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

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

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

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

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$3 I$

$2 I$

$I$

answer is D.

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Detailed Solution

$For the matrix A = [\begin{array}{rrr} 3 - 4 4 \\ 1 - 2 4 \\ 1 - 1 3 \end{array}] we have$

$\begin{array}{l} A^{3} - β_{1} A^{2} + β_{2} A - β_{3} I = 0 \\ β_{1} = 3 - 2 + 3 = 4 \\ β_{2} = |\begin{array}{ll} - 2 & 4 \\ - 1 & 3 \end{array}| + |\begin{array}{ll} 3 & 4 \\ 1 & 3 \end{array}| + |\begin{array}{ll} 3 & - 4 \\ 1 & - 2 \end{array}| = (- 2) + 5 + (- 2) = 1 \\ β_{3} = |\begin{array}{lll} 3 & - 4 & 4 \\ 1 & - 2 & 4 \\ 1 & - 1 & 3 \end{array}| = 6 \end{array}$