MCQ
If the matrix $A=\left(\begin{array}{cc}0 & 2 \\ K & -1\end{array}\right)$ satisfies $A\left(A^{3}+3 I\right)=2 I$ then the value of $\mathrm{K}$ is :
- ✓$\frac{1}{2}$
- B$-\frac{1}{2}$
- C$-1$
- D$1$
$A^{4}+3 I A=2 I$
$\Rightarrow A^{4}=2 I-3 A$
Also characteristic equation of $\mathrm{A}$ is
$|\mathrm{A}-\lambda \mathrm{I}|=0$
$\Rightarrow\left|\begin{array}{cc}0-\lambda & 2 \\ \mathrm{k} & -1-\lambda\end{array}\right|=0$
$\Rightarrow \lambda+\lambda^{2}-2 \mathrm{k}=0$
$\Rightarrow \mathrm{A}+\mathrm{A}^{2}=2 \mathrm{~K} \cdot \mathrm{I}$
$\Rightarrow \mathrm{A}^{2}=2 \mathrm{KI}-\mathrm{A}$
$\Rightarrow \mathrm{A}^{4}=4 \mathrm{~K}^{2} \mathrm{I}+\mathrm{A}^{2}-4 \mathrm{AK}$
$\text { Put } \mathrm{A}^{2}=2 \mathrm{KI}-\mathrm{A}$
$\text { and } \mathrm{A}^{4}=2 \mathrm{I}-3 \mathrm{~A}$
$2 \mathrm{I}-3 \mathrm{~A}=4 \mathrm{~K}^{2} \mathrm{I}+2 \mathrm{KI}-\mathrm{A}-4 \mathrm{AK}$
$\Rightarrow \mathrm{I}\left(2-2 \mathrm{~K}-4 \mathrm{~K}^{2}\right)=\mathrm{A}(2-4 \mathrm{~K})$
$\Rightarrow-2 \mathrm{I}\left(2 \mathrm{~K}^{2}+\mathrm{K}-1\right)=2 \mathrm{~A}(1-2 \mathrm{~K})$
$\Rightarrow-2 \mathrm{I}(2 \mathrm{~K}-1)(\mathrm{K}+1)=2 \mathrm{~A}(1-2 \mathrm{~K})$
$\Rightarrow(2 \mathrm{~K}-1)(2 \mathrm{~A})-2 \mathrm{I}(2 \mathrm{~K}-1)(\mathrm{K}+1)=0$
$\Rightarrow(2 \mathrm{~K}-1)[2 \mathrm{~A}-2 \mathrm{I}(\mathrm{K}+1)]=0$
$\Rightarrow \mathrm{K}=\frac{1}{2}$
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$\int_0^1 {(1 + {{\cos }^8}x)(a{x^2} + bx + c)\,dx} = \int_0^2 {(1 + {{\cos }^8}x)(a{x^2} + bx + c)\,dx} $
Then the quadratic equation $a{x^2} + bx + c = 0$ has