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Question
`A=[[3,-5],[-4,2]]` then find A2 − 5A − 14I. Hence, obtain A3
Solution
\[A = \begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix}\]\[A^2 = \begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix}\begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix}\]
\[ = \begin{bmatrix}9 + 20 & - 15 - 10 \\ - 12 - 8 & 20 + 4\end{bmatrix}\]
\[ = \begin{bmatrix}29 & - 25 \\ - 20 & 24\end{bmatrix}\]\[A^2 - 5A - 14I = \begin{bmatrix}29 & - 25 \\ - 20 & 24\end{bmatrix} - 5\begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix} - 14\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ = \begin{bmatrix}29 - 15 - 14 & - 25 + 25 - 0 \\ - 20 + 20 - 0 & 24 - 10 - 14\end{bmatrix}\]
\[ = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\]
Therefore,`A^2-5A-14 l =0 ............................(1)`
`"Permultiplying the (1) by A, we get"`
`A(A^2-5A-14l)=A.0`
`⇒A^3-5A^2-14A=0`
`⇒A^3=5A^2+14A`
\[\therefore A^3 = 5\begin{bmatrix}29 & - 25 \\ - 20 & 24\end{bmatrix} + 14\begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix}\]
\[ = \begin{bmatrix}145 + 42 & - 125 - 70 \\ - 100 - 56 & 120 + 28\end{bmatrix}\]
\[ = \begin{bmatrix}187 & - 195 \\ - 156 & 148\end{bmatrix}\]
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