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.
Show that inverse of square matrix exists. Then it is unique.
Let $A=\begin{bmatrix}0 & 2 & 2 \\-1 & 3 & 2 \\1 & 0 & 5\end{bmatrix}$. Find $A^{-1}$.
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 $$A^{-1}=\dfrac{Adj\,\,A}{|A|}$$ $$Adj\,\,A={{\left[ \begin{matrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \\ \end{matrix} \right]}^{t}}$$ $$A_{11}=(-1)^{1+1}\left| \begin{matrix} 3 & 2 \\ 0 & 5 \\ \end{matrix} \right|$$ $$A_{11}=15$$ $$A_{12}=(-1)^{1+2}\left| \begin{matrix} -1 & 2 \\ 1 & 5 \\ \end{matrix} \right|$$ $$A_{12}=7$$ $$A_{13}=(-1)^{1+3}\left| \begin{matrix} -1 & 3 \\ 1 & 0 \\ \end{matrix} \right|$$ $$A_{13}=-3$$ $$A_{21}=(-1)^{2+1}\left| \begin{matrix} 2 & 2 \\ 0 & 5 \\ \end{matrix} \right|$$ $$A_{21}=-10$$ $$A_{22}=(-1)^{2+2}\left| \begin{matrix} 0 & 2 \\ 1 & 5 \\ \end{matrix} \right|$$ $$A_{22}=-2$$ $$A_{23}=(-1)^{2+3}\left| \begin{matrix} 0 & 2 \\ 1 & 0 \\ \end{matrix} \right|$$ $$A_{23}=2$$ $$A_{31}=(-1)^{3+1}\left| \begin{matrix} 2 & 2 \\ 3 & 2 \\ \end{matrix} \right|$$ $$A_{31}=-2$$ $$A_{32}=(-1)^{3+2}\left| \begin{matrix} 0 & 2 \\ -1 & 2 \\ \end{matrix} \right|$$ $$A_{32}=-2$$ $$A_{33}=(-1)^{3+3}\left| \begin{matrix} 0 & 2 \\ -1 & 3 \\ \end{matrix} \right|$$ $$A_{33}=2$$ $$Adj\,\,A=\left[ \begin{matrix} 15 & -10 & -2 \\ 7 & -2 & -2 \\ -3 & 2 & 2 \\ \end{matrix} \right]$$ $$|A|=0-2(-5-2)+2(-3)$$ $$=14-6$$ $$|A|=8$$ $$A^{-1}=\dfrac{1}{8}\left[ \begin{matrix} 15 & -10 & -2 \\ 7 & -2 & -2 \\ -3 & 2 & 2 \\ \end{matrix} \right]$$ $$A^{-1}=\left[ \begin{matrix} \dfrac{15}{8} & \dfrac{-10}{8} & -\dfrac{2}{8} \\ \dfrac{7}{4} & -\dfrac{2}{4} & -\dfrac{2}{4} \\ -\dfrac{3}{4} & \dfrac{2}{4} & \dfrac{2}{4} \\ \end{matrix} \right]$$