# Question 16 & 17, Exercise 2.2

Solutions of Questions 16 & 17 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Let $A=\begin{bmatrix}3 & -1 \\4 & 2\end{bmatrix}$. Show that $|A^{-1}|=\dfrac{1}{|A|}$.

Given $$A=\left[ \begin{matrix} 3 & -1 \\ 4 & 2 \\ \end{matrix} \right]$$ $$|A|=6+4$$ $$\Rightarrow |A|=10\ldots (1)$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$AdjA=\left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{10}\left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ \end{matrix} \right]$$ $$=\left[ \begin{matrix} \dfrac{2}{10} & \dfrac{1}{10} \\ -\dfrac{4}{10} & \dfrac{3}{10} \\ \end{matrix} \right]$$ $$A^{-1}=\left[ \begin{matrix} \dfrac{1}{5} & \dfrac{1}{10} \\ -\dfrac{2}{5} & \dfrac{3}{10} \\ \end{matrix} \right]$$ $$|A^{-1}|=\dfrac{3}{50}+\dfrac{2}{50}$$ $$|A^{-1}|=\dfrac{1}{10}$$ By using (1), above expression gives,
$$|A^{-1}|=\dfrac{1}{|A|}$$

Verify that $( AB )^{-1}=B^{-1}A^{-1}$. If $A=\begin{bmatrix}2 & 3 \\1 & 0\end{bmatrix}$,$B= \begin{bmatrix} -1 & 1 \\2 & 3\end{bmatrix}$.

Given $$A=\left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]$$, $$B=\left[ \begin{matrix} -1 & 1 \\ 2 & 3 \\ \end{matrix} \right]$$ $$|A|=-3$$ $$AdjA=\left[ \begin{matrix} 0 & -3 \\ -1 & 2 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$A^{-1}=\dfrac{1}{-3}\left[ \begin{matrix} 0 & -3 \\ -1 & 2 \\ \end{matrix} \right]$$ $$A^{-1}=\left[ \begin{matrix} 0 & 1 \\ \dfrac{1}{3} & -\dfrac{2}{3} \\ \end{matrix} \right]$$ $$|B|=-5$$ $$AdjB=\left[ \begin{matrix} 3 & -1 \\ -2 & -1 \\ \end{matrix} \right]$$ $$B^{-1}=\dfrac{1}{|B|}AdjB$$ $$B^{-1}=\dfrac{1}{-5}\left[ \begin{matrix} 3 & -1 \\ -2 & -1 \\ \end{matrix} \right]$$ $$B^{-1}=\left[ \begin{matrix} -\dfrac{3}{5} & \dfrac{1}{5} \\ \dfrac{2}{5} & \dfrac{1}{5} \\ \end{matrix} \right]$$ $$B^{-1}A^{-1}=\left[ \begin{matrix} -\dfrac{3}{5} & \dfrac{1}{5} \\ \dfrac{2}{5} & \dfrac{1}{5} \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 1 \\ \dfrac{1}{3} & -\dfrac{2}{3} \\ \end{matrix} \right]$$ $$B^{-1}A^{-1}=\left[ \begin{matrix} \dfrac{1}{15} & \dfrac{-11}{15} \\ \dfrac{1}{15} & \dfrac{4}{15} \\ \end{matrix} \right]$$ $$AB=\left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 1 \\ 2 & 3 \\ \end{matrix} \right]$$ $$=\left[ \begin{matrix} -2+6 & 2+9 \\ -1 & 1 \\ \end{matrix} \right]$$ $$AB=\left[ \begin{matrix} 4 & 11 \\ -1 & 1 \\ \end{matrix} \right]$$ $$\Rightarrow |AB|=4+11$$ $$\Rightarrow \,\,|AB|=15$$ $$AdjAB=\left[ \begin{matrix} 1 & -11 \\ 1 & 4 \\ \end{matrix} \right]$$ $$( AB )^{-1}=\dfrac{1}{|AB|}AdjAB$$ $$( AB )^{-1}=\dfrac{1}{15}\left[ \begin{matrix} 1 & -11 \\ 1 & 4 \\ \end{matrix} \right]$$ $$\Rightarrow ( AB )^{-1}=\left[ \begin{matrix} \dfrac{1}{15} & -\dfrac{11}{15} \\ \dfrac{1}{15} & \dfrac{4}{15} \\ \end{matrix} \right]$$ $$\Rightarrow ( AB )^{-1}=B^{-1}A^{-1}$$