Question
If $X=\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right]$, show that $6 X-X^2=91$ Where $I$ is the unit matrix.
✓
Answer
$
\begin{aligned}
& X=\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right] \\
& x^2=x \times x=\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
16-1 & 4+2 \\
-4-2 & -1+4
\end{array}\right] \\
& =\left[\begin{array}{ll}
15 & 6 \\
-6 & 3
\end{array}\right] \\
& \text { L.H.S. } 6 \text { X- } x^2=6\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right]-\left[\begin{array}{cc}
15 & 6 \\
-6 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
24 & 6 \\
-6 & 12
\end{array}\right]-\left[\begin{array}{cc}
15 & 6 \\
-6 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
24-15 & 6-6 \\
-6-6 & 12-3
\end{array}\right] \\
& =\left[\begin{array}{ll}
9 & 0 \\
0 & 9
\end{array}\right] \\
& =9\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =91 \\
& =\text { R.H.S. }
\end{aligned}
$
Hence proved.
\begin{aligned}
& X=\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right] \\
& x^2=x \times x=\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
16-1 & 4+2 \\
-4-2 & -1+4
\end{array}\right] \\
& =\left[\begin{array}{ll}
15 & 6 \\
-6 & 3
\end{array}\right] \\
& \text { L.H.S. } 6 \text { X- } x^2=6\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right]-\left[\begin{array}{cc}
15 & 6 \\
-6 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
24 & 6 \\
-6 & 12
\end{array}\right]-\left[\begin{array}{cc}
15 & 6 \\
-6 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
24-15 & 6-6 \\
-6-6 & 12-3
\end{array}\right] \\
& =\left[\begin{array}{ll}
9 & 0 \\
0 & 9
\end{array}\right] \\
& =9\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =91 \\
& =\text { R.H.S. }
\end{aligned}
$
Hence proved.
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