Mathematics - Matrix Question with Solution | TestHub

$PQ=kI$
$\left|P\right|\left|Q\right|=k^{\mathrm{3 \; \; }}$( as order is 3 )
$\Rightarrow$$\left|P\right|=2k \ne 0 \Rightarrow P$is an invertible matrix
$\because$$PQ=kI$
$\therefore$$Q=k{P}^{-1} I$$=\mathrm{K}\frac{1}{\left|\mathrm{P}\right|}\mathrm{adj\; P}$
$\therefore$$Q=\frac{adj.P}{2}$
$\because$$q_{23}= -\frac{k}{8}$$(\therefore$cofactor$\mathrm{}{\mathrm{P}}_{32}=\mathrm{}-\left(3\mathrm{\alpha }+4\right))$
$\Rightarrow 12\mathrm{\alpha }+16=\mathrm{K\; \&}\frac{\mathrm{adjP}}{2}={\frac{{\left[P_{32}\right]}^{}}{2}}^T-q_{23}$
$\therefore$undefined ...... (i)
$\mathrm{Also}\left|\mathrm{P}\right|=2\mathrm{k}\Rightarrow \mathrm{k}=10+6\mathrm{\alpha }$......(ii)
$\mathrm{Solving\; these\; two}\Rightarrow 12\mathrm{\alpha }+16=10+6\mathrm{\alpha }\Rightarrow \mathrm{\alpha }=-1$&$\mathrm{K}=4$
$\therefore$$4\alpha -k+8=0$$\left(\Rightarrow 4(-1)-(4)+8=0\Rightarrow -4-4+8=0\Rightarrow -8+8=0\right)$
And$\det \left(P\left(adj\mathrm{(Q}\right)\right)=\left|P\right| \left|adj\mathrm{(Q)}\right|=2k. {\left(\frac{k^2}{2}\right)}^2=\frac{k^5}{2}=2^9$
Also$\text{det}\left(\text{Q}\left(\text{adj}\left(\text{P}\right)\right)\right)=\left|\text{Q}\right|\left|\text{adj}\left(\text{P}\right)\right|$$=\frac{{\text{k}}^2}{\text{2}}{\left(\text{2k}\right)}^2={\text{2k}}^{\text{4}}={\text{2}}^9$