If A=1−245 and f(t)=t2−3t+7 then f(A)+36−12−9 is equal to

A2=−7−122417 Now, fA=A2−3A+7=−7−122417−31−245+71001=−3−6129∴fA+36−12−9=−3−6129+36−12−9=0000

If A=1−245 and f(t)=t2−3t+7 then f(A)+36−12−9 is equal to

A2=−7−122417 Now, fA=A2−3A+7=−7−122417−31−245+71001=−3−6129∴fA+36−12−9=−3−6129+36−12−9=0000

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If $A = [\begin{array}{cc} 1 & - 2 \\ 4 & 5 \end{array}] and f (t) = t^{2} - 3 t + 7$ then $f (A) + [\begin{array}{cc} 3 & 6 \\ - 12 & - 9 \end{array}]$ is equal to

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$[\begin{array}{ll} 0 0 \\ 0 0 \end{array}]$

$[\begin{array}{ll} 1 0 \\ 0 1 \end{array}]$

$[\begin{array}{ll} 0 1 \\ 1 0 \end{array}]$

$[\begin{array}{ll} 1 1 \\ 0 0 \end{array}]$

answer is B.

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Detailed Solution

$\begin{array}{l} A^{2} = [\begin{array}{cc} - 7 & - 12 \\ 24 & 17 \end{array}] Now, f (A) = A^{2} - 3 A + 7 \\ = [\begin{array}{cc} - 7 & - 12 \\ 24 & 17 \end{array}] - 3 [\begin{array}{cc} 1 & - 2 \\ 4 & 5 \end{array}] + 7 [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{cc} - 3 & - 6 \\ 12 & 9 \end{array}] \\ ∴ f (A) + [\begin{array}{cc} 3 & 6 \\ - 12 & - 9 \end{array}] = [\begin{array}{cc} - 3 & - 6 \\ 12 & 9 \end{array}] + [\begin{array}{cc} 3 & 6 \\ - 12 & - 9 \end{array}] = [\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}] \end{array}$