} & \textcolor{red}{\cancel{61}}\end{vmatrix}}
7
{\text{AB}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2}(7)} = -7
{c_{21} = 47}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{67}} & \textcolor{green}{87} \\[5pt] \textcolor{red}{\cancel{47}} & \textcolor{red}{\cancel{61}}\end{vmatrix}}
8
{\text{AB}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1}(8)} = -8
{c_{22} = 61}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{67} & \textcolor{red}{\cancel{87}} \\[5pt] \textcolor{red}{\cancel{47}} & \textcolor{red}{\cancel{61}}\end{vmatrix}}
6
{\text{AB}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2}(6)} = 6
Now, the adjoint of the matrix A is calculated as below:
{= \begin{bmatrix} \text{AB}_{11} & \text{AB}_{21} \\[5pt] \text{AB}_{12} & \text{AB}_{22} \end{bmatrix}}
{= \begin{bmatrix} 61 & -87 \\[5pt] -47 & 67 \end{bmatrix}}
∴ The adjoint of the matrix {\begin{bmatrix} 67 & 87 \\[5pt] 47 & 61 \end{bmatrix}} is {\begin{bmatrix} 61 & -87 \\[5pt] -47 & 67 \end{bmatrix}.} ———-❻
Now,
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac11\begin{bmatrix} 5 & -7 \\[5pt] -2 & 3 \end{bmatrix}}
{= \begin{bmatrix} 5 & -7 \\[5pt] -2 & 3 \end{bmatrix}}
{\text{B}^{-1}}
{= \dfrac{1}{\text{|B|}}(adj \text{ B})}
{= \dfrac{1}{-2}\begin{bmatrix} 9 & -8 \\[5pt] -7 & 6 \end{bmatrix}}
{= \dfrac{-1}{2}\begin{bmatrix} 9 & -8 \\[5pt] -7 & 6 \end{bmatrix}}
To prove that {\text{(AB)}^{-1} = \text{B}^{-1}\text{A}^{-1}}
L.H.S.
{= \text{(AB)}^{-1}}
{= \dfrac{1}{\text{|AB|}}(adj \text{ AB})}
{= \dfrac{1}{-2}\begin{bmatrix} 61 & -87 \\[5pt] -47 & 67 \end{bmatrix}}
{= \dfrac{-1}{2}\begin{bmatrix} 61 & -87 \\[5pt] -47 & 67 \end{bmatrix}}
R.H.S.
{= \text{B}^{-1}\text{A}^{-1}}
{= \dfrac{-1}{2}\begin{bmatrix} 9 & -8 \\[5pt] -7 & 6 \end{bmatrix} \begin{bmatrix} 5 & -7 \\[5pt] -2 & 3 \end{bmatrix}}
{= \dfrac{-1}{2}\begin{bmatrix} 9(5) + (-8)(-2) & 9(-7) + (-8)3 \\[5pt] (-7)5 + 6(-2) & (-7)(-7) + 6(3) \end{bmatrix}}
{= \dfrac{-1}{2}\begin{bmatrix} 45 + 16 & -63 - 24 \\[5pt] -35 - 12 & 49 + 18 \end{bmatrix}}
{= \dfrac{-1}{2}\begin{bmatrix} 61 & -87 \\[5pt] -47 & 67 \end{bmatrix}}
∴ As L.H.S. = R.H.S., it is proved that {\text{(AB)}^{-1} = \text{A}^{-1}\text{B}^{-1}}
13. If {\text{A} = \begin{bmatrix} 3 & 1 \\[5pt] -1 & 2 \end{bmatrix},} show that {\text{A}^2 - 5\text{A} + 7\text{I = O.}} Hence find {\text{A}^{-1}.}
We have,
{\text{A}^2}
= A.A.
{= \begin{bmatrix} 3 & 1 \\[5pt] -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\[5pt] -1 & 2 \end{bmatrix}}
{= \begin{bmatrix} 3(3) + 1(-1) & 3(1) + 1(2) \\[5pt] (-1)3 + 2(-1) & (-1)1 + 2(2) \end{bmatrix}}
{= \begin{bmatrix} 9 - 1 & 3 + 2 \\[5pt] -3 - 2 & -1 + 4 \end{bmatrix}}
{= \begin{bmatrix} 8 & 5 \\[5pt] -5 & 3 \end{bmatrix}}
Now,
L.H.S.
{= \text{A}^2 - 5\text{A} + 7\text{I}}
{= \begin{bmatrix} 8 & 5 \\[5pt] -5 & 3 \end{bmatrix} - 5 \begin{bmatrix} 3 & 1 \\[5pt] -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\[5pt] 0 & 1 \end{bmatrix}}
{= \begin{bmatrix} 8 & 5 \\[5pt] -5 & 3 \end{bmatrix} - \begin{bmatrix} 5(3) & 5(1) \\[5pt] 5(-1) & 5(2) \end{bmatrix} + \begin{bmatrix} 7(1) & 7(0) \\[5pt] 7(0) & 7(1) \end{bmatrix}}
{= \begin{bmatrix} 8 & 5 \\[5pt] -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\[5pt] -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\[5pt] 0 & 7 \end{bmatrix}}
{= \begin{bmatrix} 8 - 15 + 7 & 5 - 5 + 0 \\[5pt] -5 - (-5) + 0 & 3 - 10 + 7 \end{bmatrix}}
{= \begin{bmatrix} 0 & 0 \\[5pt] 0 & 0 \end{bmatrix}}
= O
= R.H.S.
∴ It is proved that {\text{A}^2 - 5\text{A} + 7\text{I} = \text{O}}
Now, we have
{\text{A}^2 - 5\text{A} + 7\text{I} = \text{O}}
{⇒ 7\text{I} = -\text{A}^2 + 5\text{A}}
⇒ 7I = -A.A + 5.A
Post multiplying both sides with {\text{A}^{-1}}
{7\text{I.A}^{-1}}
{= -\text{A.A(A}^{-1}) + 5\text{A.A}^{-1}}
{⇒ 7\text{A}^{-1}}
{= -\text{A.(AA}^{-1}) + 5\text{I}}
{= -\text{A.I} + 5\text{I}}
= -A + 5I
{= -\begin{bmatrix} 3 & 1 \\[5pt] -1 & 2 \end{bmatrix} + 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}}
{= \begin{bmatrix} -3 & -1 \\[5pt] 1 & -2 \end{bmatrix} + \begin{bmatrix} 5(1) & 5(0) \\ 5(0) & 5(1) \end{bmatrix}}
{= \begin{bmatrix} -3 + 5 & -1 + 0 \\[5pt] 1 + 0 & -2 + 5 \end{bmatrix}}
{= \begin{bmatrix} 2 & -1 \\[5pt] 1 & 3 \end{bmatrix}}
{⇒ \text{A}^{-1}}
{= \dfrac17\begin{bmatrix} 2 & -1 \\[5pt] 1 & 3 \end{bmatrix}}
∴ We have {= \dfrac17\begin{bmatrix} 2 & -1 \\[5pt] 1 & 3 \end{bmatrix}}
14. For the matrix {\text{A} = \begin{bmatrix} 3 & 2 \\[5pt] 1 & 1 \end{bmatrix},} find the numbers a and b such that {\text{A}^2 + a\text{A} + b\text{I} = O.}
Given that
{\text{A} = \begin{bmatrix} 3 & 2 \\[5pt] 1 & 1 \end{bmatrix}}
Now,
{\text{A}^2}
= A.A
{= \begin{bmatrix} 3 & 2 \\[5pt] 1 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \\[5pt] 1 & 1 \end{bmatrix}}
{= \begin{bmatrix} 3(3) + 2(1) & 3(2) + 2(1) \\[5pt] 1(3) + 1(1) & 1(2) + 1(1) \end{bmatrix}}
{= \begin{bmatrix} 9 + 2 & 6 + 2 \\[5pt] 3 + 1 & 2 + 1 \end{bmatrix}}
{= \begin{bmatrix} 11 & 8 \\[5pt] 4 & 3 \end{bmatrix}}
{a\text{A}}
{= a \begin{bmatrix} 3 & 2 \\[5pt] 1 & 1 \end{bmatrix}}
{= \begin{bmatrix} 3a & 2a \\[5pt] a & a \end{bmatrix}}
{bI}
{= b \begin{bmatrix} 1 & 0 \\[5pt] 1 & 0 \end{bmatrix}}
{=\begin{bmatrix} b & 0 \\[5pt] b & 0 \end{bmatrix}}
Substituting into the given equation {\text{A}^2 + a\text{A} + b\text{I = O}}, we have,
{\begin{bmatrix} 11 & 8 \\[5pt] 4 & 3 \end{bmatrix} + \begin{bmatrix} 3a & 2a \\[5pt] a & a \end{bmatrix} + \begin{bmatrix} b & 0 \\[5pt] 0 & b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\[5pt] 0 & 0 \end{bmatrix}}
{⇒ \begin{bmatrix} 11 + 3a + b & 8 + 2a + 0 \\[5pt] 4 + a + 0 & 3 + a + b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\[5pt] 0 & 0 \end{bmatrix}}
{⇒ \begin{bmatrix} 11 + 3a + b & 8 + 2a \\[5pt] 4 + a & 3 + a + b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\[5pt] 0 & 0 \end{bmatrix}}
As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
{11 + 3a + b}
= 0 ———-❶
{8 + 2a}
= 0 ———-❷
{4 + a}
= 0 ———-❸
{3 + a + b}
= 0 ———-❹
Solving ❸, we have,
4 + a = 0
⇒ a = -4 ———-❺
Substituting ❺ in ❹, we have,
{3 + (-4) + b = 0}
{⇒ -1 + b = 0}
{⇒ b = 1}
∴ We have, {a = -4} and {b = 1}
15. For the matrix {\text{A} = \begin{bmatrix} 1 & 1 & 1 \\[5pt] 1 & 2 & -3 \\[5pt] 2 & -1 & 3 \end{bmatrix}}
Show that {\text{A}^3 - 6\text{A}^2 + 5\text{A} + \text{11I = O}.} Hence find {\text{A}^{-1}.}
To solve this problem, we’ll first find {\text{A}^2} and then using that, we’ll find {\text{A}^3.}
{\text{A}^2}
= A.A
{= \begin{bmatrix} 1 & 1 & 1 \\[5pt] 1 & 2 & -3 \\[5pt] 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\[5pt] 1 & 2 & -3 \\[5pt] 2 & -1 & 3 \end{bmatrix}}
{= \begin{bmatrix} 1(1) + 1(1) + 1(2) & 1(1) + 1(2) + 1(-1) & 1(1) + 1(-3) + 1(3) \\[5pt] 1(1) + 2(1) + (-3)2 & 1(1) + 2(2) + (-3)(-1) & 1(1) + 2(-3) + (-3)3 \\[5pt] 2(1) + (-1)1 + 3(2) & 2(1) + (-1)2 + 3(-1) & 2(1) + (-1)(-3) + 3(3) \end{bmatrix}}
{= \begin{bmatrix} 1 + 1 + 2 & 1 + 2 - 1 & 1 - 3 + 3 \\[5pt] 1 + 2 - 6 & 1 + 4 + 3 & 1 - 6 - 9 \\[5pt] 2 - 1 + 6 & 2 - 2 - 3 & 2 + 3 + 9 \end{bmatrix}}
{= \begin{bmatrix} 4 & 2 & 1 \\[5pt] -3 & 8 & -14 \\[5pt] 7 & -3 & 14 \end{bmatrix}} ———-❶
Now,
{\text{A}^3}
{= \text{A}^2.\text{A}}
{= \begin{bmatrix} 4 & 2 & 1 \\[5pt] -3 & 8 & -14 \\[5pt] 7 & -3 & 14 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\[5pt] 1 & 2 & -3 \\[5pt] 2 & -1 & 3 \end{bmatrix}} (from ❶)
{= \begin{bmatrix} 4(1) + 2(1) + 1(2) & 4(1) + 2(2) + 1(-1) & 4(1) + 2(-3) + 1(3) \\[5pt] (-3)1 + 8(1) + (-14)2 & (-3)1 + 8(2) + (-14)(-1) & (-3)1 + 8(-3) + (-14)3 \\[5pt] 7(1) + (-3)1 + 14(2) & 7(1) + (-3)2 + 14(-1) & 7(1) + (-3)(-3) + 14(3) \end{bmatrix}}
{= \begin{bmatrix} 4 + 2 + 2 & 4 + 4 - 1 & 4 - 6 + 3 \\[5pt] -3 + 8 - 28 & -3 + 16 + 14 & -3 - 24 - 42 \\[5pt] 7 - 3 + 28 & 7 - 6 - 14 & 7 + 9 + 42 \end{bmatrix}}
{= \begin{bmatrix} 8 & 7 & 1 \\[5pt] -23 & 27 & -69 \\[5pt] 32 & -13 & 58 \end{bmatrix}} ———-❷
Substituting the value {\text{A}^2} from ❶ and the value of {\text{A}^3} from ❷ into L.H.S., we have,
L.H.S.
{= \text{A}^3 - 6\text{A}^2 + 5\text{A} + 11I}
{= \begin{bmatrix} 8 & 7 & 1 \\[5pt] -23 & 27 & -69 \\[5pt] 32 & -13 & 58 \end{bmatrix} - 6 \begin{bmatrix} 4 & 2 & 1 \\[5pt] -3 & 8 & -14 \\[5pt] 7 & -3 & 14 \end{bmatrix} + 5 \begin{bmatrix} 1 & 1 & 1 \\[5pt] 1 & 2 & -3 \\[5pt] 2 & -1 & 3 \end{bmatrix} + 11 \begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{bmatrix}}
{= \begin{bmatrix} 8 & 7 & 1 \\[5pt] -23 & 27 & -69 \\[5pt] 32 & -13 & 58 \end{bmatrix} - \begin{bmatrix} 6(4) & 6(2) & 6(1) \\[5pt] 6(-3) & 6(8) & 6(-14) \\[5pt] 6(7) & 6(-3) & 6(14) \end{bmatrix} + \begin{bmatrix} 5(1) & 5(1) & 5(1) \\[5pt] 5(1) & 5(2) & 5(-3) \\[5pt] 5(2) & 5(-1) & 5(3) \end{bmatrix} + \begin{bmatrix} 11(1) & 11(0) & 11(0) \\[5pt] 11(0) & 11(1) & 11(0) \\[5pt] 11(0) & 11(0) & 11(1) \end{bmatrix}}
{= \begin{bmatrix} 8 & 7 & 1 \\[5pt] -23 & 27 & -69 \\[5pt] 32 & -13 & 58 \end{bmatrix} - \begin{bmatrix} 24 & 12 & 6 \\[5pt] -18 & 48 & -84 \\[5pt] 42 & -18 & 84 \end{bmatrix} + \begin{bmatrix} 5 & 5 & 5 \\[5pt] 5 & 10 & -15 \\[5pt] 10 & -5 & 15 \end{bmatrix} + \begin{bmatrix} 11 & 0 & 0 \\[5pt] 0 & 11 & 0 \\[5pt] 0 & 0 & 11 \end{bmatrix}}
{= \begin{bmatrix} 8 - 24 + 5 + 11 & 7 - 12 + 5 + 0 & 1 - 6 + 5 + 0 \\[5pt] -23 - (-18) + 5 + 0 & 27 - 48 + 10 + 11 & -69 - (-84) + (-15) + 0 \\[5pt] 32 - 42 + 10 + 0 & -13 - (-18) + (-5) + 0 & 58 - 84 + 15 + 11 \end{bmatrix}}
{= \begin{bmatrix} 8 - 24 + 5 + 11 & 7 - 12 + 5 + 0 & 1 - 6 + 5 + 0 \\[5pt] -23 + 18 + 5 + 0 & 27 - 48 + 10 + 11 & -69 + 84 - 15 + 0 \\[5pt] 32 - 42 + 10 + 0 & -13 + 18 - 5 + 0 & 58 - 84 + 15 + 11 \end{bmatrix}}
{= \begin{bmatrix} 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \end{bmatrix}}
= R.H.S.
∴ It is showed that {\text{A}^3 - 6\text{A}^2 + 5\text{A} + \text{11I = O}}
Now, the given equation is
{\text{A}^3 - 6\text{A}^2 + 5\text{A} + \text{11I = O}}
Post multiplying with {\text{A}^{-1}} on each side, we have,
{\text{A}^3\text{A}^{-1} - 6\text{A}^2\text{A}^{-1} + 5\text{AA}^{-1} + \text{11IA}^{-1} = \text{OA}^{-1}}
{⇒ \text{A}^2\text{.A.A}^{-1} - 6\text{A.}\text{A.A}^{-1} + 5\text{I} + \text{11A}^{-1} = \text{O}}
{⇒ \text{A}^2\text{.(AA}^{-1}) - 6\text{A.}\text{(A.A}^{-1}) + 5\text{I} + \text{11A}^{-1} = \text{O}}
{⇒ \text{A}^2\text{.I)} - 6\text{A.I} + 5\text{I} + \text{11A}^{-1} = \text{O}}
{⇒ \text{A}^2- 6\text{A} + 5\text{I} + \text{11A}^{-1} = \text{O}}
{⇒ \text{11A}^{-1} = -\text{A}^2 + 6\text{A} - 5\text{I}}
Substituting the values of {\text{A}^2} from ❶, and the value of A as given in the problem, we get,
{\text{11A}^{-1}}
{= -\begin{bmatrix} 4 & 2 & 1 \\[5pt] -3 & 8 & -14 \\[5pt] 7 & -3 & 14 \end{bmatrix} + 6\begin{bmatrix} 1 & 1 & 1 \\[5pt] 1 & 2 & -3 \\[5pt] 2 & -1 & 3 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{bmatrix}}
{= \begin{bmatrix} -4 & -2 & -1 \\[5pt] 3 & -8 & 14 \\[5pt] -7 & 3 & -14 \end{bmatrix} + \begin{bmatrix} 6(1) & 6(1) & 6(1) \\[5pt] 6(1) & 6(2) & 6(-3) \\[5pt] 6(2) & 6(-1) & 6(3) \end{bmatrix} - \begin{bmatrix} 5(1) & 5(0) & 5(0) \\[5pt] 5(0) & 5(1) & 5(0) \\[5pt] 5(0) & 5(0) & 5(1) \end{bmatrix}}
{= \begin{bmatrix} -4 & -2 & -1 \\[5pt] 3 & -8 & 14 \\[5pt] -7 & 3 & -14 \end{bmatrix} + \begin{bmatrix} 6 & 6 & 6 \\[5pt] 6 & 12 & -18 \\[5pt] 12 & -6 & 18 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \\[5pt] 0 & 5 & 0 \\[5pt] 0 & 0 & 5 \end{bmatrix}}
{= \begin{bmatrix} -4 + 6 - 5 & -2 + 6 - 0 & -1 + 6 - 0 \\[5pt] 3 + 6 - 0 & -8 + 12 - 5 & 14 + (-18) - 0 \\[5pt] -7 + 12 - 0 & 3 + (-6) - 0 & -14 + 18 - 5 \end{bmatrix}}
{= \begin{bmatrix} -3 & 4 & 5 \\[5pt] 9 & -1 & -4 \\[5pt] 5 & -3 & -1 \end{bmatrix}}
{⇒ \text{A}^{-1}}
{= \dfrac{1}{11}\begin{bmatrix} -3 & 4 & 5 \\[5pt] 9 & -1 & -4 \\[5pt] 5 & -3 & -1 \end{bmatrix}}
∴ The value of {\text{A}^{-1}} is {\dfrac{1}{11}\begin{bmatrix} -3 & 4 & 5 \\[5pt] 9 & -1 & -4 \\[5pt] 5 & -3 & -1 \end{bmatrix}}
16. If {\text{A} = \begin{bmatrix} 2 & -1 & 1 \\[5pt] -1 & 2 & -1 \\[5pt] 1 & -1 & 2 \end{bmatrix}}
Verify that {\text{A}^3 - 6\text{A}^2 + 9\text{A - 4I = O}} and hence {\text{A}^{-1}}
To solve this problem, we’ll first find {\text{A}^2} and then using that, we’ll find {\text{A}^3.}
{\text{A}^2}
= A.A
{= \begin{bmatrix} 2 & -1 & 1 \\[5pt] -1 & 2 & -1 \\[5pt] 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 & 1 \\[5pt] -1 & 2 & -1 \\[5pt] 1 & -1 & 2 \end{bmatrix}}
{= \begin{bmatrix} 2(2) + (-1)(-1) + 1(1) & 2(-1) + (-1)2 + 1(-1) & 2(1) + (-1)(-1) + 1(2) \\[5pt] (-1)2 + 2(-1) + (-1)1 & (-1)(-1) + 2(2) + (-1)(-1) & (-1)1 + 2(-1) + (-1)2 \\[5pt] 1(1) + (-1)(-1) + 2(1) & 1(-1) + (-1)2 + 2(-1) & 1(1) + (-1)(-1) + 2(2) \end{bmatrix}}
{= \begin{bmatrix} 4 + 1 + 1 & -2 - 2 - 1 & 2 + 1 + 2 \\[5pt] -2 - 2 - 1 & 1 + 4 + 1 & -1 - 2 - 2 \\[5pt] 1 + 1 + 2 & -1 - 2 - 2 & 1 + 1 + 4 \end{bmatrix}}
{= \begin{bmatrix} 6 & -5 & 5 \\[5pt] -5 & 6 & -5 \\[5pt] 5 & -5 & 6 \end{bmatrix}} ———-❶
Now,
{\text{A}^3}
{= \text{A}^2.\text{A}}
{= \begin{bmatrix} 6 & -5 & 5 \\[5pt] -5 & 6 & -5 \\[5pt] 5 & -5 & 6 \end{bmatrix} \begin{bmatrix} 2 & -1 & 1 \\[5pt] -1 & 2 & -1 \\[5pt] 1 & -1 & 2 \end{bmatrix}}
{= \begin{bmatrix} 6(2) + (-5)(-1) + 5(1) & 6(-1) + (-5)2 + 5(-1) & 6(1) + (-5)(-1) + 5(2) \\[5pt] (-5)2 + 6(-1) + (-5)1 & (-5)(-1) + 6(2) + (-5)(-1) & (-5)1 + 6(-1) + (-5)2 \\[5pt] 5(2) + (-5)(-1) + 6(1) & 5(-1) + (-5)2 + 6(-1) & 5(1) + (-5)(-1) + 6(2) \end{bmatrix}}
{= \begin{bmatrix} 12 + 5 + 5 & -6 - 10 - 5 & 6 + 5 + 10 \\[5pt] -10 - 6 - 5 & 5 + 12 + 5 & -5 - 6 - 10 \\[5pt] 10 + 5 + 6 & -5 - 10 - 6 & 5 + 5 + 12 \end{bmatrix}}
{= \begin{bmatrix} 22 & -21 & 21 \\[5pt] -21 & 22 & -21 \\[5pt] 21 & -21 & 22 \end{bmatrix}} ———-❷
Substituting the value {\text{A}^2} from ❶ and the value of {\text{A}^3} from ❷ into L.H.S., we have,
L.H.S.
{= \text{A}^3 - 6\text{A}^2 + 9\text{A - 4I}}
{= \begin{bmatrix} 22 & -21 & 21 \\[5pt] -21 & 22 & -21 \\[5pt] 21 & -21 & 22 \end{bmatrix} - 6\begin{bmatrix} 6 & -5 & 5 \\[5pt] -5 & 6 & -5 \\[5pt] 5 & -5 & 6 \end{bmatrix} + 9\begin{bmatrix} 2 & -1 & 1 \\[5pt] -1 & 2 & -1 \\[5pt] 1 & -1 & 2 \end{bmatrix} - 4\begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{bmatrix}}
{= \begin{bmatrix} 22 & -21 & 21 \\[5pt] -21 & 22 & -21 \\[5pt] 21 & -21 & 22 \end{bmatrix} - \begin{bmatrix} 6(6) & 6(-5) & 6(5) \\[5pt] 6(-5) & 6(6) & 6(-5) \\[5pt] 6(5) & 6(-5) & 6(6) \end{bmatrix} + \begin{bmatrix} 9(2) & 9(-1) & 9(1) \\[5pt] 9(-1) & 9(2) & 9(-1) \\[5pt] 9(1) & 9(-1) & 9(2) \end{bmatrix} - \begin{bmatrix} 4(1) & 4(0) & 4(0) \\[5pt] 4(0) & 4(1) & 4(0) \\[5pt] 4(0) & 4(0) & 4(1) \end{bmatrix}}
{= \begin{bmatrix} 22 & -21 & 21 \\[5pt] -21 & 22 & -21 \\[5pt] 21 & -21 & 22 \end{bmatrix} - \begin{bmatrix} 36 & -30 & 30 \\[5pt] -30 & 36 & -30 \\[5pt] 30 & -30 & 36 \end{bmatrix} + \begin{bmatrix} 18 & -9 & 9 \\[5pt] -9 & 18 & -9 \\[5pt] 9 & -9 & 18 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\[5pt] 0 & 4 & 0 \\[5pt] 0 & 0 & 4 \end{bmatrix}}
{= \begin{bmatrix} 22 - 36 + 18 - 4 & -21 - (-30) + (-9) - 0 & 21 - 30 + 9 - 0 \\[5pt] -21 - (-30) + (-9) - 0 & 22 - 36 + 18 - 4 & -21 - (-30) + (-9) - 0 \\[5pt] 21 - 30 + 9 - 0 & -21 - (-30) + (-9) - 0 & 22 - 36 + 18 - 4 \end{bmatrix}}
{= \begin{bmatrix} 22 - 36 + 18 - 4 & -21 + 30 - 9 - 0 & 21 - 30 + 9 - 0 \\[5pt] -21 + 30 - 9 - 0 & 22 - 36 + 18 - 4 & -21 + 30 - 9 - 0 \\[5pt] 21 - 30 + 9 - 0 & -21 + 30 - 9 - 0 & 22 - 36 + 18 - 4 \end{bmatrix}}
{= \begin{bmatrix} 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \end{bmatrix}}
= O
= R.H.S.
∴ It is showed that {= \text{A}^3 - 6\text{A}^2 + 9\text{A - 4I = O}}
Now, the given equation is {\text{A}^3 - 6\text{A}^2 + 9\text{A - 4I = O}}
Post multiplying with {\text{A}^{-1}} on each side, we have,
{\text{A}^3.\text{A}^{-1} - 6\text{A}^2.\text{A}^{-1} + 9\text{A.A}^{-1} - 4\text{I.A}^{-1} = \text{O.A}^{-1}}
{⇒ \text{A}^2.\text{A.A}^{-1} - 6\text{A.A.A}^{-1} + 9\text{I} - 4\text{A}^{-1} = \text{O}}
{⇒ \text{A}^2.\text{(A.A}^{-1}) - 6\text{A.(A.A}^{-1}) + 9\text{I} - 4\text{A}^{-1} = \text{O}}
{⇒ \text{A}^2.\text{I} - 6\text{A.I} + 9\text{I} - 4\text{A}^{-1} = \text{O}}
{⇒ \text{A}^2 - 6\text{A + 9I - 4A}^{-1} = \text{O}}
{⇒ -4\text{A}^{-1} = -\text{A}^2 + 6A - 9I}
{⇒ 4\text{I} = \text{A}^2 - 6\text{A + 9I}}
Substituting the value {\text{A}^2} from ❶, and the value of A as given in the problem, we get,
{4\text{A}^{-1}}
{= \begin{bmatrix} 6 & -5 & 5 \\[5pt] -5 & 6 & -5 \\[5pt] 5 & -5 & 6 \end{bmatrix} - 6\begin{bmatrix} 2 & -1 & 1 \\[5pt] -1 & 2 & -1 \\[5pt] 1 & -1 & 2 \end{bmatrix} + 9\begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{bmatrix}}
{= \begin{bmatrix} 6 & -5 & 5 \\[5pt] -5 & 6 & -5 \\[5pt] 5 & -5 & 6 \end{bmatrix} - \begin{bmatrix} 6(2) & 6(-1) & 6(1) \\[5pt] 6(-1) & 6(2) & 6(-1) \\[5pt] 6(1) & 6(-1) & 6(2) \end{bmatrix} + \begin{bmatrix} 9(1) & 9(0) & 9(0) \\[5pt] 9(0) & 9(1) & 9(0) \\[5pt] 9(0) & 9(0) & 9(1) \end{bmatrix}}
{= \begin{bmatrix} 6 & -5 & 5 \\[5pt] -5 & 6 & -5 \\[5pt] 5 & -5 & 6 \end{bmatrix} - \begin{bmatrix} 12 & -6 & 6 \\[5pt] -6 & 12 & -6 \\[5pt] 6 & -6 & 12 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\[5pt] 0 & 9 & 0 \\[5pt] 0 & 0 & 9 \end{bmatrix}}
{= \begin{bmatrix} 6 - 12 + 9 & -5 - (-6) + 0 & 5 - 6 + 0 \\[5pt] -5 - (-6) + 0 & 6 - 12 + 9 & -5 - (-6) + 0 \\[5pt] 5 - 6 + 0 & -5 - (-6) + 0 & 6 - 12 + 9 \end{bmatrix}}
{= \begin{bmatrix} 6 - 12 + 9 & -5 + 6 + 0 & 5 - 6 + 0 \\[5pt] -5 + 6 + 0 & 6 - 12 + 9 & -5 + 6 + 0 \\[5pt] 5 - 6 + 0 & -5 + 6 + 0 & 6 - 12 + 9 \end{bmatrix}}
{= \begin{bmatrix} 3 & 1 & -1 \\[5pt] 1 & 3 & 1 \\[5pt] -1 & 1 & 3 \end{bmatrix}}
{⇒ \text{A}^{-1}}
{= \dfrac14\begin{bmatrix} 3 & 1 & -1 \\[5pt] 1 & 3 & 1 \\[5pt] -1 & 1 & 3 \end{bmatrix}}
∴ The value of {\text{A}^{-1}} is {\dfrac14\begin{bmatrix} 3 & 1 & -1 \\[5pt] 1 & 3 & 1 \\[5pt] -1 & 1 & 3 \end{bmatrix}}
17. Let A be a nonsingular square matrix of order 3 × 3. Then {\big|adj \text{ A}\big|} is equal to
(A) |A|
(B) {|\text{A}|^2}
(C) {|\text{A}|^3}
(D) 3|A|
We know that,
{(adj \text{ A) A}}
{= \text{|A|I}}
{= |\text{A}|\begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{bmatrix}}
{= \begin{bmatrix} |\text{A}| & 0 & 0 \\[5pt] 0 & |\text{A}| & 0 \\[5pt] 0 & 0 & |\text{A}| \end{bmatrix}}
Considering the determinants of teh matrices on both L.H.S. and R.H.S., we have,
{|(adj \text{ A) A|}}
{= \begin{vmatrix} |\text{A}| & 0 & 0 \\[5pt] 0 & |\text{A}| & 0 \\[5pt] 0 & 0 & |\text{A}| \end{vmatrix}}
{⇒ |(adj \text{ A)| |A|}}
{= \left(|\text{A}|\right)^3\begin{vmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{vmatrix}}
{⇒ |(adj \text{ A)| |A|}}
{= \left(|\text{A}|\right)^3.1}
{⇒ |(adj \text{ A)|}}
{= \left(|\text{A}|\right)^2}
option B is the correct answer.
💡Tip: In general, if A is a square matrix of order {n,} then {|adj \text{ A}| = |\text{A}|^{n - 1}}
18. If A is an invertible matrix of order 2, then {\text{det (A}^{-1})} is equal to
(A) det (A)
(B) {\dfrac{1}{\text{det (A)}}}
(C) 1
(D) 0
We know that
{\text{A.A}^{-1}}
= I
{\big|\text{A.A}^{-1}\big|}
= |I|
{|\text{A}|.\big|\text{A}^{-1}\big|}
= 1
{\big|\text{A}^{-1}\big|}
{= \dfrac{1}{|A|}}
{\big|\text{A}^{-1}\big|}
{= \dfrac{1}{\text{det (A)}}}
option B is the correct answer.