Answer to Question #118586 in Linear Algebra for Nii Laryea

Question

Given that M =
2 −1
−3 4
and that M2 − 6M + kI = 0, find k
(Under matrix )

Answer 1

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

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

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

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

$$-6M=-6\begin{pmatrix} 2 & -1 \\ -3 & 4 \end{pmatrix}=$$

$$=\begin{pmatrix} -6(2) & -6(-1) \\ -6(-3) & -6(4) \end{pmatrix}=\begin{pmatrix} -12 & 6 \\ 18 & -24 \end{pmatrix}$$

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

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

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

$$k=5$$