# Question 19, Exercise 2.2

Solutions of Question 19 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 $A=\begin{bmatrix}2 & 3 \\-1 & 1\end{bmatrix}$. Verify that $( A^{-1})^t=( A^t)^{-1}$ .

Given $$A=\left[ \begin{matrix} 2 & 3 \\ -1 & 1 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} 2 & -1 \\ 3 & 1 \\ \end{matrix} \right]$$ $$|A^t|=5$$ $$AdjA^t=\left[ \begin{matrix} 1 & 1 \\ -3 & 2 \\ \end{matrix} \right]$$ $$( A^t)^{-1}=\dfrac{1}{|A^t|}AdjA^t$$ $$( A^t)^{-1}=\dfrac{1}{5}\left[ \begin{matrix} 1 & 1 \\ -3 & 2 \\ \end{matrix} \right]$$ $$( A^t )^{-1}=\left[ \begin{matrix} \dfrac{1}{5} & \dfrac{1}{5} \\ -\dfrac{3}{5} & \dfrac{2}{5} \\ \end{matrix} \right]$$ $$|A|=5$$ $$AdjA=\left[ \begin{matrix} 1 & -3 \\ 1 & 2 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$A^{-1}=\dfrac{1}{5}\left[ \begin{matrix} 1 & -3 \\ 1 & 2 \\ \end{matrix} \right]$$ $$A^{-1}=\left[ \begin{matrix} \dfrac{1}{5} & -\dfrac{3}{5} \\ \dfrac{1}{5} & \dfrac{2}{5} \\ \end{matrix} \right]$$ $$( A^{-1})^t=\left[ \begin{matrix} \dfrac{1}{5} & \dfrac{1}{5} \\ -\dfrac{3}{5} & \dfrac{2}{5} \\ \end{matrix} \right]$$ $$(A^{-1})^t=( A^t)^{-1}$$