Matrix Exercise asked by Narendra Kumar Sharma in the math group

Exercise:

If ${{A}^{2}}=\left( \begin{matrix}
   1 & 4  \\
   0 & 1  \\
\end{matrix} \right)$ and $B$ is the inverse of ${{A}^{2}}$ then Find ${{\left( AB \right)}^{-1}}=??$

Solution: The necessary condition for matrix to be invertible is$\det \left( {{A}^{2}} \right)\ne 0$

So$\det \left( {{A}^{2}} \right)=\det {{\left( A \right)}^{2}}=1-0=1\ne 0$ thus $\det \left( A \right)=\pm 1\ne 0$ so $B$ exits

Hence \(B={{\left( {{A}^{2}} \right)}^{-1}}={{A}^{-2}}=\left( \begin{matrix}
   1 & -4  \\
   0 & 1  \\
\end{matrix} \right)\)

${{\left( AB \right)}^{-1}}={{\left( A{{A}^{-2}} \right)}^{-1}}={{\left( A{{A}^{-1}}{{A}^{-1}} \right)}^{-1}}=I{{\left( {{A}^{-1}} \right)}^{-1}}=A$

To find $A\in {{M}_{2}}\left( \mathbb{R} \right)$ we need to assume $A=\left( \begin{matrix}
   a & b  \\
   c & d  \\
\end{matrix} \right)$ where $a,b,c,d\in \mathbb{R}$

We have $\det \left( A \right)=\pm 1\Leftrightarrow ad-bc=\pm 1$      (*)

Also ${{A}^{2}}=A.A\Leftrightarrow \left( \begin{matrix}
   a & b  \\
   c & d  \\
\end{matrix} \right).\left( \begin{matrix}
   a & b  \\
   c & d  \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 4  \\
   0 & 1  \\
\end{matrix} \right)\Leftrightarrow \left( \begin{matrix}
   {{a}^{2}}+bc & ab+bd  \\
   ac+dc & cb+{{d}^{2}}  \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 4  \\
   0 & 1  \\
\end{matrix} \right)$

So we get $ab+bd=4\Leftrightarrow b\left( a+d \right)=4\Leftrightarrow b=\frac{4}{a+d}\,\,\,,a+d\ne 0$

So $ac+dc=0\Leftrightarrow c\left( a+d \right)=0\Leftrightarrow c=0$ whenever$a+d\ne 0$

But ${{a}^{2}}+bc=1=bc+{{d}^{2}}\Leftrightarrow {{a}^{2}}={{d}^{2}}\Leftrightarrow \left| a \right|=\left| d \right|$ since $a+d\ne 0\Leftrightarrow a=d$

Also $\det \left( A \right)=ad-bc=\pm 1\Leftrightarrow {{a}^{2}}=\pm 1\Leftrightarrow a=\pm 1$

So $A=\left( \begin{matrix}
   a & b  \\
   c & d  \\
\end{matrix} \right)=\left( \begin{matrix}
   a & \frac{4}{a+d}  \\
   0 & d  \\
\end{matrix} \right)=\left( \begin{matrix}
   d & \frac{2}{d}  \\
   0 & d  \\
\end{matrix} \right)=d\left( \begin{matrix}
   1 & 2  \\
   0 & 1  \\
\end{matrix} \right)$ is true whenever \( d=1\)