# Question 7, Exercise 2.3

Solutions of Question 7 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.

Verify that $(A B)^{-1}=B^{-1} A^{-1}$ if $A=\left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right]$ and $B=\left[\begin{array}{ll}3 & 2 \\ 0 & 2\end{array}\right]$.

Solution.

Given: \begin{align*} A &= \left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right] \\ |A|& = 12 - 8 = 4\\ A^{-1} &= \dfrac{1}{4} \left[\begin{array}{ll}6 & -1 \\ -8 & 2\end{array}\right]\\ & = \left[\begin{array}{ll} \frac{6}{4} & \frac{-1}{4} \\ \frac{-8}{4} & \frac{2}{4}\end{array}\right]\\ & = \left[\begin{array}{ll} \frac{3}{2} & -\frac{1}{4} \\ -2 & \frac{1}{2}\end{array}\right] \end{align*} \begin{align*} B &= \left[\begin{array}{ll}3 & 2 \\ 0 & 2\end{array}\right]\\ |B| &= (3 \cdot 2) - (2 \cdot 0) = 6\\ B^{-1} &= \dfrac{1}{6} \left[\begin{array}{ll} 2 & -2 \\ 0 & 3 \end{array}\right]\\ & = \left[\begin{array}{ll} \frac{2}{6} & \frac{-2}{6} \\ 0 & \frac{3}{6} \end{array}\right] \\ &= \left[\begin{array}{ll} \frac{1}{3} & -\frac{1}{3} \\ 0 & \frac{1}{2} \end{array}\right] \end{align*} \begin{align*} AB &= \left[\begin{array}{ll} 2 & 1 \\ 8 & 6\end{array}\right] \left[\begin{array}{ll} 3 & 2 \\ 0 & 2 \end{array}\right]\\ & = \left[\begin{array}{ll} (2 \cdot 3 + 1 \cdot 0) & (2 \cdot 2 + 1 \cdot 2) \\ (8 \cdot 3 + 6 \cdot 0) & (8 \cdot 2 + 6 \cdot 2) \end{array}\right]\\ & = \left[\begin{array}{ll} 6 & 6 \\ 24 & 28 \end{array}\right] \\ |AB|& = (6 \cdot 28) - (6 \cdot 24) = 168 - 144 = 24 \\ (AB)^{-1} &= \frac{1}{24} \left[\begin{array}{ll} 28 & -6 \\ -24 & 6 \end{array}\right]\\ & = \left[\begin{array}{ll} \frac{28}{24} & \frac{-6}{24} \\ \frac{-24}{24} & \frac{6}{24} \end{array}\right] \\ &= \left[\begin{array}{ll} \frac{7}{6} & \frac{-1}{4} \\ -1 & \frac{1}{4} \end{array}\right] \end{align*} \begin{align*}B^{-1}A^{-1} &= \left[\begin{array}{ll} \frac{1}{3} & -\frac{1}{3} \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{ll} \frac{3}{2} & -\frac{1}{4} \\ -2 & \frac{1}{2} \end{array}\right] \\ B^{-1}A^{-1} &= \left[\begin{array}{ll} \left(\frac{1}{2} + \frac{2}{3}\right) & \left(-\frac{1}{12} - \frac{1}{6}\right) \\ -1 & \frac{1}{4} \end{array}\right] \\ &= \left[\begin{array}{ll} \frac{7}{6} & -\frac{1}{4} \\ -1 & \frac{1}{4} \end{array}\right] \end{align*} \begin{align*} (AB)^{-1} &= \left[\begin{array}{ll} \frac{7}{6} & -\frac{1}{4} \\ -1 & \frac{1}{4} \end{array}\right] \\ B^{-1}A^{-1} &= \left[\begin{array}{ll} \frac{7}{6} & -\frac{1}{4} \\ -1 & \frac{1}{4} \end{array}\right] \end{align*} Therefore, $(AB)^{-1} = B^{-1}A^{-1}$ is verified for the given matrices $A$ and $B$.

Verify that $(A B)^{-1}=B^{-1} A^{-1}$ in each of the following $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 1 \\ 2 & 1 & -3\end{array}\right]$ and $B=\left[\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right]$.

Solution.

Do yourself.

Verify that $(A B)^{-1}=B^{-1} A^{-1}$ in each of the following $A=\left[\begin{array}{ccc}2 & -i & 6 \\ 1 & 2 & i \\ -i & 1 & 6\end{array}\right]$ and $B=\left[\begin{array}{lll}3 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & 1 & 1\end{array}\right]$

Solution.

Verify that $(A B)^{-1}=B^{-1} A^{-1}$ in each of the following $A=\left[\begin{array}{ccc}1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1\end{array}\right]$ and $B=\left[\begin{array}{lll}2 & 3 & 4 \\ 1 & 0 & 2 \\ 0 & 1 & 3\end{array}\right]$

Solution.