====== Question 14 & 15, Exercise 2.2 ======
Solutions of Questions 14 & 15 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 14=====
Show that inverse of square matrix exists. Then it is unique.
====Solution====







=====Question 15=====
Let $A=\begin{bmatrix}0 & 2 & 2  \\-1 & 3 & 2  \\1 & 0 & 5\end{bmatrix}$. Find $A^{-1}$.
====Solution====
Given
$$A=\left[ \begin{matrix}
   0 & 2 & 2  \\
   -1 & 3 & 2  \\
   1 & 0 & 5  \\
\end{matrix} \right]$$
We have to find $A^{-1}$and we know that
$$A^{-1}=\dfrac{Adj\,\,A}{|A|}$$
$$Adj\,\,A={{\left[ \begin{matrix}
   A_{11} & A_{12} & A_{13}  \\
   A_{21} & A_{22} & A_{23}  \\
   A_{31} & A_{32} & A_{33}  \\
\end{matrix} \right]}^{t}}$$ 
$$A_{11}=(-1)^{1+1}\left| \begin{matrix}
   3 & 2  \\
   0 & 5  \\
\end{matrix} \right|$$ 
$$A_{11}=15$$
$$A_{12}=(-1)^{1+2}\left| \begin{matrix}
   -1 & 2  \\
   1 & 5  \\
\end{matrix} \right|$$
$$A_{12}=7$$
$$A_{13}=(-1)^{1+3}\left| \begin{matrix}
   -1 & 3  \\
   1 & 0  \\
\end{matrix} \right|$$
$$A_{13}=-3$$
$$A_{21}=(-1)^{2+1}\left| \begin{matrix}
   2 & 2  \\
   0 & 5  \\
\end{matrix} \right|$$
$$A_{21}=-10$$
$$A_{22}=(-1)^{2+2}\left| \begin{matrix}
   0 & 2  \\
   1 & 5  \\
\end{matrix} \right|$$
$$A_{22}=-2$$
$$A_{23}=(-1)^{2+3}\left| \begin{matrix}
   0 & 2  \\
   1 & 0  \\
\end{matrix} \right|$$
$$A_{23}=2$$
$$A_{31}=(-1)^{3+1}\left| \begin{matrix}
   2 & 2  \\
   3 & 2  \\
\end{matrix} \right|$$
$$A_{31}=-2$$
$$A_{32}=(-1)^{3+2}\left| \begin{matrix}
   0 & 2  \\
   -1 & 2  \\
\end{matrix} \right|$$
$$A_{32}=-2$$
$$A_{33}=(-1)^{3+3}\left| \begin{matrix}
   0 & 2  \\
   -1 & 3  \\
\end{matrix} \right|$$
$$A_{33}=2$$
$$Adj\,\,A=\left[ \begin{matrix}
   15 & -10 & -2  \\
   7 & -2 & -2  \\
   -3 & 2 & 2  \\
\end{matrix} \right]$$
$$|A|=0-2(-5-2)+2(-3)$$
$$=14-6$$
$$|A|=8$$
$$A^{-1}=\dfrac{1}{8}\left[ \begin{matrix}
   15 & -10 & -2  \\
   7 & -2 & -2  \\
   -3 & 2 & 2  \\
\end{matrix} \right]$$
$$A^{-1}=\left[ \begin{matrix}
   \dfrac{15}{8} & \dfrac{-10}{8} & -\dfrac{2}{8}  \\
   \dfrac{7}{4} & -\dfrac{2}{4} & -\dfrac{2}{4}  \\
   -\dfrac{3}{4} & \dfrac{2}{4} & \dfrac{2}{4}  \\
\end{matrix} \right]$$



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