Mathematics - Matrices Question with Solution | TestHub

Given, $A=\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}4& 3\\ 0& 1\end{array}\right]$

$\Rightarrow A^4=\left[\begin{array}{cc}4& 3\\ 0& 1\end{array}\right]\left[\begin{array}{cc}4& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}16& 15\\ 0& 1\end{array}\right]$

$\Rightarrow A^5=\left[\begin{array}{cc}16& 15\\ 0& 1\end{array}\right]\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}32& 33\\ 0& -1\end{array}\right]$

$\Rightarrow A^{10}=\left[\begin{array}{cc}32& 33\\ 0& -1\end{array}\right]\left[\begin{array}{cc}32& 33\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right] ....\left(i\right)$

Now, $2A=\left[\begin{array}{cc}4& 6\\ 0& -2\end{array}\right]$

$\Rightarrow \mathrm{Adj}\left(2A\right)={\left[\begin{array}{cc}-2& 0\\ -6& 4\end{array}\right]}^T$

$\Rightarrow \mathrm{Adj}\left(2A\right)=\left[\begin{array}{cc}-2& -6\\ 0& 4\end{array}\right]$

$\Rightarrow {\left(\mathrm{Adj}\left(2A\right)\right)}^2=\left[\begin{array}{cc}-2& -6\\ 0& 4\end{array}\right]\left[\begin{array}{cc}-2& -6\\ 0& 4\end{array}\right]=\left[\begin{array}{cc}4& -12\\ 0& 16\end{array}\right]$

$\Rightarrow {\left(\mathrm{Adj}\left(2A\right)\right)}^4=\left[\begin{array}{cc}4& -12\\ 0& 16\end{array}\right]\left[\begin{array}{cc}4& -12\\ 0& 16\end{array}\right]=\left[\begin{array}{cc}16& -240\\ 0& 256\end{array}\right]$

$\Rightarrow {\left(\mathrm{Adj}\left(2A\right)\right)}^5=\left[\begin{array}{cc}16& -240\\ 0& 256\end{array}\right]\left[\begin{array}{cc}-2& -6\\ 0& 4\end{array}\right]=\left[\begin{array}{cc}-32& -1056\\ 0& 1024\end{array}\right]$

$\Rightarrow {\left(\mathrm{Adj}\left(2A\right)\right)}^{10}=\left[\begin{array}{cc}-32& -1056\\ 0& 1024\end{array}\right]\left[\begin{array}{cc}-32& -1056\\ 0& 1024\end{array}\right]=\left[\begin{array}{cc}1024& -1047552\\ 0& 1048576\end{array}\right] .....\left(ii\right)$

From equation $\left(i\right) \& \left(ii\right)$

$\Rightarrow A^{10}-{\left(\mathrm{Adj}\left(2A\right)\right)}^{10}=\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1024& -1047552\\ 0& 1048576\end{array}\right]=\left[\begin{array}{cc}0& 1048575\\ 0& -1048575\end{array}\right]$

$\Rightarrow \det \left(A^{10}-{\left(\mathrm{Adj}\left(2A\right)\right)}^{10}\right)=\left|\begin{array}{cc}0& 1048575\\ 0& -1048575\end{array}\right|=0$

Let $E=det\left(A^4\right)+det\left(A^{10}-(Adj(2 A){)}^{10}\right)$

$\Rightarrow E={\left(det\left(A\right)\right)}^4+0$

$\Rightarrow E={\left(-2-0\right)}^4$

$\Rightarrow E=16$