Question

If A= \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\) show that A²-5A+7I=0

Answer 1

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

\(\therefore A^2-5A+7I=O\)