# Question 18, Exercise 2.2

Solutions of Question 18 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 $A$ and $B$ are non-singular matrices, then show that $( A^{-1})^{-1}=A$.

Let $A$ is $2\times 2$ non-singular matrix.
$$A=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$|A|=a_{11}a_{22}-a_{12}a_{21}$$ $$AdjA=\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$A^{-1}=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]$$ $$|A^{-1}|=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}( a_{11}a_{22}-a_{12}a_{21} )$$ $$|A^{-1}|=1$$ $$AdjA^{-1}=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$( A^{-1} )^{-1}=\dfrac{1}{|A^{-1}|}AdjA^{-1}$$ $$( A^{-1} )^{-1}=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$( A^{-1})^{-1}=A$$

$A$ and $B$ are non-singular matrices, then show that $( AB )^{-1}=B^{-1}A^{-1}$

Let $A$ and $B$ are $2\times 2$ non-singular matrices. $$A=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$B=\left[ \begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{matrix} \right]$$ $$|A|=a_{11}a_{22}-a_{12}a_{21}$$ $$AdjA=\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$A^{-1}=\dfrac{1}{a_{11}a_{22}-a_{12}a_{21}}\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]$$ $$|B|=b_{11}b_{22}-b_{12}b_{21}$$ $$AdjB=\left[ \begin{matrix} b_{22} & -b_{12} \\ -b_{21} & b_{11} \\ \end{matrix} \right]$$ $$B^{-1}=\dfrac{1}{|B|}AdjB$$ $$B^{-1}=\dfrac{1}{b_{11}b_{22}-b_{12}b_{21}}\left[ \begin{matrix} b_{22} & -b_{12} \\ -b_{21} & b_{11} \\ \end{matrix} \right]$$ $$B^{-1}A^{-1}=\dfrac{1}{( a_{11}a_{22}-a_{12}a_{21} )( b_{11}b_{22}- b_{12}b_{21} )} \left[ \begin{matrix} b_{22} & -b_{12} \\ -b_{21} & b_{11} \\ \end{matrix} \right]\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]$$ $$B^{-1}A^{-1}=\dfrac{1}{( a_{11}a_{22}-a_{12}a_{21} )(b_{11}b_{22}-b_{12}b_{21})}\left[ \begin{matrix} a_{21}b_{12}+a_{22}b_{22} & -a_{11}b_{12}-a_{12}b_{22} \\ -a_{21}b_{11}-a_{22}b_{21} & a_{11}b_{11}+a_{12}b_{21} \\ \end{matrix} \right]$$ $$AB=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]\left[ \begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{matrix} \right]$$ $$AB=\left[ \begin{matrix} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22} \\ \end{matrix} \right]$$ $$|AB|=( a_{11}b_{11}+a_{12}b_{21} )( a_{21}b_{12}+a_{22}b_{22} )-( a_{21}b_{11}+a_{22}b_{21})(a_{11}b_{12}+a_{12}b_{22} )$$ \begin{align} & =a_{11}b_{11}a_{21}b_{12}+a_{11}b_{11}a_{22}b_{22}+a_{12}b_{21}a_{21}b_{12}+a_{12}b_{21}a_{22}b_{22}- \\ & ( a_{21}b_{11}a_{11}b_{12}+a_{21}b_{11}a_{12}b_{22}+a_{22}b_{21}a_{11}b_{12}+a_{22}b_{21}a_{12}b_{22}) \\ \end{align} $$=a_{11}b_{11}a_{22}b_{22}+a_{12}b_{21}a_{21}b_{12}-( a_{21}b_{11}a_{12}b_{22}+a_{22}b_{21}a_{11}b_{12})$$ $$=a_{11}b_{11}a_{22}b_{22}-a_{22}b_{21}a_{11}b_{12}+a_{12}b_{21}a_{21}b_{12}-a_{21}b_{11}a_{12}b_{22}$$ $$|AB|=( a_{11}a_{22}-a_{12}a_{21})( b_{11}b_{22}-b_{12}b_{21})$$ $$Adj( AB )=\left[ \begin{matrix} a_{21}b_{12}+a_{22}b_{22} & -a_{11}b_{12}-a_{12}b_{22} -a_{21}b_{11}-a_{22}b_{21} & a_{11}b_{11}+a_{12}b_{21} \end{matrix} \right]$$ $$( AB )^{-1}=\dfrac{1}{|AB|}Adj( AB )$$ $$(AB)^{-1}=\dfrac{1}{( a_{11}a_{22}-a_{12}a_{21})(b_{11}b_{22}-b_{12}b_{21})} \left[ \begin{matrix} a_{21}b_{12}+a_{22}b_{22} & -a_{11}b_{12}-a_{12}b_{22} -a_{21}b_{11}-a_{22}b_{21} & a_{11}b_{11}+a_{12}b_{21} \end{matrix} \right]$$ $$(AB)^{-1}=B^{-1}A^{-1}$$