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Question
If A=then find λ, μ so that A2 = λA + μI
Solution
\[Given: A = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\]
\[Now, \]
\[ A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}7 & 12 \\ 4 & 7\end{bmatrix}\]
\[\]
` A^2 = λA + µ I`
\[ \Rightarrow \begin{bmatrix}7 & 12 \\ 4 & 7\end{bmatrix} = \lambda\begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix} + \mu\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}7 & 12 \\ 4 & 7\end{bmatrix} = \begin{bmatrix}2\lambda & 3\lambda \\ \lambda & 2\lambda\end{bmatrix} + \begin{bmatrix}\mu & 0 \\ 0 & \mu\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}7 & 12 \\ 4 & 7\end{bmatrix} = \begin{bmatrix}2\lambda + \mu & 3\lambda + 0 \\ \lambda + 0 & 2\lambda + \mu\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}7 & 12 \\ 4 & 7\end{bmatrix} = \begin{bmatrix}2\lambda + \mu & 3\lambda \\ \lambda & 2\lambda + \mu\end{bmatrix}\]
\[\]
The corresponding elements of two equal matrices are equal .
\[ \therefore 7 = 2\lambda + \mu . . . \left( 1 \right)\]
\[ 12 = 3\lambda\]
\[ \Rightarrow \lambda = \frac{12}{3} = 4\]
Putting the value of λ in eq . (1), we get
\[7 = 2\left( 4 \right) + \mu\]
\[ \Rightarrow 7 - 8 = \mu\]
\[\]
\[ \therefore \mu = - 1\]
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