If A= \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\) show that A2-5A+7I=0
Solution and Explanation
Given \(A= \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)
A2=A.A =\(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)\(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 3(3)+1(-1) & 3(1)+1(2) \\ -1(3)+2(-1) & -1(1)+2(2) \end{bmatrix}\)
= \(\begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix}\)= \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\)
\(\therefore\) LHS=A2-5A+7I
⇒ \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\)\(-5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)\(+7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
⇒ \(\begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}\)\(+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
= 0 =R.H.S
\(\therefore A^2-5A+7I=O\)
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