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If A=[0 3 -2 5] then find 'k' so that kA2=5A-6I2 |
Answer» A \(=\begin{bmatrix}0 &3\\-2&5\end{bmatrix}\) A2 \(=\begin{bmatrix}0 &3\\-2&5\end{bmatrix}\)\(\begin{bmatrix}0 &3\\-2&5\end{bmatrix}\) \(=\begin{bmatrix}-6 & 15\\-10 & 19\end{bmatrix}\) Now, 5A – 6I2 \(=5\begin{bmatrix}0 & 3\\-2 & 5\end{bmatrix}-6\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\) \(=\begin{bmatrix}0 & 15\\-10 & 25\end{bmatrix}\)\(-\begin{bmatrix}6& 0\\0 & 6\end{bmatrix}\) \(=\begin{bmatrix}-6 & 15\\-10 & 19\end{bmatrix}\) = A2 Thus k = 1. |
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