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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