If A=1-245 and f (t)=t2-3t+7, then f (A)+A=36-12-9 =

A
$(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$
B
$(\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix})$
C
$(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$
D
$(\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$

Solution:

$A^{2} = (\begin{matrix} 1 & - 2 \\ 0 & 5 \end{matrix}) (\begin{matrix} 1 & - 2 \\ 4 & 5 \end{matrix}) = (\begin{matrix} 1 - 8 & - 2 - 10 \\ 4 + 20 & - 8 + 25 \end{matrix}) = (\begin{matrix} - 7 & - 12 \\ 24 & 17 \end{matrix})$

$f (A) + (\begin{matrix} 3 & 6 \\ - 12 & - 9 \end{matrix}) = (\begin{matrix} - 7 & - 12 \\ 24 & 17 \end{matrix}) - 3 (\begin{matrix} 1 & - 2 \\ 4 & 5 \end{matrix}) + 7 (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) + (\begin{matrix} 3 & 6 \\ - 12 & - 9 \end{matrix}) = (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix})$

If $A = (\begin{matrix} 1 & - 2 \\ 4 & 5 \end{matrix}) a n d f (t) = t^{2} - 3 t + 7$ , then $f (A) + A = (\begin{matrix} 3 & 6 \\ - 12 & - 9 \end{matrix})$ =

Solution:

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