Mathematics - Matrices Question with Solution | TestHub

We have,

$A=\left[\begin{array}{ll}2& 3\\ a& 0\end{array}\right], a\in R$

Then,

$A^T=\left[\begin{array}{cc}2& a\\ 3& 0\end{array}\right]$

Now,

$P=\left(\frac{A+A^T}{2}\right)$

$\Rightarrow P=\frac{1}{2}\left\{\left[\begin{array}{ll}2& 3\\ a& 0\end{array}\right]+\left[\begin{array}{ll}2& a\\ 3& 0\end{array}\right]\right\}$

$\Rightarrow P=\left[\begin{array}{cc}2& \frac{3+a}{2}\\ \frac{a+3}{2}& 0\end{array}\right]$
And,

$Q=\frac{A-A^{\mathit{T}}}{2}$

$\Rightarrow Q=\frac{1}{2}\left\{\left[\begin{array}{ll}2& 3\\ a& 0\end{array}\right]-\left[\begin{array}{ll}2& a\\ 3& 0\end{array}\right]\right\}$

$\Rightarrow Q=\left[\begin{array}{cc}0& \frac{3-a}{2}\\ \frac{a-3}{2}& 0\end{array}\right]$

As,

$det\left(Q\right)=9$

$\Rightarrow {\left(a-3\right)}^2=36$

$\Rightarrow a=3\pm 6$

$\therefore a=9,-3$

Now,

$det\left(P\right)=-{\left(\frac{3+a}{2}\right)}^2$

For $a=-3, det\left(P\right)=0$

For $a=9$, $det\left(P\right)=-{\left(\frac{3+9}{2}\right)}^2=-36$
$\therefore$ Modulus of the sum of all possible values of $\det \left(P\right)$

$=\left|-36\right|+\left|0\right|=36$