STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

$A^{4}+3 I A=2 I$

$\Rightarrow A^{4}=2 I-3 A$

Also characteristic equation of $\mathrm{A}$ is

$|\mathrm{A}-\lambda \mathrm{I}|=0$

$\Rightarrow\left|\begin{array}{cc}0-\lambda & 2 \\ \mathrm{k} & -1-\lambda\end{array}\right|=0$

$\Rightarrow \lambda+\lambda^{2}-2 \mathrm{k}=0$

$\Rightarrow \mathrm{A}+\mathrm{A}^{2}=2 \mathrm{~K} \cdot \mathrm{I}$

$\Rightarrow \mathrm{A}^{2}=2 \mathrm{KI}-\mathrm{A}$

$\Rightarrow \mathrm{A}^{4}=4 \mathrm{~K}^{2} \mathrm{I}+\mathrm{A}^{2}-4 \mathrm{AK}$

$\text { Put } \mathrm{A}^{2}=2 \mathrm{KI}-\mathrm{A}$

$\text { and } \mathrm{A}^{4}=2 \mathrm{I}-3 \mathrm{~A}$

$2 \mathrm{I}-3 \mathrm{~A}=4 \mathrm{~K}^{2} \mathrm{I}+2 \mathrm{KI}-\mathrm{A}-4 \mathrm{AK}$

$\Rightarrow \mathrm{I}\left(2-2 \mathrm{~K}-4 \mathrm{~K}^{2}\right)=\mathrm{A}(2-4 \mathrm{~K})$

$\Rightarrow-2 \mathrm{I}\left(2 \mathrm{~K}^{2}+\mathrm{K}-1\right)=2 \mathrm{~A}(1-2 \mathrm{~K})$

$\Rightarrow-2 \mathrm{I}(2 \mathrm{~K}-1)(\mathrm{K}+1)=2 \mathrm{~A}(1-2 \mathrm{~K})$

$\Rightarrow(2 \mathrm{~K}-1)(2 \mathrm{~A})-2 \mathrm{I}(2 \mathrm{~K}-1)(\mathrm{K}+1)=0$

$\Rightarrow(2 \mathrm{~K}-1)[2 \mathrm{~A}-2 \mathrm{I}(\mathrm{K}+1)]=0$

$\Rightarrow \mathrm{K}=\frac{1}{2}$