# Question 1, Exercise 2.2

Solutions of Question 1 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.

If $A=\begin{bmatrix}1 & 3 & 1 \\-1 & 2 & 0 \\2 & 0 & -2 \end{bmatrix}$ , then find $A_{11},A_{21},A_{23},A_{31},A_{32},A_{33}.$ Also find $|A|.$

Given $$A=\left[ \begin{matrix} 1 & 3 & 1 \\ -1 & 2 & 0 \\ 2 & 0 & -2 \\ \end{matrix} \right]$$ $${{A}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix} 2 & 0 \\ 0 & -2 \\ \end{matrix} \right|$$ $$\implies{{A}_{11}}=-4$$ $${{A}_{21}}={{\left( -1 \right)}^{2+1}}\left| \begin{matrix} 3 & 0 \\ 1 & -2 \\ \end{matrix} \right|$$ $$\implies{{A}_{21}}=6$$ $${{A}_{23}}={{\left( -1 \right)}^{2+3}}\left| \begin{matrix} 1 & 3 \\ 2 & 0 \\ \end{matrix} \right|$$ $$\implies{{A}_{23}}=6$$ $${{A}_{31}}={{\left( -1 \right)}^{3+1}}\left| \begin{matrix} 3 & 1 \\ 2 & 0 \\ \end{matrix} \right|$$ $$\implies{{A}_{31}}=-2$$ $${{A}_{32}}={{\left( -1 \right)}^{3+2}}\left| \begin{matrix} 1 & 1 \\ -1 & 0 \\ \end{matrix} \right|$$ $$\implies{{A}_{32}}=-1$$ $${{A}_{33}}={{\left( -1 \right)}^{3+3}}\left| \begin{matrix} 1 & 3 \\ -1 & 2 \\ \end{matrix} \right|$$ $$\implies{{A}_{33}}=5$$ Now \begin{align}|A|&={{a}_{11}}{{A}_{11}}+{{a}_{12}}{{A}_{12}}+{{a}_{13}}{{A}_{13}}\\ &=1\left( -4 \right)+3\left( -2 \right)+1\left( -4 \right) \\ \implies|A|&=-14\end{align}