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Question
If A=, find k such that A2 = kA − 2I2
Solution
\[Given: A = \begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix}\]
\[Now, \]
\[ A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix}\begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}9 - 8 & - 6 + 4 \\ 12 - 8 & - 8 + 4\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}1 & - 2 \\ 4 & - 4\end{bmatrix}\]
\[\]
\[ A^2 = kA - 2 I_2 \]
\[ \Rightarrow \begin{bmatrix}1 & - 2 \\ 4 & - 4\end{bmatrix} = k\begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix} - 2\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}1 & - 2 \\ 4 & - 4\end{bmatrix} = \begin{bmatrix}3k & - 2k \\ 4k & - 2k\end{bmatrix} - \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}1 & - 2 \\ 4 & - 4\end{bmatrix} = \begin{bmatrix}3k - 2 & - 2k - 0 \\ 4k - 0 & - 2k - 2\end{bmatrix}\]
\[\]
The corresponding elements of two equal matrices are equal .
\[ \therefore 1 = 3k - 2 \]
\[ \Rightarrow 1 + 2 = 3k \]
\[ \Rightarrow 3 = 3k \]
\[ \Rightarrow k = 1\]
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