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If $A=\left[\begin{array}{cc}1 & -2 \\ 4 & 5\end{array}\right]$ and $f(t)=t^2-3 t+7$, then…
If $A=\left[\begin{array}{cc}1 & -2 \\ 4 & 5\end{array}\right]$ and $f(t)=t^2-3 t+7$, then $f(A)+\left[\begin{array}{cc}3 & 6 \\ -12 & -9\end{array}\right]$ is equal to
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$ $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ $\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$ Solution
Given that,
$
A=\left[\begin{array}{cc}
1 & -2 \\
4 & 5
\end{array}\right] \text { and } f(t)=t^2-3 t+7
$
Now,
$
\begin{aligned}
A^2 & =\left[\begin{array}{cc}
1 & -2 \\
4 & 5
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
4 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
-7 & -12 \\
24 & 17
\end{array}\right]
\end{aligned}
$
Now,
$
\begin{aligned}
& f(A)=A^2-3 A+7 \\
& =\left[\begin{array}{cc}
-7 & -12 \\
24 & 17
\end{array}\right]-3\left[\begin{array}{cc}
1 & -2 \\
4 & 5
\end{array}\right]+7\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
-3 & -6 \\
12 & 9
\end{array}\right]
\end{aligned}
$
$\begin{aligned} \therefore f(A)+\left[\begin{array}{cc}3 & 6 \\ -12 & -9\end{array}\right] & =\left[\begin{array}{cc}-3 & -6 \\ 12 & 9\end{array}\right]+\left[\begin{array}{cc}3 & 6 \\ -12 & -9\end{array}\right] \\ & =\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\end{aligned}$
Asked in: AP EAMCET 2008
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