Solveeit Logo
Login

Question

Mathematics Question on Matrices

If A= [31 −12]\begin{bmatrix} 3 & 1 \\\ -1 & 2 \end{bmatrix} show that A2-5A+7I=0

Answer

Given A=[31 −12]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}

= [3(3)+1(−1)3(1)+1(2) −1(3)+2(−1)−1(1)+2(2)]\begin{bmatrix} 3(3)+1(-1) & 3(1)+1(2) \\\ -1(3)+2(-1) & -1(1)+2(2) \end{bmatrix}

= [9−13+2 −3−2−1+4]\begin{bmatrix} 9-1 & 3+2 \\\ -3-2 & -1+4 \end{bmatrix}= [85 −53]\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}

⇒ [00 00]\begin{bmatrix} 0 & 0 \\\ 0 & 0 \end{bmatrix}

= 0 =R.H.S

∴A2−5A+7I=O\therefore A^2-5A+7I=O