If A=2−3−41, then adj⁡3A2+12A is equal to

We have __A=2−3−41∴A2=A⋅A=2−3−412−3−41=4+12−6−3−8−412+1=16−9−1213 Now, 3A2+12A=316−9−1213+122−3−41=48−27−3639+24−36−4812=72−63−8451∴ adj A2+12A=51638472

If A=2−3−41, then adj⁡3A2+12A is equal to

We have __A=2−3−41∴A2=A⋅A=2−3−412−3−41=4+12−6−3−8−412+1=16−9−1213 Now, 3A2+12A=316−9−1213+122−3−41=48−27−3639+24−36−4812=72−63−8451∴ adj A2+12A=51638472

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$If A = [\begin{array}{cc} 2 & - 3 \\ - 4 & 1 \end{array}], then adj (3 A^{2} + 12 A) is equal to$

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$[\begin{matrix} 72 & - 84 \\ - 63 & 51 \end{matrix}]$

$[\begin{matrix} 51 & 63 \\ 84 & 72 \end{matrix}]$

$[\begin{matrix} 51 & 84 \\ 63 & 72 \end{matrix}]$

$[\begin{matrix} 72 & - 63 \\ - 84 & 51 \end{matrix}]$

answer is B.

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Detailed Solution

$We \underline{\underline{have}} A = [\begin{array}{cc} 2 & - 3 \\ - 4 & 1 \end{array}]$

$\begin{aligned} ∴ A^{2} = A \cdot A = [\begin{array}{cc} 2 & - 3 \\ - 4 & 1 \end{array}] [\begin{array}{cc} 2 & - 3 \\ - 4 & 1 \end{array}] \\ = [\begin{array}{cc} 4 + 12 & - 6 - 3 \\ - 8 - 4 & 12 + 1 \end{array}] \\ = [\begin{array}{cc} 16 & - 9 \\ - 12 & 13 \end{array}] \end{aligned}$

$\begin{aligned} Now, 3 A^{2} + 12 A = 3 [\begin{array}{cc} 16 & - 9 \\ - 12 & 13 \end{array}] + 12 [\begin{array}{cc} 2 & - 3 \\ - 4 & 1 \end{array}] \\ = [\begin{array}{cc} 48 & - 27 \\ - 36 & 39 \end{array}] + [\begin{array}{cc} 24 & - 36 \\ - 48 & 12 \end{array}] \\ = [\begin{array}{cc} 72 & - 63 \\ - 84 & 51 \end{array}] \end{aligned}$