Let A be a matrix such that AB is a scalar matrix where $B = [\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}] and \det (3 A) = 27 .$ Then $3 A^{- 1} + A^{2} =$

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$[\begin{array}{cc} 10 & - 6 \\ 0 & 4 \end{array}]$

$[\begin{array}{cc} 4 & - 6 \\ 0 & 2 \end{array}]$

$[\begin{array}{cc} 9 & - 4 \\ 0 & 3 \end{array}]$

$[\begin{array}{cc} 10 & - 6 \\ 0 & 2 \end{array}]$

answer is D.

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Detailed Solution

$A \cdot [\begin{array}{ll} 1 2 \\ o 3 \end{array}] = [\begin{array}{ll} k o \\ o k \end{array}]$

$A = [\begin{array}{ll} k 0 \\ o k \end{array}] {[\begin{array}{ll} 1 2 \\ 0 3 \end{array}]}^{- 1}$

$\Rightarrow A = [\begin{array}{ll} k 0 \\ 0 k \end{array}] [\begin{array}{cc} 1 & - 2 / 3 \\ O & 1 / 3 \end{array}]$

$\Rightarrow A = [\begin{array}{cc} k & - 2 k / 3 \\ 0 & k / 3 \end{array}]$

$| A | = \frac{k^{2}}{3}$

$k = \pm 3$

$\Rightarrow A = [\begin{array}{cc} 3 & - 2 \\ 0 & 1 \end{array}]$

$3 A^{- 1} + A^{2} = 3 (\frac{1}{3}) [\begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix}] + [\begin{matrix} 9 & - 8 \\ 0 & 1 \end{matrix}]$

$= [\begin{array}{cc} 10 & - 6 \\ 0 & 4 \end{array}]$

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