2020-05-26T18:04:58-04:00
Given that M = 2 −1
−3 4
and that M^2 − 6M + kI = 0, find k.
1
2020-05-28T18:01:34-0400
M = ( 2 − 1 − 3 4 ) M=\begin{pmatrix}
2 & -1 \\
-3 & 4
\end{pmatrix} M = ( 2 − 3 − 1 4 )
M 2 = ( 2 − 1 − 3 4 ) ( 2 − 1 − 3 4 ) = M^2=\begin{pmatrix}
2 & -1 \\
-3 & 4
\end{pmatrix}\begin{pmatrix}
2 & -1 \\
-3 & 4
\end{pmatrix}= M 2 = ( 2 − 3 − 1 4 ) ( 2 − 3 − 1 4 ) =
= ( 2 ( 2 ) − 1 ( − 3 ) 2 ( − 1 ) − 1 ( 4 ) − 3 ( 2 ) + 4 ( − 3 ) − 3 ( − 1 ) + 4 ( 4 ) ) = =\begin{pmatrix}
2(2)-1(-3) & 2(-1)-1(4) \\
-3(2)+4(-3) & -3(-1)+4(4)
\end{pmatrix}= = ( 2 ( 2 ) − 1 ( − 3 ) − 3 ( 2 ) + 4 ( − 3 ) 2 ( − 1 ) − 1 ( 4 ) − 3 ( − 1 ) + 4 ( 4 ) ) =
= ( 7 − 6 − 18 19 ) =\begin{pmatrix}
7 & -6 \\
-18 & 19
\end{pmatrix} = ( 7 − 18 − 6 19 )
M 2 − 6 M = ( 7 − 6 − 18 19 ) − ( 12 − 6 − 18 24 ) = M^2-6M=\begin{pmatrix}
7 & -6 \\
-18 & 19
\end{pmatrix}-\begin{pmatrix}
12 & -6 \\
-18 & 24
\end{pmatrix}= M 2 − 6 M = ( 7 − 18 − 6 19 ) − ( 12 − 18 − 6 24 ) =
= ( − 5 0 0 − 5 ) = − 5 ( 1 0 0 1 ) = − 5 I =\begin{pmatrix}
-5 & 0 \\
0 & -5
\end{pmatrix}=-5\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}=-5I = ( − 5 0 0 − 5 ) = − 5 ( 1 0 0 1 ) = − 5 I k = 5 k=5 k = 5
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