# Exercise 4.6 Solutions

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Exercise 4.6 Solutions
Examine the consistency of the system of equations in Exercises 1 to 6.
1.
{x + 2y = 2}
{2x + 3y = 3}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 1 & 2 \\[5pt] 2 & 3 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \end{bmatrix}} & = & {\begin{bmatrix} 2 \\[5pt] 3 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 1 & 2 \\[5pt] 2 & 3 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \end{bmatrix}} and {\text{B} = \begin{bmatrix} 2 \\[5pt] 3 \end{bmatrix}}
Now,
|A|
{= \begin{bmatrix} 1 & 2 \\[5pt] 3 & 4 \end{bmatrix}}
= 1(3) – 2(2)
= 3 – 4
= -1
≠ 0
∴ As |A| ≠ 0, the given system of equations is consistent.
Examine the consistency of the system of equations in Exercises 1 to 6.
2.
{2x - y = 5}
{x + y = 4}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 2 & -1 \\[5pt] 1 & 1 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \end{bmatrix}} & = & {\begin{bmatrix} 5 \\[5pt] 4 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 2 & -1 \\[5pt] 1 & 1 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \end{bmatrix}} and {\text{B} = \begin{bmatrix} 5 \\[5pt] 4 \end{bmatrix}}
Now,
|A|
{= \begin{bmatrix} 2 & -1 \\[5pt] 1 & 1 \end{bmatrix}}
= 2(1) – (-1)1
= 2 + 1
= 3
≠ 0
∴ As |A| ≠ 0, the given system of equations is consistent.
Examine the consistency of the system of equations in Exercises 1 to 6.
3.
{x + 3y = 5}
{2x + 6y = 8}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 1 & 3 \\[5pt] 2 & 6 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \end{bmatrix}} & = & {\begin{bmatrix} 5 \\[5pt] 8 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 1 & 3 \\[5pt] 2 & 6 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \end{bmatrix}} and {\text{B} = \begin{bmatrix} 5 \\[5pt] 8 \end{bmatrix}}
Now,
|A|
{= \begin{bmatrix} 1 & 3 \\[5pt] 2 & 6 \end{bmatrix}}
= 1(6) – 3(2)
= 6 – 6
= 0
∴ As |A| = 0, we need to calculate {(adj \text{A})\text{B}}
Now, {adj \text{A}} can be calculated as below:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{1}^{\textbf{\large{Interchange}}}} & & 3 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ 2 & & \textcolor{green}{\underbrace{6}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} 6 & 3 \\[5pt] 2 & 1 \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{2}_{\textbf{\large{Change Sign}}}} & & 1 \end{bmatrix} = \begin{bmatrix} 6 & -3 \\[5pt] -2 & 1 \end{bmatrix}}
Now,
{= \begin{bmatrix} 6 & -3 \\[5pt] -2 & 1 \end{bmatrix} \begin{bmatrix} 5 \\[5pt] 8 \end{bmatrix}}
{= \begin{bmatrix} 6(5) + (-3)8 \\[5pt] (-2)5 + 1(8) \end{bmatrix}}
{= \begin{bmatrix} 30 - 24 \\[5pt] -10 + 8 \end{bmatrix}}
{= \begin{bmatrix} -6 \\[5pt] -2 \end{bmatrix}}
≠ O
∴ As |A| = 0 and {(adj \text{ A)B}} ≠ O the given system of equations has no solution.
∴ the given system of equations is inconsistent.
Examine the consistency of the system of equations in Exercises 1 to 6.
4.
{x + y + z = 1}
{2x + 3y + 2z = 2}
{ax + ay + 2az = 4}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 1 & 1 & 1 \\[5pt] 2 & 3 & 2 \\[5pt] a & a & 2a \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} & = & {\begin{bmatrix} 1 \\[5pt] 2 \\[5pt] 4 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 1 & 1 & 1 \\[5pt] 2 & 3 & 2 \\[5pt] a & a & 2a \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} and {\text{B} = \begin{bmatrix} 1 \\[5pt] 2 \\[5pt] 4 \end{bmatrix}}
Expanding along {\text{R}_1,} we have,
|A|
{= \begin{vmatrix} 1 & 1 & 1 \\[5pt] 2 & 3 & 2 \\[5pt] a & a & 2a \end{vmatrix}}
{= 1 \begin{vmatrix} 3 & 2 \\[5pt] a & 2a \end{vmatrix} - 1 \begin{vmatrix} 2 & 2 \\[5pt] a & 2a \end{vmatrix} + 1 \begin{vmatrix} 2 & 3 \\[5pt] a & a \end{vmatrix}}
{= \left[3(2a) - 2(a)\right] - \left[2(2a) - 2(a)\right] + \left[2(a) - 3(a)\right]}
{= (6a - 2a) - (4a - 2a) + (2a - 3a)}
{= 4a - 2a + (-a)}
{= a}
≠ 0
∴ As |A| ≠ 0, the given system of equations is consistent.
Examine the consistency of the system of equations in Exercises 1 to 6.
5.
{3x - y - 2z = 2}
{2y - z = -1}
{3x - 5y = 3}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 3 & -1 & -2 \\[5pt] 0 & 2 & -1 \\[5pt] 3 & -5 & 0 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} & = & {\begin{bmatrix} 2 \\[5pt] -1 \\[5pt] 3 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 3 & -1 & -2 \\[5pt] 0 & 2 & -1 \\[5pt] 3 & -5 & 0 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} and {\text{B} = \begin{bmatrix} 2 \\[5pt] -1 \\[5pt] 3 \end{bmatrix}}
Expanding along {\text{C}_1,} we have,
|A|
{= \begin{vmatrix} 3 & -1 & -2 \\[5pt] 0 & 2 & -1 \\[5pt] 3 & -5 & 0 \end{vmatrix}}
{= 3 \begin{vmatrix} 2 & -1 \\[5pt] -5 & 0 \end{vmatrix} - 0 \begin{vmatrix} -1 & -2 \\[5pt] -5 & 0 \end{vmatrix} + 3 \begin{vmatrix} -1 & -2 \\[5pt] 2 & -1 \end{vmatrix}}
= 3[2(0) – (-1)(-5)] – 0 + 3[(-1)(-1) – 2(-2)]
= 3(0 – 5) + 3(1 + 4)
= 3(-5) + 3(5)
= -15 – 15
= 0
∴ As |A| = 0, we need to calculate {(adj \text{ A)B}}
Let’s first find the Minors of each of the elements of A and then the Cofactors. We can then find {adj \text{A}} as below:
{adj \text{A} = \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}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{a_{11} = 3}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{2} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{-5} & \textcolor{green}{0} \end{vmatrix}}
{= \begin{vmatrix} 2 & -1 \\[5pt] -5 & 0 \end{vmatrix} = \big(2 × 0 - (-1) × (-5)\big) = 0 - 5 = -5}
{a_{12} = -1}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{2}} & \textcolor{green}{-1} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{-5}} & \textcolor{green}{0} \end{vmatrix}}
{= \begin{vmatrix} 0 & -1 \\[5pt] 3 & 0 \end{vmatrix} = \big(0 × 0 - (-1) × 3\big) = 0 + 3 = 3}
{a_{13} = -2}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{2} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{-5} & \textcolor{red}{\cancel{0}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 2 \\[5pt] 3 & -5 \end{vmatrix} = \big(0 × (-5) - 2 × 3\big) = 0 - 6 = -6}
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{3}} & \textcolor{green}{-1} & \textcolor{green}{-2} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{-5} & \textcolor{green}{0} \end{vmatrix}}
{= \begin{vmatrix} -1 & -2 \\[5pt] -5 & 0 \end{vmatrix} = \big(-1 × 0 - (-2) × (-5)\big) = 0 - 10 = -10}
{a_{22} = 2}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{3} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{-2} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{-5}} & \textcolor{green}{0} \end{vmatrix}}
{= \begin{vmatrix} 3 & -2 \\[5pt] 3 & 0 \end{vmatrix} = \big(3 × 0 - (-2) × 3\big) = 0 + 6 = 6}
{a_{23} = -1}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{3} & \textcolor{green}{-1} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{-5} & \textcolor{red}{\cancel{0}} \end{vmatrix}}
{= \begin{vmatrix} 3 & -1 \\[5pt] 3 & -5 \end{vmatrix} = \big(3 × (-5) - (-1) × 3\big) = -15 + 3 = -12}
{a_{31} = 3}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{3}} & \textcolor{green}{-1} & \textcolor{green}{-2} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{2} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-5}} & \textcolor{red}{\cancel{0}} \end{vmatrix}}
{= \begin{vmatrix} -1 & -2 \\[5pt] 2 & -1 \end{vmatrix} = \big((-1) × (-1) - (-2) × 2\big) = 1 + 4 = 5}
{a_{32} = -5}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{3} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{-2} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{2}} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-5}} & \textcolor{red}{\cancel{0}} \end{vmatrix}}
{= \begin{vmatrix} 3 & -2 \\[5pt] 0 & -1 \end{vmatrix} = \big(3 × (-1) - (-2) × 0\big) = -3 - 0 = -3}
{a_{33} = 0}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{3} & \textcolor{green}{-1} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{2} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-5}} & \textcolor{red}{\cancel{0}} \end{vmatrix}}
{= \begin{vmatrix} 3 & -1 \\[5pt] 0 & 2 \end{vmatrix} = \big(3 × 2 - (-1) × 0\big) = 6 - 0 = 6}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 3}
{\text{M}_{11} = -5}
{\text{A}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × -5 = -5}
{a_{12} = -1}
{\text{M}_{12} = 3}
{\text{A}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × 3 = -3}
{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} = -10}
{\text{A}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × -10 = 10}
{a_{22} = 2}
{\text{M}_{22} = 6}
{\text{A}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × 6 = 6}
{a_{23} = -1}
{\text{M}_{23} = -12}
{\text{A}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × -12 = 12}
{a_{31} = 3}
{\text{M}_{31} = 5}
{\text{A}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × 5 = 5}
{a_{32} = -5}
{\text{M}_{32} = -3}
{\text{A}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × -3 = 3}
{a_{33} = 0}
{\text{M}_{33} = 6}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × 6 = 6}
So,
{= \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} -5 & 10 & 5 \\[5pt] -3 & 6 & 3 \\[5pt] -6 & 12 & 6 \end{bmatrix}}
Now,
{(adj \text{ A) B}}
{= \begin{bmatrix} -5 & 10 & 5 \\[5pt] -3 & 6 & 3 \\[5pt] -6 & 12 & 6 \end{bmatrix} \begin{bmatrix} 2 \\[5pt] -1 \\[5pt] 3 \end{bmatrix}}
{= \begin{bmatrix} (-5)2 + 10(-1) + 5(3) \\[5pt] (-3)2 + 6(-1) + 3(3) \\[5pt] (-6)2 + 12(-1) + 6(3) \end{bmatrix}}
{= \begin{bmatrix} -10 - 10 + 15 \\[5pt] -6 - 6 + 9 \\[5pt] -12 - 12 + 18 \end{bmatrix}}
{= \begin{bmatrix} -5 \\[5pt] -3 \\[5pt] -6 \end{bmatrix}}
≠ O
∴ As |A| = 0 and {(adj \text{A) B}} ≠ O the given system of equations has no solution.
∴ the given system of equations is inconsistent.
Examine the consistency of the system of equations in Exercises 1 to 6.
6.
{5x - y + 4z = 5}
{2x + 3y + 5z = 2}
{5x - 2y + 6z = -1}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 5 & -1 & 4 \\[5pt] 2 & 3 & 5 \\[5pt] 5 & -2 & 6 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} & = & {\begin{bmatrix} 5 \\[5pt] 2 \\[5pt] -1 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 5 & -1 & 4 \\[5pt] 2 & 3 & 5 \\[5pt] 5 & -2 & 6 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} and {\text{B} = \begin{bmatrix} 5 \\[5pt] 2 \\[5pt] -1 \end{bmatrix}}
Expanding along {\text{R}_1,} we have,
|A|
{= \begin{vmatrix} 5 & -1 & 4 \\[5pt] 2 & 3 & 5 \\[5pt] 5 & -2 & 6 \end{vmatrix}}
{= 5 \begin{vmatrix} 3 & 5 \\[5pt] -2 & 6 \end{vmatrix} - (-1) \begin{vmatrix} 2 & 5 \\[5pt] 5 & 6 \end{vmatrix} + 4 \begin{vmatrix} 2 & 3 \\[5pt] 5 & -2 \end{vmatrix}}
= 5[3(6) – 5(-2)] + 1[2(6) – 5(5)] + 4[2(-2) – 3(5)]
= 5(18 + 10) + (12 – 25) + 3(-4 – 15)
= 5(28) + (-13) + 4(-19)
= 140 – 13 – 76
= 51
≠ 0
∴ As |A| ≠ 0, the given system of equations is consistent.
Solve system of linear equations, using matrix method, in Exercises 7 to 14.
7.
{5x + 2y = 4}
{7x + 3y = 5}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 5 & 2 \\[5pt] 7 & 3 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \end{bmatrix}} & = & {\begin{bmatrix} 4 \\[5pt] 5 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 5 & 2 \\[5pt] 7 & 3 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \end{bmatrix}} and {\text{B} = \begin{bmatrix} 4 \\[5pt] 5 \end{bmatrix}}
Now, we have,
|A|
{= \begin{bmatrix} 5 & 2 \\[5pt] 7 & 3 \end{bmatrix}}
= 5(3) – 2(7)
= 15 – 14
= 1 ———-❶
≠ 0
∴ As |A| ≠ 0, A is nonsingular matrix and so has a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
{\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A})}
Now, {adj \text{ A}} can be calculated as below:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{5}^{\textbf{\large{Interchange}}}} & & 2 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ 7 & & \textcolor{green}{\underbrace{3}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} 3 & 2 \\[5pt] 7 & 5 \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{7}_{\textbf{\large{Change Sign}}}} & & 5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\[5pt] -7 & 5 \end{bmatrix}}
Thus,
{\text{A}^{-1}}
{= \dfrac11 \begin{bmatrix} 3 & -2 \\[5pt] -7 & 5 \end{bmatrix}} (from ❶)
{= \begin{bmatrix} 3 & -2 \\[5pt] -7 & 5 \end{bmatrix}} ———-❷
Substituting the value of {\text{A}^{-1}} from ❷, we have,
X
{= \text{A}^{-1}\text{B}}
{= \begin{bmatrix} 3 & -2 \\[5pt] -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \\[5pt] 5\end{bmatrix}}
{= \begin{bmatrix} 3(4) + (-2)5 \\[5pt] (-7)4 + 5(5) \end{bmatrix}}
{= \begin{bmatrix} 12 - 10 \\[5pt] -28 + 25 \end{bmatrix}}
{= \begin{bmatrix} 2 \\[5pt] -3 \end{bmatrix}}
∴ The solution of given system of linear equations is {x = 2} and {y = -3}
Solve system of linear equations, using matrix method, in Exercises 7 to 14.
8.
{2x - y = -2}
{3x + 4y = 3}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 2 & -1 \\[5pt] 3 & 4 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \end{bmatrix}} & = & {\begin{bmatrix} -2 \\[5pt] 3 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 2 & -1 \\[5pt] 3 & 4 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \end{bmatrix}} and {\text{B} = \begin{bmatrix} -2 \\[5pt] 3 \end{bmatrix}}
Now, we have,
|A|
{= \begin{bmatrix} 2 & -1 \\[5pt] 3 & 4 \end{bmatrix}}
= 2(4) – (-1)3
= 8 + 3
= 11 ———-❶
≠ 0
∴ As |A| ≠ 0, A is nonsingular matrix and so has a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
{\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A})}
Now, {adj \text{ A}} can be calculated as below:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{2}^{\textbf{\large{Interchange}}}} & & -1 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ 3 & & \textcolor{green}{\underbrace{4}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} 4 & -1 \\[5pt] 3 & 2 \end{bmatrix}}
● Change the sign of the elements in the secondary diagonal {\begin{bmatrix} 4 & & \textcolor{red}{\overbrace{-1}^{\textbf{\large{Change Sign}}}} \\[10pt] & \textcolor{red}{\Huge⤢} & \\ \textcolor{red}{\underbrace{3}_{\textbf{\large{Change Sign}}}} & & 2 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\[5pt] -3 & 2 \end{bmatrix}}
Thus,
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{11} \begin{bmatrix} 4 & 1 \\[5pt] -3 & 2 \end{bmatrix}} (from ❶) ———-❷
Substituting the value of {\text{A}^{-1}} from ❷, we have,
X
{= \text{A}^{-1}\text{B}}
{= \dfrac{1}{11} \begin{bmatrix} 4 & 1 \\[5pt] -3 & 2 \end{bmatrix} \begin{bmatrix} -2 \\[5pt] 3 \end{bmatrix}}
{= \dfrac{1}{11} \begin{bmatrix} 4(-2) + 1(3) \\[5pt] (-3)(-2) + 2(3) \end{bmatrix}}
{= \dfrac{1}{11} \begin{bmatrix} -8 + 3 \\[5pt] 6 + 6 \end{bmatrix}}
{= \dfrac{1}{11} \begin{bmatrix} -5 \\[5pt] 12 \end{bmatrix}}
{= \begin{bmatrix} \dfrac{1}{11} × (-5) \\[10pt] \dfrac{1}{11} × 12 \end{bmatrix}}
{= \begin{bmatrix} \dfrac{-5}{11} \\[10pt] \dfrac{12}{11} \end{bmatrix}}
∴ The solution of given system of linear equations is {x = \dfrac{-5}{11}} and {y = \dfrac{12}{11}}
Solve system of linear equations, using matrix method, in Exercises 7 to 14.
9.
{4x - 3y = 3}
{3x - 5y = 7}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 4 & -3 \\[5pt] 3 & -5 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \end{bmatrix}} & = & {\begin{bmatrix} 3 \\[5pt] 7 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 4 & -3 \\[5pt] 3 & -5 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \end{bmatrix}} and {\text{B} = \begin{bmatrix} 3 \\[5pt] 7 \end{bmatrix}}
Now, we have,
|A|
{= \begin{bmatrix} 4 & -3 \\[5pt] 3 & -5 \end{bmatrix}}
= 4(-5) – (-3)3
= -20 + 9
= -11 ———-❶
≠ 0
∴ As |A| ≠ 0, A is nonsingular matrix and so has a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
{\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A})}
Now, {adj \text{ A}} can be calculated as below:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{4}^{\textbf{\large{Interchange}}}} & & -3 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ 3 & & \textcolor{green}{\underbrace{-5}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} -5 & -3 \\[5pt] 3 & 4 \end{bmatrix}}
● Change the sign of the elements in the secondary diagonal {\begin{bmatrix} -5 & & \textcolor{red}{\overbrace{-3}^{\textbf{\large{Change Sign}}}} \\[10pt] & \textcolor{red}{\Huge⤢} & \\ \textcolor{red}{\underbrace{3}_{\textbf{\large{Change Sign}}}} & & 4 \end{bmatrix} = \begin{bmatrix} -5 & 3 \\[5pt] -3 & 4 \end{bmatrix}}
Thus,
{\text{A}^{-1}}
{= \dfrac{1}{-11} \begin{bmatrix} -5 & 3 \\[5pt] -3 & 4 \end{bmatrix}} (from ❶)
{= -\dfrac{1}{11}\begin{bmatrix} -5 & 3 \\[5pt] -3 & 4 \end{bmatrix}} ———-❷
Substituting the value of {\text{A}^{-1}} from ❷, we have,
X
{= \text{A}^{-1}\text{B}}
{= -\dfrac{1}{11}\begin{bmatrix} -5 & 3 \\[5pt] -3 & 4 \end{bmatrix} \begin{bmatrix} 3 \\[5pt] 7 \end{bmatrix}}
{= -\dfrac{1}{11} \begin{bmatrix} (-5)3 + 3(7) \\[5pt] (-3)3 + 4(7) \end{bmatrix}}
{= -\dfrac{1}{11} \begin{bmatrix} -15 + 21 \\[5pt] -9 + 28 \end{bmatrix}}
{= -\dfrac{1}{11} \begin{bmatrix} 6 \\[5pt] 19 \end{bmatrix}}
{= \begin{bmatrix} -\dfrac{1}{11} × 6 \\[10pt] -\dfrac{1}{11} × 19 \end{bmatrix}}
{= \begin{bmatrix} \dfrac{-6}{11} \\[10pt] \dfrac{-19}{11} \end{bmatrix}}
∴ The solution of given system of linear equations is {x = \dfrac{-6}{11}} and {y = \dfrac{-19}{11}}
Solve system of linear equations, using matrix method, in Exercises 7 to 14.
10.
{5x + 2y = 3}
{3x + 2y = 5}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 5 & 2 \\[5pt] 3 & 2 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \end{bmatrix}} & = & {\begin{bmatrix} 3 \\[5pt] 5 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 5 & 2 \\[5pt] 3 & 2 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \end{bmatrix}} and {\text{B} = \begin{bmatrix} 3 \\[5pt] 5 \end{bmatrix}}
Now, we have,
|A|
{= \begin{bmatrix} 5 & 2 \\[5pt] 3 & 2 \end{bmatrix}}
= 5(2) – 2(3)
= 10 – 6
= 4 ———-❶
≠ 0
∴ As |A| ≠ 0, A is nonsingular matrix and so has a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
{\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A})}
Now, {adj \text{ A}} can be calculated as below:
● Interchange the elements in the primary diagonal {\begin{bmatrix} \textcolor{green}{\overbrace{5}^{\textbf{\large{Interchange}}}} & & 2 \\[10pt] & \textcolor{green}{\Huge⤡} & \\ 3 & & \textcolor{green}{\underbrace{2}_{\textbf{\large{Interchange}}}} \end{bmatrix} = \begin{bmatrix} 2 & 2 \\[5pt] 3 & 5 \end{bmatrix}}
● Change the sign of the elements in the secondary diagonal {\begin{bmatrix} 2 & & \textcolor{red}{\overbrace{2}^{\textbf{\large{Change Sign}}}} \\[10pt] & \textcolor{red}{\Huge⤢} & \\ \textcolor{red}{\underbrace{3}_{\textbf{\large{Change Sign}}}} & & 5 \end{bmatrix} = \begin{bmatrix} 2 & -2 \\[5pt] -3 & 5 \end{bmatrix}}
Thus,
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{4} × \begin{bmatrix} 2 & -2 \\[5pt] -3 & 5 \end{bmatrix}} (from ❶)
Substituting the value of {\text{A}^{-1}} from ❷, we have,
X
{= \text{A}^{-1}\text{B}}
{= \dfrac{1}{4} × \begin{bmatrix} 2 & -2 \\[5pt] -3 & 5 \end{bmatrix} \begin{bmatrix} 3 \\[5pt] 5 \end{bmatrix}}
{= \dfrac{1}{4} × \begin{bmatrix} 2(3) + (-2)5 \\[5pt] (-3)3 + 5(5) \end{bmatrix}}
{= \dfrac{1}{4} × \begin{bmatrix} -4 \\[5pt] 16 \end{bmatrix}}
{= \begin{bmatrix} \dfrac14 × (-4) \\[10pt] \dfrac14 × 16 \end{bmatrix}}
{= \begin{bmatrix} -1 \\[10pt] 4 \end{bmatrix}}
∴ The solution of given system of linear equations is {x = -1} and {y = 4}
Solve system of linear equations, using matrix method, in Exercises 7 to 14.
11.
{2x + y + z = 1}
{x - 2y - z = \dfrac{3}{2}}
{3y - 5z = 9}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 2 & 1 & 1 \\[5pt] 1 & -2 & -1 \\[5pt] 0 & 3 & -5 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} & = & {\begin{bmatrix} 1 \\[5pt] \dfrac32 \\[10pt] 9\end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 2 & 1 & 1 \\[5pt] 1 & -2 & -1 \\[5pt] 0 & 3 & -5 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} and {\text{B} = \begin{bmatrix} 1 \\[5pt] \dfrac32 \\[10pt] 9 \end{bmatrix}}
Now, let’s expand along {\text{C}_1,} to find |A|. Thus, we have,
|A|
{= \begin{bmatrix} 2 & 1 & 1 \\[5pt] 1 & -2 & -1 \\[5pt] 0 & 3 & -5 \end{bmatrix}}
{= 2 \begin{bmatrix} -2 & -1 \\[5pt] 3 & -5 \end{bmatrix} - 1 \begin{bmatrix} 1 & 1 \\[5pt] 3 & -5 \end{bmatrix} + 0 \begin{bmatrix} 1 & 1 \\[5pt] -2 & -1 \end{bmatrix}}
= 2[(-2)(-5) – (-1)3] – 1[1(-5) – 1(3)] + 0
= 2(10 + 3) – 1(-5 – 3) + 0
= 2(13) – 1(-8)
= 26 + 8
= 34 ———-❶
≠ 0
∴ As |A| ≠ 0, A is nonsingular matrix and so has a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
{\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A})}
Now, {adj \text{ A}} can be calculated as below:
Let’s first find the Minors of each of the elements of A and then the Cofactors. We can then find {adj \text{ A}} as below:
{adj \text{A} = \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}}
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{1}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{-2} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{3} & \textcolor{green}{-5} \end{vmatrix}}
{= \begin{vmatrix} -2 & -1 \\[5pt] 3 & -5 \end{vmatrix} = \big((-2) × (-5) - (-1) × 3\big) = 10 + 3 = 13}
{a_{12} = 1}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{-2}} & \textcolor{green}{-1} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{3}} & \textcolor{green}{-5} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 0 & -5 \end{vmatrix} = \big(1 × (-5) - (-1) × 0\big) = -5 - 0 = -5}
{a_{13} = 1}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{-2} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{3} & \textcolor{red}{\cancel{-5}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -2 \\[5pt] 0 & 3 \end{vmatrix} = \big(1 × 3 - (-2) × 0\big) = 3 - 0 = 3}
{a_{21} = 1}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{1} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{3} & \textcolor{green}{-5} \end{vmatrix}}
{= \begin{vmatrix} 1 & 1 \\[5pt] 3 & -5 \end{vmatrix} = \big(1 × (-5) - 1 × 3\big) = -5 - 3 = -8}
{a_{22} = -2}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{1}} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{3}} & \textcolor{green}{-5} \end{vmatrix}}
{= \begin{vmatrix} 2 & 1 \\[5pt] 0 & -5 \end{vmatrix} = \big(2 × (-5) - 1 × 0\big) = -10 - 0 = -10}
{a_{23} = -1}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{green}{1} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{3} & \textcolor{red}{\cancel{-5}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 1 \\[5pt] 0 & 3 \end{vmatrix} = (2 × 3 - 1 × 0) = 6 - 0 = 6}
{a_{31} = 0}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{1} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{-2} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-5}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 1 \\[5pt] -2 & -1 \end{vmatrix} = \big(1 × (-1) - 1 × (-2)\big) = -1 + 2 = 1}
{a_{32} = 3}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{1}} & \textcolor{green}{1} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{-2}} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-5}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 1 \\[5pt] 1 & -1 \end{vmatrix} = \big(2 × (-1) - 1 × 1\big) = -2 - 1 = -3}
{a_{33} = -5}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{green}{1} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{-2} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{-5}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 1 \\[5pt] 1 & -2 \end{vmatrix} = \big(2 × (-2) - 1 × 1\big) = -4 - 1 = -5}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 2}
{\text{M}_{11} = 13}
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = 1(13) = 13}
{a_{12} = 1}
{\text{M}_{12} = -5}
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)(-5) = 5}
{a_{13} = 1}
{\text{M}_{13} = 3}
{\text{A}_{13} = (-1)^{1 + 3}\text{M}_{13} = 1(3) = 3}
{a_{21} = 1}
{\text{M}_{21} = -8}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)(-8) = 8}
{a_{22} = -2}
{\text{M}_{22} = -10}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(-10) = -10}
{a_{23} = -1}
{\text{M}_{23} = 6}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)6 = -6}
{a_{31} = 0}
{\text{M}_{31} = 1}
{\text{A}_{31} = (-1)^{3 + 1}\text{M}_{31} = 1(1) = 1}
{a_{32} = 3}
{\text{M}_{32} = -3}
{\text{A}_{32} = (-1)^{3 + 2}\text{M}_{32} = (-1)(-3) = 3}
{a_{33} = -5}
{\text{M}_{33} = -5}
{\text{A}_{33} = (-1)^{3 + 3}\text{M}_{33} = 1(-5) = -5}
So,
{= \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} 13 & 8 & 1 \\[5pt] 5 & -10 & 3 \\[5pt] 3 & -6 & -5 \end{bmatrix}} ———-❷
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{34} \begin{bmatrix} 13 & 8 & 1 \\[5pt] 5 & -10 & 3 \\[5pt] 3 & -6 & -5 \end{bmatrix}} (from ❶ and ❷)
As we know,
X
{= \text{A}^{-1}\text{B}}
{⇒ \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}}
{= \dfrac{1}{34} \begin{bmatrix} 13 & 8 & 1 \\[5pt] 5 & -10 & 3 \\[5pt] 3 & -6 & -5 \end{bmatrix} \begin{bmatrix} 1 \\[5pt] \dfrac32 \\[10pt] 9 \end{bmatrix}}
{= \dfrac{1}{34} \begin{bmatrix} 13(1) + 8\left(\dfrac32\right) + 1(9) \\[10pt] 5(1) + (-10)\left(\dfrac32\right) + 3(9) \\[10pt] 3(1) + (-6)\left(\dfrac32\right) + (-5)9 \end{bmatrix}}
{= \dfrac{1}{34} \begin{bmatrix} 13 + 12 + 9 \\[5pt] 5 - 15 + 27 \\[5pt] 3 - 9 - 45 \end{bmatrix}}
{= \dfrac{1}{34} \begin{bmatrix} 34 \\[5pt] 17 \\[5pt] -51 \end{bmatrix}}
{= \begin{bmatrix} \dfrac{1}{34} × 34 \\[10pt] \dfrac{1}{34} × 17 \\[10pt] \dfrac{1}{34} × (-51) \end{bmatrix}}
{= \begin{bmatrix} 1 \\[5pt] \dfrac12 \\[10pt] \dfrac{-3}{2} \end{bmatrix}}
∴ The solution of given system of linear equations is {x = 1,} {y = \dfrac12} and {z = \dfrac{-3}{2}}
Solve system of linear equations, using matrix method, in Exercises 7 to 14.
12.
{x - y + z = 4}
{2x + y - 3z = 0}
{x + y + z = 2}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 1 & -1 & 1 \\[5pt] 2 & 1 & -3 \\[5pt] 1 & 1 & 1 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} & = & {\begin{bmatrix} 4 \\[5pt] 0 \\[5pt] 2 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 1 & -1 & 1 \\[5pt] 2 & 1 & -3 \\[5pt] 1 & 1 & 1 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} and {\text{B} = \begin{bmatrix} 4 \\[5pt] 0 \\[5pt] 2 \end{bmatrix}}
Now, let’s expand along {\text{R}_1,} to find |A|. Thus, we have,
|A|
{= \begin{bmatrix} 2 & 1 & 1 \\[5pt] 1 & -2 & -1 \\[5pt] 0 & 3 & -5 \end{bmatrix}}
{= 1 \begin{bmatrix} 1 & -3 \\[5pt] 1 & 1 \end{bmatrix} - (-1) \begin{bmatrix} 2 & -3 \\[5pt] 1 & 1 \end{bmatrix} + 1 \begin{bmatrix} 2 & 1 \\[5pt] 1 & 1 \end{bmatrix}}
= 1[1(1) – (-3)1] + 1[2(1) – (-3)1] + 1[2(1) – 1(1)]
= 1(1 + 3) + 1(2 + 3) + 1(2 – 1)
= 1 × 4 + 1 × 5 + 1 × 1
= 4 + 5 + 1
= 10 ———-❶
≠ 0
∴ As |A| ≠ 0, A is nonsingular matrix and so has a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
{\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A})}
Now, {adj \text{ A}} can be calculated as below:
Let’s first find the Minors of each of the elements of A and then the Cofactors. We can then find {adj \text{ A}} as below:
{adj \text{A} = \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}}
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{1}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{1} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{1} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & -3 \\[5pt] 1 & 1 \end{vmatrix} = 1(1) - (-3)1 = 1 + 3 = 4}
{a_{12} = -1}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{1}} & \textcolor{green}{-3} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{1}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 2 & -3 \\[5pt] 1 & 1 \end{vmatrix} = 2(1) - (-3)1 = 2 + 3 = 5}
{a_{13} = 1}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{2} & \textcolor{green}{1} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{1} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 1 \\[5pt] 1 & 1 \end{vmatrix} = 2(1) - 1(1) = 2 - 1 = 1}
{a_{21} = 2}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{1} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} -1 & 1 \\[5pt] 1 & 1 \end{vmatrix} = (-1)1 - 1(1) = -1 - 1 = -2}
{a_{22} = 1}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{1}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & 1 \\[5pt] 1 & 1 \end{vmatrix} = 1(1) - 1(1) = 1 - 1 = 0}
{a_{23} = -3}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{1} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 1 & 1 \end{vmatrix} = 1(1) - (-1)1\big) = 1 + 1 = 2}
{a_{31} = 1}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{1} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} -1 & 1 \\[5pt] 1 & -3 \end{vmatrix} = (-1)(-3) - 1(1)\big) = 3 - 1 = 2}
{a_{32} = 1}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{1} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{1}} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 1 \\[5pt] 2 & -3 \end{vmatrix} = 1(-3) - 1(2) = -3 - 2 = -5}
{a_{33} = 1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{2} & \textcolor{green}{1} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 2 & 1 \end{vmatrix} = 1(1) - (-1)2 = 1 + 2 = 3}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 4}
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = 1(4) = 4}
{a_{12} = -1}
{\text{M}_{12} = 5}
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)5 = -5}
{a_{13} = 1}
{\text{M}_{13} = 1}
{\text{A}_{13} = (-1)^{1 + 3}\text{M}_{13} = 1(1) = 1}
{a_{21} = 2}
{\text{M}_{21} = -2}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)(-2) = 2}
{a_{22} = 1}
{\text{M}_{22} = 0}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(0) = 0}
{a_{23} = -3}
{\text{M}_{23} = 2}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)2 = -2}
{a_{31} = 1}
{\text{M}_{31} = 2}
{\text{A}_{31} = (-1)^{3 + 1}\text{M}_{31} = 1(2) = 2}
{a_{32} = 1}
{\text{M}_{32} = -5}
{\text{A}_{32} = (-1)^{3 + 2}\text{M}_{32} = (-1)(-5) = 5}
{a_{33} = 1}
{\text{M}_{33} = 3}
{\text{A}_{33} = (-1)^{3 + 3}\text{M}_{33} = 1(3) = 3}
So,
{= \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} 4 & 2 & 2 \\[5pt] -5 & 0 & 5 \\[5pt] 1 & 2 & 3 \end{bmatrix}} ———-❷
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\[5pt] -5 & 0 & 5 \\[5pt] 1 & 2 & 3 \end{bmatrix}} (from ❶ and ❷)
As we know,
X
{= \text{A}^{-1}\text{B}}
{⇒ \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\[5pt] -5 & 0 & 5 \\[5pt] 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\[5pt] 0 \\[10pt] 2 \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 4(4) + 2(0) + 2(2) \\[5pt] (-5)4 + 0(0) + 5(2) \\[5pt] 1(4) + 2(0) + 3(2) \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 16 + 0 + 4 \\[5pt] -20 + 0 + 10 \\[5pt] 4 + 0 + 6 \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 20 \\[5pt] -10 \\[5pt] 10 \end{bmatrix}}
{= \begin{bmatrix} \dfrac{1}{10} × 20 \\[10pt] \dfrac{1}{10} × (-10) \\[10pt] \dfrac{1}{10} × 10 \end{bmatrix}}
{= \begin{bmatrix} 2 \\[5pt] -1 \\[5pt] 1 \end{bmatrix}}
∴ The solution of given system of linear equations is {x = 2,} {y = -1} and {z = 1}
Solve system of linear equations, using matrix method, in Exercises 7 to 14.
13.
{2x + 3y + 3z = 5}
{x - 2y + z = -4}
{3x - y - 2z = 3}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 2 & 3 & 3 \\[5pt] 1 & -2 & 1 \\[5pt] 3 & -1 & -2 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} & = & {\begin{bmatrix} 5 \\[5pt] -4 \\[5pt] 3 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 2 & 3 & 3 \\[5pt] 1 & -2 & 1 \\[5pt] 3 & -1 & -2 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} and {\text{B} = \begin{bmatrix} 5 \\[5pt] -4 \\[5pt] 3 \end{bmatrix}}
Now, let’s expand along {\text{R}_1,} to find |A|. Thus, we have,
|A|
{= \begin{bmatrix} 2 & 1 & 1 \\[5pt] 1 & -2 & -1 \\[5pt] 0 & 3 & -5 \end{bmatrix}}
{= 1 \begin{bmatrix} 1 & -3 \\[5pt] 1 & 1 \end{bmatrix} - (-1) \begin{bmatrix} 2 & -3 \\[5pt] 1 & 1 \end{bmatrix} + 1 \begin{bmatrix} 2 & 1 \\[5pt] 1 & 1 \end{bmatrix}}
= 1[1(1) – (-3)1] + 1[2(1) – (-3)1] + 1[2(1) – 1(1)]
= 1(1 + 3) + 1(2 + 3) + 1(2 – 1)
= 1(4) + 1(5) + 1(1)
= 4 + 5 + 1
= 10 ———-❶
≠ 0
∴ As |A| ≠ 0, A is nonsingular matrix and so has a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
{\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A})}
Now, {adj \text{ A}} can be calculated as below:
Let’s first find the Minors of each of the elements of A and then the Cofactors. We can then find {adj \text{ A}} as below:
{adj \text{A} = \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}}
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{1}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{1} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{1} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & -3 \\[5pt] 1 & 1 \end{vmatrix} = 1(1) - (-3)1 = 1 + 3 = 4}
{a_{12} = -1}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{1}} & \textcolor{green}{-3} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{1}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 2 & -3 \\[5pt] 1 & 1 \end{vmatrix} = 2(1) - (-3)1\big) = 2 + 3 = 5}
{a_{13} = 1}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{2} & \textcolor{green}{1} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{1} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 1 \\[5pt] 1 & 1 \end{vmatrix} = 2(1) - 1(1) = 2 - 1 = 1}
{a_{21} = 2}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{1} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} -1 & 1 \\[5pt] 1 & 1 \end{vmatrix} = (-1)1 - 1(1) = -1 - 1 = -2}
{a_{22} = 1}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{1}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & 1 \\[5pt] 1 & 1 \end{vmatrix} = 1(1) - 1(1) = 1 - 1 = 0}
{a_{23} = -3}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{1} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 1 & 1 \end{vmatrix} = 1(1) - (-1)1\big) = 1 + 1 = 2}
{a_{31} = 1}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{1} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} -1 & 1 \\[5pt] 1 & -3 \end{vmatrix} = (-1)(-3) - 1(1) = 3 - 1 = 2}
{a_{32} = 1}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{1} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{1}} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 1 \\[5pt] 2 & -3 \end{vmatrix} = 1(-3) - 1(2) = -3 - 2 = -5}
{a_{33} = 1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{2} & \textcolor{green}{1} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 2 & 1 \end{vmatrix} = 1(1) - (-1)2\big) = 1 + 2 = 3}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 4}
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = 1(4) = 4}
{a_{12} = -1}
{\text{M}_{12} = 5}
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)5 = -5}
{a_{13} = 1}
{\text{M}_{13} = 1}
{\text{A}_{13} = (-1)^{1 + 3}\text{M}_{13} = 1(1) = 1}
{a_{21} = 2}
{\text{M}_{21} = -2}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)(-2) = 2}
{a_{22} = 1}
{\text{M}_{22} = 0}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(0) = 0}
{a_{23} = -3}
{\text{M}_{23} = 2}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)2 = -2}
{a_{31} = 1}
{\text{M}_{31} = 2}
{\text{A}_{31} = (-1)^{3 + 1}\text{M}_{31} = 1(2) = 2}
{a_{32} = 1}
{\text{M}_{32} = -5}
{\text{A}_{32} = (-1)^{3 + 2}\text{M}_{32} = (-1)(-5) = 5}
{a_{33} = 1}
{\text{M}_{33} = 3}
{\text{A}_{33} = (-1)^{3 + 3}\text{M}_{33} = 1(3) = 3}
So,
{= \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} 4 & 2 & 2 \\[5pt] -5 & 0 & 5 \\[5pt] 1 & 2 & 3 \end{bmatrix}} ———-❷
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\[5pt] -5 & 0 & 5 \\[5pt] 1 & 2 & 3 \end{bmatrix}} (from ❶ and ❷)
As we know,
X
{= \text{A}^{-1}\text{B}}
{⇒ \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\[5pt] -5 & 0 & 5 \\[5pt] 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\[5pt] 0 \\[10pt] 2 \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 4(4) + 2(0) + 2(2) \\[5pt] (-5)4 + 0(0) + 5(2) \\[5pt] 1(4) + 2(0) + 3(2) \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 16 + 0 + 4 \\[5pt] -20 + 0 + 10 \\[5pt] 4 + 0 + 6 \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 20 \\[5pt] -10 \\[5pt] 10 \end{bmatrix}}
{= \begin{bmatrix} \dfrac{1}{10} × 20 \\[10pt] \dfrac{1}{10} × (-10) \\[10pt] \dfrac{1}{10} × 10 \end{bmatrix}}
{= \begin{bmatrix} 2 \\[5pt] -1 \\[5pt] 1 \end{bmatrix}}
∴ The solution of given system of linear equations is {x = 2,} {y = -1} and {z = 1}
Solve system of linear equations, using matrix method, in Exercises 7 to 14.
14.
{x - y + 2z = 7}
{3x + 4y - 5z = -5}
{2x - y + 3z = 12}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 1 & -1 & 2 \\[5pt] 3 & 4 & -5 \\[5pt] 2 & -1 & 3 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} & = & {\begin{bmatrix} 7 \\[5pt] -5 \\[5pt] 12 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 1 & -1 & 2 \\[5pt] 3 & 4 & -5 \\[5pt] 2 & -1 & 3 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} and {\text{B} = \begin{bmatrix} 7 \\[5pt] -5 \\[5pt] 12 \end{bmatrix}}
Now, let’s expand along {\text{R}_1,} to find |A|. Thus, we have,
|A|
{= \begin{bmatrix} 1 & -1 & 2 \\[5pt] 3 & 4 & -5 \\[5pt] 2 & -1 & 3 \end{bmatrix}}
{= 1 \begin{bmatrix} 4 & -5 \\[5pt] -1 & 3 \end{bmatrix} - (-1) \begin{bmatrix} 3 & -5 \\[5pt] 2 & 3 \end{bmatrix} + 2 \begin{bmatrix} 3 & 4 \\[5pt] 2 & -1 \end{bmatrix}}
= 1[4(3) – (-5)(-1)] + 1[3(3) – (-5)2] + 2[3(-1) – 4(2)]
= 1(12 – 5) + 1(9 + 10) + 2(-3 – 8)
= 1(7) + 1(19) + 2(-11)
= 7 + 19 – 22
= 4 ———-❶
≠ 0
∴ As |A| ≠ 0, A is nonsingular matrix and so has a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
{\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A})}
Now, {adj \text{ A}} can be calculated as below:
Let’s first find the Minors of each of the elements of A and then the Cofactors. We can then find {adj \text{A}} as below:
{adj \text{ A} = \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}}
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}{4} & \textcolor{green}{-5} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{-1} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 4 & -5 \\[5pt] -1 & 3 \end{vmatrix} = 4(3) - (-5)(-1) = 12 - 5 = 7}
{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{4}} & \textcolor{green}{-5} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 3 & -5 \\[5pt] 2 & 3 \end{vmatrix} = 3(3) - (-5)2 = 9 + 10 = 19}
{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}{4} & \textcolor{red}{\cancel{-5}} \\[5pt] \textcolor{green}{2} & \textcolor{green}{-1} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 3 & 4 \\[5pt] 2 & -1 \end{vmatrix} = 3(-1) - 4(2) = -3 - 8 = -11}
{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{4}} & \textcolor{red}{\cancel{-5}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{-1} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} -1 & 2 \\[5pt] -1 & 3 \end{vmatrix} = (-1)3 - 2(-1) = -3 + 1 = -1}
{a_{22} = 4}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-5}} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] 2 & 3 \end{vmatrix} = 1(3) - 2(2) = 3 - 4 = -1}
{a_{23} = -5}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-5}} \\[5pt] \textcolor{green}{2} & \textcolor{green}{-1} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 2 & -1 \end{vmatrix} = 1(-1) - (-1)2 = -1 + 2 = 1}
{a_{31} = 2}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-1} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{4} & \textcolor{green}{-5} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} -1 & 2 \\[5pt] 4 & -5 \end{vmatrix} = (-1)(-5) - 2(4) = 5 - 8 = -3}
{a_{32} = -1}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-1}} & \textcolor{green}{2} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{4}} & \textcolor{green}{-5} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] 3 & -5 \end{vmatrix} = 1(-5) - 2(3) = -5 - 6 = -11}
{a_{33} = 3}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-1} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{4} & \textcolor{red}{\cancel{-5}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -1 \\[5pt] 3 & 4 \end{vmatrix} = 1(4) - (-1)3 = 4 + 3 = 7}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 7}
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = 1(7) = 7}
{a_{12} = -1}
{\text{M}_{12} = 19}
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)19 = -19}
{a_{13} = 2}
{\text{M}_{13} = -11}
{\text{A}_{13} = (-1)^{1 + 3}\text{M}_{13} = 1(-11) = -11}
{a_{21} = 3}
{\text{M}_{21} = -1}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)(-1) = 1}
{a_{22} = 4}
{\text{M}_{22} = -1}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(-1) = -1}
{a_{23} = -5}
{\text{M}_{23} = 1}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)1 = -1}
{a_{31} = 2}
{\text{M}_{31} = -3}
{\text{A}_{31} = (-1)^{3 + 1}\text{M}_{31} = 1(-3) = -3}
{a_{32} = -1}
{\text{M}_{32} = -11}
{\text{A}_{32} = (-1)^{3 + 2}\text{M}_{32} = (-1)(-11) = 11}
{a_{33} = 3}
{\text{M}_{33} = 7}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1(7) = 7}
So,
{= \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} 7 & 1 & -3 \\[5pt] -19 & -1 & 11 \\[5pt] -11 & -1 & 7 \end{bmatrix}} ———-❷
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\[5pt] -19 & -1 & 11 \\[5pt] -11 & -1 & 7 \end{bmatrix}} (from ❶ and ❷)
As we know,
X
{= \text{A}^{-1}\text{B}}
{⇒ \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}}
{= \dfrac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\[5pt] -19 & -1 & 11 \\[5pt] -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \\[5pt] -5 \\[10pt] 12 \end{bmatrix}}
{= \dfrac{1}{4} \begin{bmatrix} 7(7) + 1(-5) + (-3)12 \\[5pt] (-19)7 + (-1)(-5) + 11(12) \\[5pt] (-11)7 + (-1)(-5) + 7(12) \end{bmatrix}}
{= \dfrac{1}{4} \begin{bmatrix} 49 - 5 - 36 \\[5pt] -133 + 5 + 132 \\[5pt] -77 + 5 + 84 \end{bmatrix}}
{= \dfrac{1}{4} \begin{bmatrix} 8 \\[5pt] 4 \\[5pt] 12 \end{bmatrix}}
{= \begin{bmatrix} \dfrac{1}{4} × 8 \\[10pt] \dfrac{1}{4} × 4 \\[10pt] \dfrac{1}{4} × 12 \end{bmatrix}}
{= \begin{bmatrix} 2 \\[5pt] 1 \\[5pt] 3 \end{bmatrix}}
∴ The solution of given system of linear equations is {x = 2,} {y = 1} and {z = 3}
15. If {\text{A} = \begin{bmatrix} 2 & -3 & 5 \\[5pt] 3 & 2 & -4 \\[5pt] 1 & 1 & -2 \end{bmatrix},} find {\text{A}^{-1}.} Using {\text{A}^{-1}} solve the system of equations
{\begin{array}{c} 2x - 3y + 5z = 11 \\[5pt] 3x + 2y - 4z = -5 \\[5pt] x + y - 2z = -3 \end{array}}
Given that,
{\text{A} = \begin{bmatrix} 2 & -3 & 5 \\[5pt] 3 & 2 & -4 \\[5pt] 1 & 1 & -2 \end{bmatrix}}
Now, let’s expand along {\text{R}_1,} to find |A|. Thus, we have,
|A|
{= \begin{bmatrix} 2 & -3 & 5 \\[5pt] 3 & 2 & -4 \\[5pt] 1 & 1 & -2 \end{bmatrix}}
{= 2 \begin{bmatrix} 2 & -4 \\[5pt] 1 & -2 \end{bmatrix} - (-3) \begin{bmatrix} 3 & -4 \\[5pt] 1 & -2 \end{bmatrix} + 5 \begin{bmatrix} 3 & 2 \\[5pt] 1 & 1 \end{bmatrix}}
= 2[2(-2) – (-4)1] + 3[3(-2) – (-4)1] + 5[3(1) – 2(1)]
= 2(-4 + 4) + 3(-6 + 4) + 5(3 – 2)
= 2(0) + 3(-2) + 5(1)
= 0 – 6 + 5
= -1 ———-❶
≠ 0
∴ As |A| ≠ 0, A is nonsingular matrix and so has a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
{\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A})}
Now, {adj \text{ A}} can be calculated as below:
Let’s first find the Minors of each of the elements of A and then the Cofactors. We can then find {adj \text{ A}} as below:
{adj \text{A} = \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}}
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{-3}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{2} & \textcolor{green}{-4} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{1} & \textcolor{green}{-2} \end{vmatrix}}
{= \begin{vmatrix} 2 & -4 \\[5pt] 1 & -2 \end{vmatrix} = 2(-2) - (-4)1 = -4 + 4 = 0}
{a_{12} = -3}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-3}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{2}} & \textcolor{green}{-4} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{1}} & \textcolor{green}{-2} \end{vmatrix}}
{= \begin{vmatrix} 3 & -4 \\[5pt] 1 & -2 \end{vmatrix} = 3(-2) - (-4)1 = -6 + 4 = -2}
{a_{13} = 5}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-3}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{2} & \textcolor{red}{\cancel{-4}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{1} & \textcolor{red}{\cancel{-2}} \end{vmatrix}}
{= \begin{vmatrix} 3 & 2 \\[5pt] 1 & 1 \end{vmatrix} = 3(1) - 2(1) = 3 - 2 = 1}
{a_{21} = 3}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{-3} & \textcolor{green}{5} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-4}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{1} & \textcolor{green}{-2} \end{vmatrix}}
{= \begin{vmatrix} -3 & 5 \\[5pt] 1 & -2 \end{vmatrix} = (-3)(-2) - 5(1) = 6 - 5 = 1}
{a_{22} = 2}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{-3}} & \textcolor{green}{5} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-4}} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{1}} & \textcolor{green}{-2} \end{vmatrix}}
{= \begin{vmatrix} 2 & 5 \\[5pt] 1 & -2 \end{vmatrix} = 2(-2) - 5(1) = -4 - 5 = -9}
{a_{23} = -4}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{green}{-3} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-4}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{1} & \textcolor{red}{\cancel{-2}} \end{vmatrix}}
{= \begin{vmatrix} 2 & -3 \\[5pt] 1 & 1 \end{vmatrix} = 2(1) - (-3)1 = 2 + 3 = 5}
{a_{31} = 1}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{-3} & \textcolor{green}{5} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{2} & \textcolor{green}{-4} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-2}} \end{vmatrix}}
{= \begin{vmatrix} -3 & 5 \\[5pt] 2 & -4 \end{vmatrix} = (-3)(-4) - 5(2) = 12 - 10 = 2}
{a_{32} = 1}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{-3}} & \textcolor{green}{5} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{2}} & \textcolor{green}{-4} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-2}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 5 \\[5pt] 3 & -4 \end{vmatrix} = 2(-4) - 5(3) = -8 - 15 = -23}
{a_{33} = -2}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{green}{-3} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{2} & \textcolor{red}{\cancel{-4}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-2}} \end{vmatrix}}
{= \begin{vmatrix} 2 & -3 \\[5pt] 3 & 2 \end{vmatrix} = 2(2) - (-3)3 = 4 + 9 = 13}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 2}
{\text{M}_{11} = 0}
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = 1(0) = 0}
{a_{12} = -3}
{\text{M}_{12} = -2}
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)(-2) = 2}
{a_{13} = 5}
{\text{M}_{13} = 1}
{\text{A}_{13} = (-1)^{1 + 3}\text{M}_{13} = 1(1) = 1}
{a_{21} = 3}
{\text{M}_{21} = 1}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)1 = -1}
{a_{22} = 2}
{\text{M}_{22} = -9}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(-9) = -9}
{a_{23} = -4}
{\text{M}_{23} = 5}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)5 = -5}
{a_{31} = 1}
{\text{M}_{31} = 2}
{\text{A}_{31} = (-1)^{3 + 1}\text{M}_{31} = 1(2) = 2}
{a_{32} = 1}
{\text{M}_{32} = -23}
{\text{A}_{32} = (-1)^{3 + 2}\text{M}_{32} = (-1)(-23) = 23}
{a_{33} = -2}
{\text{M}_{33} = 13}
{\text{A}_{33} = (-1)^{3 + 3}\text{M}_{33} = 1(13) = 13}
So,
{= \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 & -1 & 2 \\[5pt] 2 & -9 & 23 \\[5pt] 1 & -5 & 13 \end{bmatrix}} ———-❷
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{-1}\begin{bmatrix} 0 & -1 & 2 \\[5pt] 2 & -9 & 23 \\[5pt] 1 & -5 & 13 \end{bmatrix}} (from ❶ and ❷)
{= -1\begin{bmatrix} 0 & -1 & 2 \\[5pt] 2 & -9 & 23 \\[5pt] 1 & -5 & 13 \end{bmatrix}}
{= \begin{bmatrix} 0 & 1 & -2 \\[5pt] -2 & 9 & -23 \\[5pt] -1 & 5 & -13 \end{bmatrix}}
Thus, we have,
{\text{A}^{-1} = \begin{bmatrix} 0 & 1 & -2 \\[5pt] -2 & 9 & -23 \\[5pt] -1 & 5 & -13 \end{bmatrix}} ———-❸
Now, the system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 2 & -3 & 5 \\[5pt] 3 & 2 & -4 \\[5pt] 1 & 1 & -2 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} & = & {\begin{bmatrix} 11 \\[5pt] -5 \\[5pt] -3 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 2 & -3 & 5 \\[5pt] 3 & 2 & -4 \\[5pt] 1 & 1 & -2 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} and {\text{B} = \begin{bmatrix} 11 \\[5pt] -5 \\[5pt] -3 \end{bmatrix}}
As we know,
X
{= \text{A}^{-1}\text{B}}
{⇒ \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}}
{= \begin{bmatrix} 0 & 1 & -2 \\[5pt] -2 & 9 & -23 \\[5pt] -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\[5pt] -5 \\[5pt] -3 \end{bmatrix}} (from ❸)
{= \begin{bmatrix} 0(11) + 1(-5) + (-2)(-3) \\[5pt] (-2)11 + 9(-5) + (-23)(-3) \\[5pt] (-1)11 + 5(-5) + (-13)(-3) \end{bmatrix}}
{= \begin{bmatrix} 0 - 5 + 6 \\[5pt] -22 - 45 + 69 \\[5pt] -11 - 25 + 39 \end{bmatrix}}
{= \begin{bmatrix} 1 \\[5pt] 2 \\[5pt] 3 \end{bmatrix}}
∴ The solution of given system of linear equations is {x = 1,} {y = 2} and {z = 3}
16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹ 70. Find cost of each item per kg by matrix method.
Let the unit cost per kg of the three items is as given below.
Item
Cost per kg
Onion
= x
Wheat
= y
Rice
= z
So, each of the given cost statements can be represented with the following equations.
Cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60: {4x + 3y + 2z = 60}
Cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90: {2x + 4y + 6z = 90}
Cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹ 70: {6x + 2y + 3z = 70}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 4 & 3 & 2 \\[5pt] 2 & 4 & 6 \\[5pt] 6 & 2 & 3 \end{bmatrix}} & {\begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} & = & {\begin{bmatrix} 60 \\[5pt] 90 \\[5pt] 70 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 4 & 3 & 2 \\[5pt] 2 & 4 & 6 \\[5pt] 6 & 2 & 3 \end{bmatrix},} {\text{X} = \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}} and {\text{B} = \begin{bmatrix} 60 \\[5pt] 90 \\[5pt] 70 \end{bmatrix}}
Now, let’s expand along {\text{R}_1,} to find |A|. Thus, we have,
|A|
{= \begin{bmatrix} 4 & 3 & 2 \\[5pt] 2 & 4 & 6 \\[5pt] 6 & 2 & 3 \end{bmatrix}}
{= 4\begin{bmatrix} 4 & 6 \\[5pt] 2 & 3 \end{bmatrix} - 3\begin{bmatrix} 2 & 6 \\[5pt] 6 & 3 \end{bmatrix} + 2 \begin{bmatrix} 2 & 4 \\[5pt] 6 & 2 \end{bmatrix}}
= 4[4(3) – 6(2)] – 3[2(3) – 6(6)] + 2[2(2) – 4(6)]
= 4(12 – 12) – 3(6 – 36) + 2(4 – 24)
= 4(0) – 3(-30) + 2(-20)
= 0 + 90 – 40
= 50 ———-❶
≠ 0
∴ As |A| ≠ 0, A is nonsingular matrix and the given system of equations have a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
{\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A})}
Now, {adj \text{ A}} can be calculated as below:
Let’s first find the Minors of each of the elements of A and then the Cofactors. We can then find {adj \text{ A}} as below:
{adj \text{ A} = \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}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{a_{11} = 4}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{4} & \textcolor{green}{6} \\[5pt] \textcolor{red}{\cancel{6}} & \textcolor{green}{2} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 4 & 6 \\[5pt] 2 & 3 \end{vmatrix} = 4(3) - 6(2) = 12 - 12 = 0}
{a_{12} = 3}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{4}} & \textcolor{green}{6} \\[5pt] \textcolor{green}{6} & \textcolor{red}{\cancel{2}} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 2 & 6 \\[5pt] 6 & 3 \end{vmatrix} = 2(3) - 6(6) = 6 - 36 = -30}
{a_{13} = 2}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{2} & \textcolor{green}{4} & \textcolor{red}{\cancel{6}} \\[5pt] \textcolor{green}{6} & \textcolor{green}{2} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 4 \\[5pt] 6 & 2 \end{vmatrix} = 2(2) - 4(6) = 4 - 24 = -20}
{a_{21} = 2}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{4}} & \textcolor{green}{3} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{6}} \\[5pt] \textcolor{red}{\cancel{6}} & \textcolor{green}{2} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 3 & 2 \\[5pt] 2 & 3 \end{vmatrix} = 3(3) - 2(2) = 9 - 4 = 5}
{a_{22} = 4}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{4} & \textcolor{red}{\cancel{3}} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{6}} \\[5pt] \textcolor{green}{6} & \textcolor{red}{\cancel{2}} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 4 & 2 \\[5pt] 6 & 3 \end{vmatrix} = 4(3) - 2(6) = 12 - 12 = 0}
{a_{23} = 6}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{4} & \textcolor{green}{3} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{6}} \\[5pt] \textcolor{green}{6} & \textcolor{green}{2} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 4 & 3 \\[5pt] 6 & 2 \end{vmatrix} = 4(2) - 3(6) = 8 - 18 = -10}
{a_{31} = 6}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{4}} & \textcolor{green}{3} & \textcolor{green}{2} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{green}{4} & \textcolor{green}{6} \\[5pt] \textcolor{red}{\cancel{6}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 3 & 2 \\[5pt] 4 & 6 \end{vmatrix} = 3(6) - 2(4) = 18 - 8 = 10}
{a_{32} = 2}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{4} & \textcolor{red}{\cancel{3}} & \textcolor{green}{2} \\[5pt] \textcolor{green}{2} & \textcolor{red}{\cancel{4}} & \textcolor{green}{6} \\[5pt] \textcolor{red}{\cancel{6}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 4 & 2 \\[5pt] 2 & 6 \end{vmatrix} = 4(6) - 2(2) = 24 - 4 = 20}
{a_{33} = 3}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{4} & \textcolor{green}{3} & \textcolor{red}{\cancel{2}} \\[5pt] \textcolor{green}{2} & \textcolor{green}{4} & \textcolor{red}{\cancel{6}} \\[5pt] \textcolor{red}{\cancel{6}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 4 & 3 \\[5pt] 2 & 4 \end{vmatrix} = 4(4) - 3(2) = 16 - 6 = 10}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 4}
{\text{M}_{11} = 0}
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = 1(0) = 0}
{a_{12} = 3}
{\text{M}_{12} = -30}
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)(-30) = 30}
{a_{13} = 2}
{\text{M}_{13} = -20}
{\text{A}_{13} = (-1)^{1 + 3}\text{M}_{13} = 1(-20) = -20}
{a_{21} = 2}
{\text{M}_{21} = 5}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)5 = -5}
{a_{22} = 4}
{\text{M}_{22} = 0}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(0) = 0}
{a_{23} = 6}
{\text{M}_{23} = -10}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)(-10) = 10}
{a_{31} = 6}
{\text{M}_{31} = 10}
{\text{A}_{31} = (-1)^{3 + 1}\text{M}_{31} = 1(10) = 10}
{a_{32} = 2}
{\text{M}_{32} = 20}
{\text{A}_{32} = (-1)^{3 + 2}\text{M}_{32} = (-1)20 = -20}
{a_{33} = 3}
{\text{M}_{33} = 10}
{\text{A}_{33} = (-1)^{3 + 3}\text{M}_{33} = 1(10) = 10}
So,
{= \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 & -5 & 10 \\[5pt] 30 & 0 & -20 \\[5pt] -20 & 10 & 10 \end{bmatrix}} ———-❷
{\text{A}^{-1}}
{= \dfrac{1}{\text{|A|}}(adj \text{ A})}
{= \dfrac{1}{50} \begin{bmatrix} 0 & -5 & 10 \\[5pt] 30 & 0 & -20 \\[5pt] -20 & 10 & 10 \end{bmatrix}} (from ❶ and ❷)
Taking 5 common,
{\text{A}^{-1}}
{= \dfrac{1}{50} × 5 × \begin{bmatrix} 0 & -1 & 2 \\[5pt] 6 & 0 & -4 \\[5pt] -4 & 2 & 2 \end{bmatrix}}
{= \dfrac{1}{10} × \begin{bmatrix} 0 & -1 & 2 \\[5pt] 6 & 0 & -4 \\[5pt] -4 & 2 & 2 \end{bmatrix}}
As we know,
X
{= \text{A}^{-1}\text{B}}
{⇒ \begin{bmatrix} x \\[5pt] y \\[5pt] z \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 0 & -1 & 2 \\[5pt] 6 & 0 & -4 \\[5pt] -4 & 2 & 2 \end{bmatrix} \begin{bmatrix} 60 \\[5pt] 90 \\[5pt] 70 \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 0(60) + (-1)90 + 2(70) \\[5pt] 6(60) + 0(90) + (-4)70 \\[5pt] (-4)60 + 2(90) + 2(70) \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 0 - 90 + 140 \\[5pt] 360 + 0 - 280 \\[5pt] -240 + 180 + 140 \end{bmatrix}}
{= \dfrac{1}{10} \begin{bmatrix} 50 \\[5pt] 80 \\[5pt] 80 \end{bmatrix}}
{= \begin{bmatrix} \dfrac{1}{10} × 50 \\[10pt] \dfrac{1}{10} × 80 \\[10pt] \dfrac{1}{10} × 80 \end{bmatrix}}
{= \begin{bmatrix} 5 \\[5pt] 8 \\[5pt] 8 \end{bmatrix}}
∴ The solution of given system of linear equations is {x = 5,} {y = 8} and {z = 8}
This implies that
Cost of Onions per kg
₹ 5
Cost of Wheat per kg
₹ 8
Cost of Rice per kg
₹ 8