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

Moderate

A

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

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

$[\begin{matrix} 4 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 4 \end{matrix}]$
D

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

Solution

$\begin{array}{l} (A - I)^{2} = [\begin{array}{lll} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{array}] [\begin{array}{lll} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{array}] = [\begin{array}{lll} 4 & 8 & 4 \\ 4 & 8 & 4 \\ 4 & 8 & 4 \end{array}] \\ (A - 5 I) (A - I)^{2} = [\begin{array}{ccc} - 3 & 2 & 1 \\ 1 & - 2 & 1 \\ 1 & 2 & - 3 \end{array}] [\begin{array}{lll} 4 & 8 & 4 \\ 4 & 8 & 4 \\ 4 & 8 & 4 \end{array}] \\ = [\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] \\ = 0 \\ \det A = 2 (6 - 2) - 2 (2 - 1) + 1 (2 - 3) = 8 - 2 - 1 = 5 \\ A^{- 1} = \frac{1}{5} (Adj A) = \frac{1}{5} [\begin{array}{ccc} 4 & - 2 & - 1 \\ - 1 & 3 & - 1 \\ - 1 & - 2 & 4 \end{array}] \end{array}$

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