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Miscellaneous Exercise 4(B) · Q198

Q.If A=[3−5−42]A=\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}, show that A2−5A−14I=0A^2-5A-14I=0.

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A=[3−5−42]A=\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}.

A2=[3(3)+(−5)(−4)3(−5)+(−5)(2)−4(3)+2(−4)−4(−5)+2(2)]=[9+20−15−10−12−820+4]=[29−25−2024]A^2=\begin{bmatrix}3(3)+(-5)(-4) & 3(-5)+(-5)(2)\\-4(3)+2(-4) & -4(-5)+2(2)\end{bmatrix}=\begin{bmatrix}9+20 & -15-10\\-12-8 & 20+4\end{bmatrix}=\begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}.

5A=[15−25−2010]5A=\begin{bmatrix}15 & -25\\-20 & 10\end{bmatrix}, 14I=[140014]14I=\begin{bmatrix}14 & 0\\0 & 14\end{bmatrix}. …

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