If A=2012131−10, then A2−5A+6I=

A2−5A+6I=2012131−102012131−10−1005105155−50+600060006=5−129−250−1−2−1005105155−50+600060006=−5−1−3−1−7−10−54−2+600060006=1−1−3−1−1−10−544

If A=2012131−10, then A2−5A+6I=

A2−5A+6I=2012131−102012131−10−1005105155−50+600060006=5−129−250−1−2−1005105155−50+600060006=−5−1−3−1−7−10−54−2+600060006=1−1−3−1−1−10−544

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If $A = [\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{array}], then A^{2} - 5 A + 6 I =$

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$[\begin{array}{ccc} 1 & - 1 & - 5 \\ - 1 & - 1 & 4 \\ - 3 & - 10 & 4 \end{array}]$

$[\begin{array}{ccc} 1 & - 1 & - 3 \\ - 1 & - 1 & - 10 \\ - 5 & 4 & 4 \end{array}]$

$0$

$I$

answer is B.

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Detailed Solution

$A^{2} - 5 A + 6 I = [\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{array}] [\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{array}] - [\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & - 5 & 0 \end{array}] + [\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}]$

$\begin{array}{l} = [\begin{array}{ccc} 5 & - 1 & 2 \\ 9 & - 2 & 5 \\ 0 & - 1 & - 2 \end{array}] - [\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & - 5 & 0 \end{array}] + [\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}] \\ = [\begin{array}{ccc} - 5 & - 1 & - 3 \\ - 1 & - 7 & - 10 \\ - 5 & 4 & - 2 \end{array}] + [\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}] = [\begin{array}{ccc} 1 & - 1 & - 3 \\ - 1 & - 1 & - 10 \\ - 5 & 4 & 4 \end{array}] \end{array}$