If A=2 2 11 3 11 2 2 then A−1+(A−5I)(A−I)2=

|A−λI|=0, i.e (λ−5)(λ−1)2=0⇒(A−5I)(A−I)2=0,A−1=adj⁡Adet⁡A A=2 2 11 3 11 2 2=8-2-1=5 AdjA=4−2−1−13−1−1−24

If A=2 2 11 3 11 2 2 then A−1+(A−5I)(A−I)2=

|A−λI|=0, i.e (λ−5)(λ−1)2=0⇒(A−5I)(A−I)2=0,A−1=adj⁡Adet⁡A A=2 2 11 3 11 2 2=8-2-1=5 AdjA=4−2−1−13−1−1−24

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If $A = [\begin{array}{lll} 2 2 1 \\ 1 3 1 \\ 1 2 2 \end{array}] then A^{- 1} + (A - 5 I) (A - I)^{2} =$

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$\frac{1}{5} [\begin{array}{ccc} 4 & 2 & - 1 \\ - 1 & 3 & 1 \\ - 1 & 2 & 4 \end{array}]$

$\frac{1}{3} [\begin{array}{ccc} 4 & 2 & - 1 \\ - 1 & 3 & 1 \\ - 1 & 2 & 4 \end{array}]$

$\frac{1}{3} [\begin{array}{ccc} 4 & - 2 & - 1 \\ - 1 & 3 & - 1 \\ - 1 & - 2 & 4 \end{array}]$

$\frac{1}{5} [\begin{array}{ccc} 4 & - 2 & - 1 \\ - 1 & 3 & - 1 \\ - 1 & - 2 & 4 \end{array}]$

answer is B.

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Detailed Solution

$\begin{aligned} | A - λ I | = 0, i.e (λ - 5) (λ - 1)^{2} = 0 \\ \Rightarrow (A - 5 I) (A - I)^{2} = 0, A^{- 1} = \frac{adj A}{\det A} \end{aligned} |A| = |\begin{array}{l} 2 2 1 \\ 1 3 1 \\ 1 2 2 \end{array}| = 8 - 2 - 1 = 5 A d j A = [\begin{array}{ccc} 4 & - 2 & - 1 \\ - 1 & 3 & - 1 \\ - 1 & - 2 & 4 \end{array}]$