Answer to Question #118711 in Calculus for Max

Question

Given that M =

2 −1
−3 4 !
and that M^2 − 6M + kI = 0, find k.

Answer 1

$M=\begin{pmatrix} 2 & -1 \\ -3 & 4 \end{pmatrix}$

$M^2=\begin{pmatrix} 2 & -1 \\ -3 & 4 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -3 & 4 \end{pmatrix}=$

$=\begin{pmatrix} 2(2)-1(-3) & 2(-1)-1(4) \\ -3(2)+4(-3) & -3(-1)+4(4) \end{pmatrix}=$

$=\begin{pmatrix} 7 & -6 \\ -18 & 19 \end{pmatrix}$

$M^2-6M=\begin{pmatrix} 7 & -6 \\ -18 & 19 \end{pmatrix}-\begin{pmatrix} 12 & -6 \\ -18 & 24 \end{pmatrix}=$

$=\begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix}=-5\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=-5I$

$M^2-6M+kI=-5I+kI=0$

$k=5$