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$A = [\begin{array}{lll} 3 - 4 4 \\ 1 - 2 4 \\ 1 - 1 3 \end{array}] t h e n A^{3} - 4 A^{2} + A + 8 I =$

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Correct option is D

For the matrix A=3 −4 41 −2 41 −1 3 we have A3−β1A2+β2A−β3I=0β1=3−2+3=4β2=−24−13+3413+3−41−2=(−2)+5+(−2)=1β3=3−441−241−13=6∴ we have A3−4A2+A+6I=0⇒A3−4A2+A+8I=2I

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A=3 −4 41 −2 41 −1 3 then A3−4A2+A+8I=

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$A = [\begin{array}{lll} 3 - 4 4 \\ 1 - 2 4 \\ 1 - 1 3 \end{array}] t h e n A^{3} - 4 A^{2} + A + 8 I =$