====== 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.
=====Question 5(i)=====
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 $(3, 1, 2)$.
=====Question 5(ii)=====
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:
$$( -\frac{1}{7}, \frac{1}{7}, 0 )$$
=====Question 5(iii)=====
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.
=====Question 5(iv)=====
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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[[math-11-nbf:sol:unit02:ex2-6-p4|< Question 4]]
[[math-11-nbf:sol:unit02:ex2-6-p6|Question 6 >]]