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\( A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1\end{array}\right] \) then \( A^{2} \) is equal toAb \( -A \)C \( 2 A \)d \( -2 A \) |
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Answer» Correct option is (c) 2A \(A = \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix}\) \(A^2 = \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix} \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix} \) \( = \begin{bmatrix}2&0&2\\0&0&0\\2&0&2\end{bmatrix}\) \(=2 \begin{bmatrix}1&0&1\\0&0&0\\1&0&1\end{bmatrix}\) \(= 2A\) |
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