If the matrix A=begin{bmatrix} 0 & 2 K & -1 | vedclass

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(A) Given matrix $A = \begin{bmatrix} 0 & 2 \ K & -1 \end{bmatrix}$.The given equation is $A(A^{3} + 3I) = 2I$,which implies $A^{4} + 3A = 2I$,or $A^

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$\frac{1}{2}$

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(A) Given matrix $A = \begin{bmatrix} 0 & 2 \ K & -1 \end{bmatrix}$.The given equation is $A(A^{3} + 3I) = 2I$,which implies $A^{4} + 3A = 2I$,or $A^{4} = 2I - 3A$.The characteristic equation of $A$ is given by $|A - \lambda I| = 0$.$\begin{vmatrix} -\lambda & 2 \ K & -1 - \lambda \end{vmatrix} = 0 \Rightarrow \lambda(\lambda + 1) - 2K = 0 \Rightarrow \lambda^{2} + \lambda - 2K = 0$.By the Cayley-Hamilton theorem,$A^{2} + A - 2KI = 0$,so $A^{2} = 2KI - A$.Now,$A^{4} = (A^{2})^{2} = (2KI - A)^{2} = 4K^{2}I - 4KA + A^{2}$.Substituting $A^{2} = 2KI - A$ into the expression for $A^{4}$:$A^{4} = 4K^{2}I - 4KA + (2KI - A) = (4K^{2} + 2K)I - (4K + 1)A$.Equating the two expressions for $A^{4}$:$2I - 3A = (4K^{2} + 2K)I - (4K + 1)A$.Rearranging terms:$(4K + 1 - 3)A = (4K^{2} + 2K - 2)I$.$(4K - 2)A = (4K^{2} + 2K - 2)I$.$2(2K - 1)A = 2(2K^{2} + K - 1)I$.$2(2K - 1)A = 2(2K - 1)(K + 1)I$.If $2K - 1 
eq 0$,then $A = (K + 1)I$,which implies $\begin{bmatrix} 0 & 2 \ K & -1 \end{bmatrix} = \begin{bmatrix} K+1 & 0 \ 0 & K+1 \end{bmatrix}$.This leads to $K+1 = 0$ and $2 = 0$,which is impossible.Thus,we must have $2K - 1 = 0$,which gives $K = \frac{1}{2}$.

If the matrix $A=\begin{bmatrix} 0 & 2 \\ K & -1 \end{bmatrix}$ satisfies $A(A^{3}+3I)=2I$,then the value of $K$ is:

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