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(B) Given $A = \begin{bmatrix} \lambda & 1 \ -1 & -\lambda \end{bmatrix}$.We need to find $\lambda$ such that $A^2 = O$,where $O$ is the zero matrix.

Vedclass · Accepted Answer

$\pm 1$

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(B) Given $A = \begin{bmatrix} \lambda & 1 \ -1 & -\lambda \end{bmatrix}$.We need to find $\lambda$ such that $A^2 = O$,where $O$ is the zero matrix.$A^2 = A \cdot A = \begin{bmatrix} \lambda & 1 \ -1 & -\lambda \end{bmatrix} \begin{bmatrix} \lambda & 1 \ -1 & -\lambda \end{bmatrix}$$A^2 = \begin{bmatrix} (\lambda)(\lambda) + (1)(-1) & (\lambda)(1) + (1)(-\lambda) \ (-1)(\lambda) + (-\lambda)(-1) & (-1)(1) + (-\lambda)(-\lambda) \end{bmatrix}$$A^2 = \begin{bmatrix} \lambda^2 - 1 & \lambda - \lambda \ -\lambda + \lambda & -1 + \lambda^2 \end{bmatrix} = \begin{bmatrix} \lambda^2 - 1 & 0 \ 0 & \lambda^2 - 1 \end{bmatrix}$.Since $A^2 = O$,we have $\begin{bmatrix} \lambda^2 - 1 & 0 \ 0 & \lambda^2 - 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Comparing the elements,we get $\lambda^2 - 1 = 0$.$\lambda^2 = 1 \Rightarrow \lambda = \pm 1$.

If $A = \begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix}$,then for what value of $\lambda$ is $A^2 = O$?

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