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Miscellaneous Exercise 4(B) · Q199

Q.If A=[2−1−12]A=\begin{bmatrix}2 & -1\\-1 & 2\end{bmatrix}, show that A2−4A+3I=0A^2-4A+3I=0.

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A=[2−1−12]A=\begin{bmatrix}2 & -1\\-1 & 2\end{bmatrix}.

A2=[2(2)+(−1)(−1)2(−1)+(−1)(2)−1(2)+2(−1)−1(−1)+2(2)]=[4+1−2−2−2−21+4]=[5−4−45]A^2=\begin{bmatrix}2(2)+(-1)(-1) & 2(-1)+(-1)(2)\\-1(2)+2(-1) & -1(-1)+2(2)\end{bmatrix}=\begin{bmatrix}4+1 & -2-2\\-2-2 & 1+4\end{bmatrix}=\begin{bmatrix}5 & -4\\-4 & 5\end{bmatrix}. …

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