Question 6, Exercise 2.3
Solutions of Question 6 of Exercise 2.3 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 6
If $A=\left[\begin{array}{ccc}2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6\end{array}\right]$ then find $A^{-1}$ and hence show that $A A^{-1}=A^{-1} A=I_{3}$.
Solution.
\begin{align*} A &= \begin{bmatrix} 2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6 \end{bmatrix} \end{align*}
To find the inverse $ A^{-1} $ of the matrix $ A $, we will use the method of row reduction (Gaussian elimination).
\begin{align*}
A \mid I & = \left[\begin{array}{ccc|ccc}
2 & 1 & -3 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
2 & 1 & 6 & 0 & 0 & 1
\end{array}\right] \\
&=\left[\begin{array}{ccc|ccc}
2 & 1 & -3 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 9 & -1 & 0 & 1
\end{array}\right] \quad R_3 - R_1 \\
&=\left[\begin{array}{ccc|ccc}
2 & 1 & -3 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -\frac{1}{9} & 0 & \frac{1}{9}
\end{array}\right] \quad \frac{1}{9} R_3\\
&=\left[\begin{array}{ccc|ccc}
2 & 1 & 0 & 1 - 3 \left(\frac{1}{9}\right) & 0 & 3 \left(\frac{1}{9}\right) \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -\frac{1}{9} & 0 & \frac{1}{9}
\end{array}\right] \quad R_1 + 3 R_3\\
&=\left[\begin{array}{ccc|ccc}
2 & 1 & 0 & 1 - \frac{1}{3} & 0 & \frac{1}{3} \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -\frac{1}{9} & 0 & \frac{1}{9}
\end{array}\right] \quad R_1 + 3 R_3\\
&=\left[\begin{array}{ccc|ccc}
2 & 1 & 0 & \frac{2}{3} & 0 & \frac{1}{3} \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -\frac{1}{9} & 0 & \frac{1}{9}
\end{array}\right]\\
&=\left[\begin{array}{ccc|ccc}
2 & 0 & 0 & \frac{2}{3} & -1 & \frac{1}{3} \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -\frac{1}{9} & 0 & \frac{1}{9}
\end{array}\right]\quad R_1 - R_2 \\
&=\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & \frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -\frac{1}{9} & 0 & \frac{1}{9}
\end{array}\right] \quad \frac{1}{2} R_1\\
\end{align*}
Now, \begin{align*} A^{-1} &= \begin{bmatrix}
\frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\
0 & 1 & 0 \\
-\frac{1}{9} & 0 & \frac{1}{9}
\end{bmatrix} \end{align*}
To show $A A^{-1} = I_3\quad \text{and }\quad A^{-1} A = I_3$
\begin{align*}
A A^{-1} &= \begin{bmatrix}
2 & 1 & -3 \\
0 & 1 & 0 \\
2 & 1 & 6
\end{bmatrix} \begin{bmatrix}
\frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\
0 & 1 & 0 \\
-\frac{1}{9} & 0 & \frac{1}{9}
\end{bmatrix} \\
& = \begin{bmatrix}
\frac{2}{3} + \frac{3}{9} & -1 + 1 & \frac{1}{3} - \frac{1}{3} \\
0 & 1 & 0 \\
\frac{2}{3} - \frac{2}{3} & -1 + 1 & \frac{1}{3} + \frac{2}{3}
\end{bmatrix} \\
&= \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} = I_3 \\
\text{Now}\quad A^{-1} A \\
A^{-1} A &= \begin{bmatrix}
\frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\
0 & 1 & 0 \\
-\frac{1}{9} & 0 & \frac{1}{9}
\end{bmatrix} \begin{bmatrix}
2 & 1 & -3 \\
0 & 1 & 0 \\
2 & 1 & 6
\end{bmatrix}\\
& = \begin{bmatrix}
\frac{2}{3} + \frac{1}{3} & \frac{1}{3} - \frac{1}{2} + \frac{1}{6} & -1 + 1 \\
0 & 1 & 0 \\
-\frac{2}{9} + \frac{2}{9} & -\frac{1}{9} + \frac{1}{9} & \frac{1}{3} + \frac{2}{3}
\end{bmatrix}\\
&= \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} = I_3 \end{align*}
Thus, we have shown that $ A A^{-1} = A^{-1} A = I_3 $.
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