If A=1−245 and f(t)=t2−3t+7, then f(A)+3 6−12 −9=

A2=1−2451−245=1−8−2−104+20−8+25=−7−122417f(A)+36−12−9=(A2-3A+7)+36−12−9 =−7−122417−31−245+71001+36−12−9=0000

If A=1−245 and f(t)=t2−3t+7, then f(A)+3 6−12 −9=

A2=1−2451−245=1−8−2−104+20−8+25=−7−122417f(A)+36−12−9=(A2-3A+7)+36−12−9 =−7−122417−31−245+71001+36−12−9=0000

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

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$(\begin{array}{ll} 1 0 \\ 0 1 \end{array})$

$(\begin{array}{ll} 0 0 \\ 0 0 \end{array})$

$(\begin{array}{ll} 0 1 \\ 1 0 \end{array})$

$(\begin{array}{ll} 1 1 \\ 0 0 \end{array})$

answer is B.

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Detailed Solution

$\begin{array}{l} A^{2} = (\begin{array}{rr} 1 & - 2 \\ 4 & 5 \end{array}) (\begin{array}{rr} 1 & - 2 \\ 4 & 5 \end{array}) = (\begin{array}{rr} 1 - 8 & - 2 - 10 \\ 4 + 20 & - 8 + 25 \end{array}) = (\begin{array}{rr} - 7 & - 12 \\ 24 & 17 \end{array}) \\ f (A) + (\begin{array}{rr} 3 & 6 \\ - 12 & - 9 \end{array}) = (A^{2} - 3 A + 7) + (\begin{array}{rr} 3 & 6 \\ - 12 & - 9 \end{array}) \\ = (\begin{array}{rr} - 7 & - 12 \\ 24 & 17 \end{array}) - 3 (\begin{array}{rr} 1 & - 2 \\ 4 & 5 \end{array}) + 7 (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}) + (\begin{array}{rr} 3 & 6 \\ - 12 & - 9 \end{array}) = (\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}) \end{array}$