# Exercise 4.4 Solutions

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Exercise 4.4 Solutions
Write Minors and Cofactors of the elements of following determinants:
1. (i) \begin{vmatrix} 2 & -4 \\[5pt] 0 & 3 \end{vmatrix} (ii) \begin{vmatrix} a & c \\[5pt] b & d \end{vmatrix}
(i) To write Minors and Cofactors of the elements of the determinants \begin{vmatrix} 2 & -4 \\[5pt] 0 & 3 \end{vmatrix}
Minor of an element {a_{ij}} of a determinant is the determinant obtained by deleting its {i^{th}} row and {j^{th}} column in which element {a_{ij}} lies and is denoted by {\text{M}_{ij}}
Cofactor of an element {a_{ij},} denoted by {\text{A}_{ij}} is defined by {\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij},} where {\text{M}_{ij}} is minor of {a_{ij}}
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{-4}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{3}\end{vmatrix}}
3
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1}(3)} = 3
{a_{12} = -4}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-4}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{3}}\end{vmatrix}}
0
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2}(0)} = 0
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{-4} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{3}}\end{vmatrix}}
-4
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1}(-4)} = 4
{a_{22} = 3}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{-4}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{3}}\end{vmatrix}}
2
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2}(2)} = 2
∴ The Minors and Cofactors of the elements of the determinant \begin{vmatrix} 2 & -4 \\[5pt] 0 & 3 \end{vmatrix} are {\text{M}_{11} = 3,} {\text{M}_{12} = 0,} {\text{M}_{21} = -4,} {\text{M}_{22} = 2,} {\text{A}_{11} = 3,} {\text{A}_{12} = 0,} {\text{A}_{21} = 4,} {\text{A}_{22} = 2,}
💡Tip: The minor of any element in a 2 × 2 matrix is the element diagonally opposite to the given element.
(ii) To write Minors and Cofactors of the elements of the determinants \begin{vmatrix} a & b \\[5pt] c & d \end{vmatrix}
Minor of an element {a_{ij}} of a determinant is the determinant obtained by deleting its {i^{th}} row and {j^{th}} column in which element {a_{ij}} lies and is denoted by {\text{M}_{ij}}
Cofactor of an element {a_{ij},} denoted by {\text{A}_{ij}} is defined by {\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij},} where {\text{M}_{ij}} is minor of {a_{ij}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
Cofactor
{\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij}}
{a_{11} = a}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{a}} & \textcolor{red}{\cancel{c}} \\[5pt] \textcolor{red}{\cancel{b}} & \textcolor{green}{d}\end{vmatrix}}
d
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1}a} = d
{a_{12} = c}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{a}} & \textcolor{red}{\cancel{c}} \\[5pt] \textcolor{green}{b} & \textcolor{red}{\cancel{d}}\end{vmatrix}}
b
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2}b} = -b
{a_{21} = b}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{a}} & \textcolor{green}{c} \\[5pt] \textcolor{red}{\cancel{b}} & \textcolor{red}{\cancel{d}}\end{vmatrix}}
c
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1}c} = -c
{a_{22} = d}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{a} & \textcolor{red}{\cancel{c}} \\[5pt] \textcolor{red}{\cancel{b}} & \textcolor{red}{\cancel{d}}\end{vmatrix}}
a
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2}d} = a
∴ The Minors and Cofactors of the elements of the determinant \begin{vmatrix} a & b \\[5pt] c & d \end{vmatrix} are {\text{M}_{11} = d,} {\text{M}_{12} = b,} {\text{M}_{21} = c,} {\text{M}_{22} = a,} {\text{A}_{11} = d,} {\text{A}_{12} = -b,} {\text{A}_{21} = -c,} {\text{A}_{22} = a,}
💡Tip: The minor of any element in a 2 × 2 matrix is the element diagonally opposite to the given element.
Write Minors and Cofactors of the elements of following determinants:
2. (i) {\begin{vmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{vmatrix}} (ii) {\begin{vmatrix} 1 & 0 & 4 \\[5pt] 3 & 5 & -1 \\[5pt] 0 & 1 & 2 \end{vmatrix}}
(i) To find the Minors and Cofactors of the elements of the determinant {\begin{vmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{vmatrix}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{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}{1} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{0} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 1(1) – 0(0)
= 1 – 0
= 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{1}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{0}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 0(1) – 0(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}{1} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{0} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 1 \\[5pt] 0 & 0 \end{vmatrix}}
= 0(0) – 1(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{1}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{0} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 0(1) – 0(0)
= 0 – 0
= 0
{a_{22} = 1}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{0}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 1(1) – 0(0)
= 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{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{0} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 0 \end{vmatrix}}
= 1(0) – 0(0)
= 0 – 0
= 0
{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}{1} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] 1 & 0 \end{vmatrix}}
= 0(0) – 0(1)
= 0 – 0
= 0
{a_{32} = 0}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{1}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 0 \end{vmatrix}}
= 1(0) – 0(0)}
= 0 – 0
= 0
{a_{33} = 1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{1} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 1(1) – 0(0)
= 1 – 0
= 1
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 = \textcolor{green}{\textbf{1}}}
{a_{12} = 0}
{\text{M}_{12} = 0}
{\text{A}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × 0 = \textcolor{green}{\textbf{0}}}
{a_{13} = 0}
{\text{M}_{13} = 0}
{\text{A}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × 0 = \textcolor{green}{\textbf{0}}}
{a_{21} = 0}
{\text{M}_{21} = 0}
{\text{A}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × 0 = \textcolor{green}{\textbf{0}}}
{a_{22} = 1}
{\text{M}_{22} = 1}
{\text{A}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × 1 = \textcolor{green}{\textbf{1}}}
{a_{23} = 0}
{\text{M}_{23} = 0}
{\text{A}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × 0 = \textcolor{green}{\textbf{0}}}
{a_{31} = 0}
{\text{M}_{31} = 0}
{\text{A}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × 0 = \textcolor{green}{\textbf{0}}}
{a_{32} = 0}
{\text{M}_{32} = 0}
{\text{A}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × 0 = \textcolor{green}{\textbf{0}}}
{a_{33} = 1}
{\text{M}_{33} = 1}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × 1 = \textcolor{green}{\textbf{1}}}
∴ The Minors and Cofactors of the elements of the determinant {\begin{vmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{vmatrix}} are {\text{M}_{11} = 1,} {\text{M}_{12} = 0,} {\text{M}_{13} = 0,} {\text{M}_{21} = 0,} {\text{M}_{22} = 1,} {\text{M}_{23} = 0,} {\text{M}_{31} = 0,} {\text{M}_{32} = 0,} {\text{M}_{33} = 1,} {\text{A}_{11} = 1,} {\text{A}_{12} = 0,} {\text{A}_{13} = 0,} {\text{A}_{21} = 0,} {\text{A}_{22} = 1,} {\text{A}_{23} = 0,} {\text{A}_{31} = 0,} {\text{A}_{32} = 0,} {\text{A}_{33} = 1,}
💡Tip: The minor of the elements of the determinant of an identity matrix is same as the element i.e {\text{M}_{ij} = a_{ij}}.
💡Tip: The cofactor of the elements of the determinant of an identity matrix is same as the element i.e {\text{A}_{ij} = \text{M}_{ij} = a_{ij}}.
(ii) To find the Minors and Cofactors of the elements of the determinant {\begin{vmatrix} 1 & 0 & 4 \\[5pt] 3 & 5 & -1 \\[5pt] 0 & 1 & 2 \end{vmatrix}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{5} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{1} & \textcolor{green}{2} \end{vmatrix}}
{= \begin{vmatrix} 5 & -1 \\[5pt] 1 & 2 \end{vmatrix}}
= 5(2) – (-1)1
= 10 + 1
= 11
{a_{12} = 0}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{5}} & \textcolor{green}{-1} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{1}} & \textcolor{green}{2} \end{vmatrix}}
{= \begin{vmatrix} 3 & -1 \\[5pt] 0 & 2 \end{vmatrix}}
= 3(2) – (-1)0
= 6 – 0
= 6
{a_{13} = 0}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{5} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{1} & \textcolor{red}{\cancel{2}} \end{vmatrix}}
{= \begin{vmatrix} 3 & 5 \\[5pt] 0 & 1 \end{vmatrix}}
= 3(1) – 5(0)
= 3 – 0
= 3
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{4} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{5}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{1} & \textcolor{green}{2} \end{vmatrix}}
{= \begin{vmatrix} 0 & 4 \\[5pt] 1 & 2 \end{vmatrix}}
= 0(2) – 4(1)
= 0 – 4
= -4
{a_{22} = 1}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{4} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{5}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{1}} & \textcolor{green}{2} \end{vmatrix}}
{= \begin{vmatrix} 1 & 4 \\[5pt] 0 & 2 \end{vmatrix}}
= 1(2) – 4(0)
= 2 – 0
= 2
{a_{23} = 0}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{5}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{1} & \textcolor{red}{\cancel{2}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 1(1) – 0(0)
= 1 – 0
= 1
{a_{31} = 0}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{4} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{5} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{4}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 4 \\[5pt] 5 & -1 \end{vmatrix}}
= 0(-1) – 4(5)
= 0 – 20
= -20
{a_{32} = 0}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{4} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{5}} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{4}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 4 \\[5pt] 3 & -1 \end{vmatrix}}
= 1(-1) – 4(3)
= -1 – 12
= -13
{a_{33} = 1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{5} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{2}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 3 & 5 \end{vmatrix}}
= 1(5) – 0(3)
= 5 – 0
= 5
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 11}
{\text{A}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × 11 = \textcolor{green}{\textbf{11}}}
{a_{12} = 0}
{\text{M}_{12} = 6}
{\text{A}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × 6 = \textcolor{green}{\textbf{-6}}}
{a_{13} = 4}
{\text{M}_{13} = 3}
{\text{A}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × 3 = \textcolor{green}{\textbf{3}}}
{a_{21} = 3}
{\text{M}_{21} = -4}
{\text{A}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × -4 = \textcolor{green}{\textbf{4}}}
{a_{22} = 5}
{\text{M}_{22} = 2}
{\text{A}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × 2 = \textcolor{green}{\textbf{2}}}
{a_{23} = -1}
{\text{M}_{23} = 1}
{\text{A}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × 1 = \textcolor{green}{\textbf{-1}}}
{a_{31} = 0}
{\text{M}_{31} = -20}
{\text{A}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × -20 = \textcolor{green}{\textbf{-20}}}
{a_{32} = 1}
{\text{M}_{32} = -13}
{\text{A}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × -13 = \textcolor{green}{\textbf{13}}}
{a_{33} = 2}
{\text{M}_{33} = 5}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × 5 = \textcolor{green}{\textbf{5}}}
∴ The Minors and Cofactors of the elements of the determinant {\begin{vmatrix} 1 & 0 & 4 \\[5pt] 3 & 5 & -1 \\[5pt] 0 & 1 & 2 \end{vmatrix}} are {\text{M}_{11} = 11,} {\text{M}_{12} = 6,} {\text{M}_{13} = 3,} {\text{M}_{21} = -4,} {\text{M}_{22} = 2,} {\text{M}_{23} = 1,} {\text{M}_{31} = -20,} {\text{M}_{32} = -13,} {\text{M}_{33} = 5,} {\text{A}_{11} = 11,} {\text{A}_{12} = -6,} {\text{A}_{13} = 3,} {\text{A}_{21} = 4,} {\text{A}_{22} = 2,} {\text{A}_{23} = -1,} {\text{A}_{31} = -20,} {\text{A}_{32} = 13,} {\text{A}_{33} = 5,}
3. Using Cofactors of elements of third column, evaluate {Δ = \begin{vmatrix} 5 & 3 & 8 \\[5pt] 2 & 0 & 1 \\[5pt] 1 & 2 & 3 \end{vmatrix}}
To find the Cofactors, we first find the Minors of the elements of the second row and then deduce the Cofactors from the Minors as given below.
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{a_{21} = 2}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{5}} & \textcolor{green}{3} & \textcolor{green}{8} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{2} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 3 & 8 \\[5pt] 2 & 3 \end{vmatrix}}
= 3(3) – 8(2) = 9 – 16 = -7
= 9 – 16
= -7
{a_{22} = 0}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{5} & \textcolor{red}{\cancel{3}} & \textcolor{green}{8} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{2}} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 5 & 8 \\[5pt] 1 & 3 \end{vmatrix}}
= 5(3) – 8(1)
= 15 – 8
= 7
{a_{23} = 1}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{5} & \textcolor{green}{3} & \textcolor{red}{\cancel{8}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{2} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 5 & 3 \\[5pt] 1 & 2 \end{vmatrix}}
= 5(2) – 3(1)
= 10 – 3
= 7
Now, the Cofactors of the elements of the second row are found as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{21} = 2}
{\text{M}_{21} = -7}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)(-7) = 7}
{a_{22} = 0}
{\text{M}_{22} = 7}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(7) = 7}
{a_{23} = 1}
{\text{M}_{23} = 7}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)7 = -7}
Now, the determinants of the matrix can be calculated from the elements of the second row, using the formula,
Δ
{= a_{21}\text{A}_{21} + a_{22}\text{A}_{22} + a_{23}\text{A}_{22}}
= 2(7) + 0(7) + 1(-7)
= 14 + 0 – 7
= 7
∴ The determinant {Δ = \begin{vmatrix} 5 & 3 & 8 \\[5pt] 2 & 0 & 1 \\[5pt] 1 & 2 & 3 \end{vmatrix}} evaluated using Cofactors of elements of second row is 7
4. Using cofactors of elements of third column evaluate {Δ = \begin{vmatrix} 1 & x & yz \\[5pt] 1 & y & yz \\[5pt] 1 & z & xy \end{vmatrix}}
To find the Cofactors, we first find the Minors of the elements of the third column and then deduce the Cofactors from the Minors as given below.
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{a_{13} = yz}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{x}} & \textcolor{red}{\cancel{yz}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{y} & \textcolor{red}{\cancel{zx}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{z} & \textcolor{red}{\cancel{xy}} \end{vmatrix}}
{= \begin{vmatrix} 1 & y \\[5pt] 1 & z \end{vmatrix}}
{= 1(z) - (y)1}
{= z - y}
{a_{23} = zx}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{x} & \textcolor{red}{\cancel{yz}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{y}} & \textcolor{red}{\cancel{zx}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{z} & \textcolor{red}{\cancel{xy}} \end{vmatrix}}
{= \begin{vmatrix} 1 & x \\[5pt] 1 & z \end{vmatrix}}
{= 1(z) - x(1)}
{= z - x}
{a_{33} = xy}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{x} & \textcolor{red}{\cancel{yz}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{y} & \textcolor{red}{\cancel{zx}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{z}} & \textcolor{red}{\cancel{xy}} \end{vmatrix}}
{= \begin{vmatrix} 1 & x \\[5pt] 1 & y \end{vmatrix}}
{= 1(y) - x(1)}
{= y - x}
Now, the Cofactors of the elements of the third column are found as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{13} = 2}
{\text{M}_{13} = z - y}
{\text{A}_{13} = (-1)^{1 + 3} × \text{M}_{21} = 1(z - y) = z - y}
{a_{23} = 0}
{\text{M}_{23} = z - x}
{\text{A}_{23} = (-1)^{2 + 3} × \text{M}_{22} = (-1)(z - x) = x - z}
{a_{33} = 1}
{\text{M}_{33} = y - z}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{23} = 1(y - z) = y - z}
Now, the determinants of the matrix can be calculated from the elements of the third column, using the formula,
Δ
{= a_{13}\text{A}_{13} + a_{23}\text{A}_{23} + a_{33}\text{A}_{33}}
{= yz(z - y) + zx(x - z) + xy(y - x)}
{= z\left[y(z - y) + x(x - z)\right] + xy(y - x)}
{= z(yz - y^2 + x^2 - xz) + xy(y - x)}
{= z(x^2 - y^2 -xz + yz) + xy(y - x)}
{= z\left[(x + y)(x - y) - z(x - y)\right] + xy(y - x)}
{= (x - y)\left[z(x + y - z) - xy\right]}
{= (x - y)(xz + yz - z^2 - xy)}
{= (x - y)(yz - z^2 - xy + xz)}
{= (x - y)\left[z(y - z) - x(y - z)\right]}
= {(x - y)(y - z)(x - y)}
∴ The determinant {Δ = \begin{vmatrix} 1 & x & yz \\[5pt] 1 & y & zx \\[5pt] 1 & z & xy \end{vmatrix}} evaluated using Cofactors of elements of third column is {(x - y)(y - z)(x - y)}
5. If {Δ = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\[5pt] a_{21} & a_{22} & a_{23} \\[5pt] a_{31} & a_{32} & a_{33} \end{vmatrix}} and {\text{A}_{ij}} is Cofactors of {a_{ij},} then value of Δ is given by
(A) {a_{11}\text{A}_{31} + a_{12}\text{A}_{32} + a_{13}\text{A}_{33}}
(B) {a_{11}\text{A}_{11} + a_{12}\text{A}_{21} + a_{13}\text{A}_{31}}
(C) {a_{21}\text{A}_{11} + a_{22}\text{A}_{12} + a_{23}\text{A}_{13}}
(D) {a_{11}\text{A}_{11} + a_{21}\text{A}_{21} + a_{31}\text{A}_{31}}
We know that the determinant Δ is equal to the sum of the product of elements of any row (or column) with their corresponding Cofactors.
We also know that if elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero.
option A is the sum of the product of elements of first row and the Cofactors of the elements of the third row. So, it will be zero. so, (option A is not correct ❌)
option B is the sum of the product of elements of first row and the Cofactors of the elements of the third column. So, (option B is not correct ❌)
option C is the sum of the product of elements of second row and the Cofactors of the elements of the first row. So, (option C is not correct ❌)
option D is the sum of the product of elements of first column and the Cofactors of the elements of the first column. So, (option D) is the correct answer.