Question 5, Exercise 2.6

Solutions of Question 5 of Exercise 2.6 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Solve the system of linear equation by using Cramer's rule.
$x_{1}+x_{2}+2 x_{3}=8$
$-x_{1}-2 x_{2}+3 x_{3}=1$
$3 x_{1}-7 x_{2}+4 x_{3}=10$

Solution. The above system may be written as $A X=B$; where, \begin{align*} &A = \begin{bmatrix} 1 & 1 & 2 \\ -1 & -2 & 3 \\ 3 & -7 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 1 \\ 10 \end{bmatrix}\\ |A| &= \begin{vmatrix} 1 & 1 & 2 \\ -1 & -2 & 3 \\ 3 & -7 & 4 \end{vmatrix}\\ &= 1(-8 +21) - 1(-4 - 9) + 2(7 +6)\\ &= 13 + 13 + 26 = 52\end{align*} So, $A$ is non-singular. \begin{align*} x_1 &= \frac{A_1}{|A|}\\ &=\frac{ \begin{vmatrix} 8 & 1 & 2 \\ 1 & -2 & 3 \\ 10 & -7 & 4 \end{vmatrix}}{52}\\ &=\frac{8(-8 + 21) - 1(4 - 30) + 2(-7 + 20)}{52}\\ &=\frac{104 + 26 + 26}{52}\\ & = \frac{156}{52} = 3 \end{align*} \begin{align*} x_2& = \frac{|A_2|}{|A|} \\ &=\frac{\begin{vmatrix} 1 & 8 & 2 \\ -1 & 1 & 3 \\ 3 & 10 & 4 \end{vmatrix}}{52}\\ &=\frac{1(4 - 30) - 8(-4 - 9) + 2(-10 - 3)}{52}\\ &=\frac{-26 + 104 - 26}{52}\\ &= \frac{52}{52} = 1 \end{align*} \begin{align*} x_3 &= \frac{|A_3|}{|A|}\\ &=\frac{ \begin{vmatrix} 1 & 1 & 8 \\ -1 & -2 & 1 \\ 3 & -7 & 10 \end{vmatrix}}{52}\\ &=\frac{1(-20 + 7) - 1(-10 - 3) + 8(7 +6)}{52}\\ &=\frac{-13 + 13 + 104}{52}\\ &= \frac{104}{52} = 2 \end{align*} Thus, the solution set is $(3, 1, 2)$.

Solve the system of linear equation by using Cramer's rule.
$2 x_{1}+2 x_{2}+x_{3}=0$
$-2 x_{1}+5 x_{2}+2 x_{3}=1$
$8 x_{1}+x_{2}+4 x_{3}=-1$

Solution. The above system maybe written as $AX = B $, where: \begin{align*} &A = \begin{bmatrix} 2 & 2 & 1 \\ -2 & 5 & 2 \\ 8 & 1 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}\\ |A| &= \begin{vmatrix} 2 & 2 & 1 \\ -2 & 5 & 2 \\ 8 & 1 & 4 \end{vmatrix}\\ &=2 \left( 20 - 2 \right) - 2 \left( -8 - 16 \right) + \left( -2 - 40 \right)\\ &=36 + 48 - 42\\ &=42\neq 0\end{align*} $A$ is non-singular. NOw \begin{align*} x_1 &= \frac{|A_1|}{|A|}\\ &=\frac{\begin{vmatrix} 0 & 2 & 1 \\ 1 & 5 & 2 \\ -1 & 1 & 4 \end{vmatrix}}{42}\\ &=\frac{ 0-2 ( 4 + 2) + 1(1 + 5)}{42}\\ &=\frac{ -12 + 6}{42}\\ &=-\frac{6}{42}\\ &=-\frac{1}{7}\end{align*} Now\begin{align*} x_2 &=\frac{|A_2|}{|A|}\\ &= \frac{ \begin{vmatrix} 2 & 0 & 1 \\ -2 & 1 & 2 \\ 8 & -1 & 4 \end{vmatrix}}{42}\\ &=\frac{2 ( 4 + 2 )-0 + 1(2 - 8)}{42}\\ &=\frac{ 12 - 6}{42}\\ &=\frac{6}{42}\\ & = \frac{1}{7} \end{align*} \begin{align*} x_3 &= \frac{|A_3|}{|A|}\\ &=\frac{\begin{vmatrix} 2 & 2 & 0 \\ -2 & 5 & 1 \\ 8 & 1 & -1 \end{vmatrix}}{42}\\ &=\frac{2 ( -5 - 1 ) - 2 ( 2 - 8 )+0}{42}\\ &=\frac{-12 + 12}{42}\\ &= \frac{0}{42} = 0 \end{align*} The solution set for the given system of equations using Cramer's rule is: $$( -\frac{1}{7}, \frac{1}{7}, 0 )$$

Solve the system of linear equation by using Cramer's rule.
$-2 x_{2}+3 x_{3}=1$
$3 x_{1}+6 x_{2}-3 x_{3}=-2$
$6 x_{1}+6 x_{2}+3 x_{3}=5$

Solution. The above system maybe written as $AX = B $, where: \begin{align*} &A = \begin{bmatrix} 0 & -2 & 3 \\ 3 & 6 & -3 \\ 6 & 6 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ -2 \\ 5 \end{bmatrix}\\ |A| &= \begin{vmatrix} 0 & -2 & 3 \\ 3 & 6 & -3 \\ 6 & 6 & 3 \end{vmatrix}\\ &=0+ 2(9 + 18) + 3(18 - 36)\\ &= 54 - 54 = 0\end{align*} $A$ is singular,so solution is not possible.

Solve the system of linear equation by using Cramer's rule.
$2 x_{1}+x_{2}+3 x_{3}=1$
$x_{1}-2 x_{2}+x_{3}=2$
$3 x_{1}-4 x_{2}-x_{3}=4$

Solution. Do yourself.

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