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.

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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