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.
Question 18(i)
If $A$ and $B$ are non-singular matrices, then show that $( A^{-1})^{-1}=A$.
Solution
Let $A$ is $2\times 2$ non-singular matrix.
$$A=\left[ \begin{matrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{matrix} \right]$$
$$|A|=a_{11}a_{22}-a_{12}a_{21}$$
$$AdjA=\left[ \begin{matrix}
a_{22} & -a_{12} \\
-a_{21} & a_{11} \\
\end{matrix} \right]$$
$$A^{-1}=\dfrac{1}{|A|}AdjA$$
$$A^{-1}=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}\left[ \begin{matrix}
a_{22} & -a_{12} \\
-a_{21} & a_{11} \\
\end{matrix} \right]$$
$$|A^{-1}|=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}( a_{11}a_{22}-a_{12}a_{21} )$$
$$|A^{-1}|=1$$
$$AdjA^{-1}=\left[ \begin{matrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{matrix} \right]$$
$$( A^{-1} )^{-1}=\dfrac{1}{|A^{-1}|}AdjA^{-1}$$
$$( A^{-1} )^{-1}=\left[ \begin{matrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{matrix} \right]$$
$$( A^{-1})^{-1}=A$$
Question 18(ii)
$A$ and $B$ are non-singular matrices, then show that $( AB )^{-1}=B^{-1}A^{-1}$
Solution
Let $A$ and $B$ are $2\times 2$ non-singular matrices.
$$A=\left[ \begin{matrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{matrix} \right]$$
$$B=\left[ \begin{matrix}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
\end{matrix} \right]$$
$$|A|=a_{11}a_{22}-a_{12}a_{21}$$
$$AdjA=\left[ \begin{matrix}
a_{22} & -a_{12} \\
-a_{21} & a_{11} \\
\end{matrix} \right]$$
$$A^{-1}=\dfrac{1}{|A|}AdjA$$
$$A^{-1}=\dfrac{1}{a_{11}a_{22}-a_{12}a_{21}}\left[ \begin{matrix}
a_{22} & -a_{12} \\
-a_{21} & a_{11} \\
\end{matrix} \right]$$
$$|B|=b_{11}b_{22}-b_{12}b_{21}$$
$$AdjB=\left[ \begin{matrix}
b_{22} & -b_{12} \\
-b_{21} & b_{11} \\
\end{matrix} \right]$$
$$B^{-1}=\dfrac{1}{|B|}AdjB $$
$$B^{-1}=\dfrac{1}{b_{11}b_{22}-b_{12}b_{21}}\left[ \begin{matrix}
b_{22} & -b_{12} \\
-b_{21} & b_{11} \\
\end{matrix} \right]$$
$$B^{-1}A^{-1}=\dfrac{1}{( a_{11}a_{22}-a_{12}a_{21} )( b_{11}b_{22}- b_{12}b_{21} )} \left[ \begin{matrix}
b_{22} & -b_{12} \\
-b_{21} & b_{11} \\
\end{matrix} \right]\left[ \begin{matrix}
a_{22} & -a_{12} \\
-a_{21} & a_{11} \\
\end{matrix} \right]$$
$$B^{-1}A^{-1}=\dfrac{1}{( a_{11}a_{22}-a_{12}a_{21} )(b_{11}b_{22}-b_{12}b_{21})}\left[ \begin{matrix}
a_{21}b_{12}+a_{22}b_{22} & -a_{11}b_{12}-a_{12}b_{22} \\
-a_{21}b_{11}-a_{22}b_{21} & a_{11}b_{11}+a_{12}b_{21} \\
\end{matrix} \right]$$
$$AB=\left[ \begin{matrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{matrix} \right]\left[ \begin{matrix}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
\end{matrix} \right]$$
$$AB=\left[ \begin{matrix}
a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\
a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22} \\
\end{matrix} \right]$$
$$|AB|=( a_{11}b_{11}+a_{12}b_{21} )( a_{21}b_{12}+a_{22}b_{22} )-( a_{21}b_{11}+a_{22}b_{21})(a_{11}b_{12}+a_{12}b_{22} )$$
$$\begin{align}
& =a_{11}b_{11}a_{21}b_{12}+a_{11}b_{11}a_{22}b_{22}+a_{12}b_{21}a_{21}b_{12}+a_{12}b_{21}a_{22}b_{22}- \\
& ( a_{21}b_{11}a_{11}b_{12}+a_{21}b_{11}a_{12}b_{22}+a_{22}b_{21}a_{11}b_{12}+a_{22}b_{21}a_{12}b_{22}) \\
\end{align}$$
$$=a_{11}b_{11}a_{22}b_{22}+a_{12}b_{21}a_{21}b_{12}-( a_{21}b_{11}a_{12}b_{22}+a_{22}b_{21}a_{11}b_{12})$$
$$=a_{11}b_{11}a_{22}b_{22}-a_{22}b_{21}a_{11}b_{12}+a_{12}b_{21}a_{21}b_{12}-a_{21}b_{11}a_{12}b_{22}$$
$$|AB|=( a_{11}a_{22}-a_{12}a_{21})( b_{11}b_{22}-b_{12}b_{21})$$
$$Adj( AB )=\left[ \begin{matrix}
a_{21}b_{12}+a_{22}b_{22} & -a_{11}b_{12}-a_{12}b_{22}
-a_{21}b_{11}-a_{22}b_{21} & a_{11}b_{11}+a_{12}b_{21}
\end{matrix} \right]$$
$$( AB )^{-1}=\dfrac{1}{|AB|}Adj( AB )$$
$$(AB)^{-1}=\dfrac{1}{( a_{11}a_{22}-a_{12}a_{21})(b_{11}b_{22}-b_{12}b_{21})}
\left[ \begin{matrix}
a_{21}b_{12}+a_{22}b_{22} & -a_{11}b_{12}-a_{12}b_{22}
-a_{21}b_{11}-a_{22}b_{21} & a_{11}b_{11}+a_{12}b_{21}
\end{matrix} \right]$$
$$(AB)^{-1}=B^{-1}A^{-1}$$
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