====== Question 7 and 8, Exercise 2.6 ======
Solutions of Question 7 and 8 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 7=====
If $A=\left[\begin{array}{ccc}3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3\end{array}\right]$; find $A^{-1}$ and hence solve the system of equations.\\ $3 x+4 y+7 z=14 ; 2 x-y+3 z=4 ; \quad x+2 y-3 z=0$.

** Solution. **

Given
\begin{align*}
A &= \begin{bmatrix}
3 & 2 & 1 \\
4 & -1 & 2 \\
7 & 3 & -3
\end{bmatrix}\\
|A| &=3(-3) - 2(-26) + 19\\
&= -9 + 52 + 19\\
&= 62 \neq 0\end{align*}
This system is consistent. Now to find $A^{-1}$, we calculate the cofactors of each element .
\begin{align*}
A_{11} &= (-1)^{1+1} \left| \begin{array}{cc}
-1 & 2 \\
3 & -3
\end{array} \right| = 3 - 6 = -3 \\
A_{12} &= (-1)^{1+2} \left| \begin{array}{cc}
4 & 2 \\
7 & -3
\end{array} \right| = -(-12 - 14) = 26 \\
A_{13} &= (-1)^{1+3} \left| \begin{array}{cc}
4 & -1 \\
7 & 3
\end{array} \right| = 12 + 7 = 19 \\
A_{21} &= (-1)^{2+1} \left| \begin{array}{cc}
2 & 1 \\
3 & -3
\end{array} \right| = -(-6 - 3) = 9 \\
A_{22} &= (-1)^{2+2} \left| \begin{array}{cc}
3 & 1 \\
7 & -3
\end{array} \right| = -9 - 7 = -16 \\
A_{23} &= (-1)^{2+3} \left| \begin{array}{cc}
3 & 2 \\
7 & 3
\end{array} \right| = -(9 - 14) = 5 \\
A_{31} &= (-1)^{3+1} \left| \begin{array}{cc}
2 & 1 \\
-1 & 2
\end{array} \right| = 4 + 1 = 5 \\
A_{32} &= (-1)^{3+2} \left| \begin{array}{cc}
3 & 1 \\
4 & 2
\end{array} \right| = -(6 - 4) = -2 \\
A_{33} &= (-1)^{3+3} \left| \begin{array}{cc}
3 & 2 \\
4 & -1
\end{array} \right| = -3 - 8 = -11 \\
A &= \begin{bmatrix}
-3 & 26 & 19 \\
9 & -16 & 5 \\
5 & -2 & -11
\end{bmatrix}\\
\text{adj}(A) &= \begin{bmatrix}
-3 & 9 & 5 \\
26 & -16 & -2 \\
19 & 5 & -11
\end{bmatrix}\\
A^{-1} &= \frac{1}{62} \begin{bmatrix}
-3 & 9 & 5 \\
26 & -16 & -2 \\
19 & 5 & -11
\end{bmatrix}\\
A^{-1} &= \begin{bmatrix}
\dfrac{-3}{62} & \dfrac{9}{62} & \dfrac{5}{62} \\
\dfrac{26}{62} & \dfrac{-16}{62} & \dfrac{-2}{62} \\
\dfrac{19}{62} & \dfrac{-5}{62} & \dfrac{-11}{62}
\end{bmatrix}\\
A^{-1} &= \begin{bmatrix}
\dfrac{-3}{62} & \dfrac{9}{62} & \dfrac{5}{62} \\
\dfrac{13}{32} & \dfrac{-8}{32} & \dfrac{-1}{32} \\
\dfrac{19}{62} & \dfrac{-5}{62} & \dfrac{-11}{62}
\end{bmatrix}
\end{align*}	
Therefore, the inverse of matrix $A $ is:
$$A^{-1} =  \begin{bmatrix}
\dfrac{-3}{62} & \dfrac{9}{62} & \dfrac{5}{62} \\
\dfrac{13}{32} & \dfrac{-8}{32} & \dfrac{-1}{32} \\
\dfrac{19}{62} & \dfrac{-5}{62} & \dfrac{-11}{62}
\end{bmatrix}$$
Now given the system of equations:
\begin{align*}
3x + 4y + 7z &= 14 \\
2x - y + 3z &= 4 \\
x + 2y - 3z &= 0
\end{align*}
The associated augmented matrix for this system is:
\begin{align*}
A_b &= \begin{bmatrix}
3 & 4 & 7 & 14 \\
2 & -1 & 3 & 4 \\
1 & 2 & -3 & 0
\end{bmatrix}\\
&\sim \text{R} = \begin{bmatrix}
1 & 2 & -3 & 0\\
2 & -1 & 3 & 4 \\
3 & 4 & 7 & 14
\end{bmatrix} \quad R_1 \text{interchange} R_3\\
&\sim \text{R} = \begin{bmatrix}
1 & 2 & -3 & 0\\
0 & -5 & 9 & 4 \\
0 & -2 & 16 & 14
\end{bmatrix} \quad R_2-2R_1\quad \text{and} R_3-3R_1\\
&\sim \text{R} = \begin{bmatrix}
1 & 2 & -3 & 0\\
0 & -5 & 9 & 4 \\
0 & -1 & 8 & 7
\end{bmatrix} \quad \frac{1}{2}R_3\\
&\sim \text{R} = \begin{bmatrix}
1 & 0 & 13 & 14\\
0 & 0 & -31 & -31 \\
0 & -1 & 8 & 7
\end{bmatrix} \quad R_1+2R_3 \text{and}\quad R_2-5R_3\\
&\sim \text{R} = \begin{bmatrix}
1 & 0 & 13 & 14\\
0 & -1 & 8 & 7\\
0 & 0 & -31 & -31 
\end{bmatrix} \quad R_2\text{interchange}\quad R_3\\
&\sim \text{R} = \begin{bmatrix}
1 & 0 & 13 & 14\\
0 & -1 & 8 & 7\\
0 & 0 & 1 & 1 
\end{bmatrix} \quad \frac{-1}{31}R_2\text{interchange}\quad R_3\\
&\sim \text{R} = \begin{bmatrix}
1 & 0 & 0 & 11\\
0 & -1 & 0 & -1\\
0 & 0 & 1 & 1 
\end{bmatrix} \quad R_2-8R_3\quad R_1-13R_3
\end{align*}
From above equation we get,
\begin{align*}
x_1&=1\\
x_2&=1\\
x_3&=1
\end{align*}
Now solutions of above equations are;
$$ \begin{bmatrix}
\dfrac{-3}{62} & \dfrac{9}{62} & \dfrac{5}{62} \\
\dfrac{13}{32} & \dfrac{-8}{32} & \dfrac{-1}{32} \\
\dfrac{19}{62} & \dfrac{-5}{62} & \dfrac{-11}{62}
\end{bmatrix}\quad 1;1;1 $$

=====Question 8=====
Determine the value of $\lambda$ for which the following system has no solution, unique solution or infinitely many solutions.\\
$x+2 y-3 z=4 ; 3 x-y+5 z=2 ; 4 x+y+\left(\lambda^{2}-14\right) z=\lambda+2$

** Solution. **

====Go to ====

<text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-6-p6|< Question 6]]</btn></text>
<text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-6-p8|Question 9 & 10 >]]</btn></text>