====== 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.

=====Question 16=====
Let $A=\begin{bmatrix}3 & -1  \\4 & 2\end{bmatrix}$. Show that $|A^{-1}|=\dfrac{1}{|A|}$.
====Solution====
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,\\
$$|A^{-1}|=\dfrac{1}{|A|}$$

=====Question 17=====
Verify that $( AB )^{-1}=B^{-1}A^{-1}$. If $A=\begin{bmatrix}2 & 3  \\1 & 0\end{bmatrix}$,$B= \begin{bmatrix}
-1 & 1  \\2 & 3\end{bmatrix}$. 
====Solution====
Given
$$A=\left[ \begin{matrix}
   2 & 3  \\
   1 & 0  \\
\end{matrix} \right]$$,
$$B=\left[ \begin{matrix}
   -1 & 1  \\
   2 & 3  \\
\end{matrix} \right]$$
$$|A|=-3$$ 
$$AdjA=\left[ \begin{matrix}
   0 & -3  \\
   -1 & 2  \\
\end{matrix} \right]$$ 
$$A^{-1}=\dfrac{1}{|A|}AdjA$$
$$A^{-1}=\dfrac{1}{-3}\left[ \begin{matrix}
   0 & -3  \\
   -1 & 2  \\
\end{matrix} \right]$$
$$A^{-1}=\left[ \begin{matrix}
   0 & 1  \\
   \dfrac{1}{3} & -\dfrac{2}{3}  \\
\end{matrix} \right]$$
$$|B|=-5$$ 
$$AdjB=\left[ \begin{matrix}
   3 & -1  \\
   -2 & -1  \\
\end{matrix} \right]$$
$$B^{-1}=\dfrac{1}{|B|}AdjB$$
$$B^{-1}=\dfrac{1}{-5}\left[ \begin{matrix}
   3 & -1  \\
   -2 & -1  \\
\end{matrix} \right]$$
$$B^{-1}=\left[ \begin{matrix}
   -\dfrac{3}{5} & \dfrac{1}{5}  \\
   \dfrac{2}{5} & \dfrac{1}{5}  \\
\end{matrix} \right]$$
$$B^{-1}A^{-1}=\left[ \begin{matrix}
   -\dfrac{3}{5} & \dfrac{1}{5}  \\
   \dfrac{2}{5} & \dfrac{1}{5}  \\
\end{matrix} \right]\left[ \begin{matrix}
   0 & 1  \\
   \dfrac{1}{3} & -\dfrac{2}{3}  \\
\end{matrix} \right]$$
$$B^{-1}A^{-1}=\left[ \begin{matrix}
   \dfrac{1}{15} & \dfrac{-11}{15}  \\
   \dfrac{1}{15} & \dfrac{4}{15}  \\
\end{matrix} \right]$$
$$AB=\left[ \begin{matrix}
   2 & 3  \\
   1 & 0  \\
\end{matrix} \right]\left[ \begin{matrix}
   -1 & 1  \\
   2 & 3  \\
\end{matrix} \right]$$ 
$$=\left[ \begin{matrix}
   -2+6 & 2+9  \\
   -1 & 1  \\
\end{matrix} \right]$$
$$AB=\left[ \begin{matrix}
   4 & 11  \\
   -1 & 1  \\
\end{matrix} \right]$$
$$\Rightarrow |AB|=4+11$$
$$\Rightarrow \,\,|AB|=15$$
$$AdjAB=\left[ \begin{matrix}
   1 & -11  \\
   1 & 4  \\
\end{matrix} \right]$$
$$( AB )^{-1}=\dfrac{1}{|AB|}AdjAB$$
$$( AB )^{-1}=\dfrac{1}{15}\left[ \begin{matrix}
   1 & -11  \\
   1 & 4  \\
\end{matrix} \right]$$
$$\Rightarrow ( AB )^{-1}=\left[ \begin{matrix}
   \dfrac{1}{15} & -\dfrac{11}{15}  \\
   \dfrac{1}{15} & \dfrac{4}{15}  \\
\end{matrix} \right]$$
$$\Rightarrow ( AB )^{-1}=B^{-1}A^{-1}$$



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