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Matrices (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Matrices (p-1)3 Marks
Question
If $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$, find $k$ so that $A^2-8 A-k l=O$, where $I$ is a $2 \times 2$ unit matrix and $O$ is null matrix of order 2 .
✓

Answer

$
\begin{aligned}
A ^2= A \cdot A & =\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right] \\
= & {\left[\begin{array}{rr}
1-0 & 0+0 \\
-1-7 & 0+49
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-8 & 49
\end{array}\right] } \\
\therefore A ^2-8 A -k I & =\left[\begin{array}{rr}
1 & 0 \\
-8 & 49
\end{array}\right]-8\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]-k\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{rr}
1 & 0 \\
-8 & 49
\end{array}\right]-\left[\begin{array}{rr}
8 & 0 \\
-8 & 56
\end{array}\right]-\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right] \\
& =\left[\begin{array}{cc}
1-8-k & 0-0-0 \\
-8+8-0 & 49-56-k
\end{array}\right] \\
& =\left[\begin{array}{cc}
-k-7 & 0 \\
0 & -k-7
\end{array}\right]
\end{aligned}
$
But, $A ^2-8 A -k I =0$
$
\therefore\left[\begin{array}{cc}
-k-7 & 0 \\
0 & -k-7
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right)
$
By equality of matrices,
$
\begin{aligned}
& - k -7=0 \\
& \therefore k =-7 .
\end{aligned}
$

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