Let and . If det (B) = 66, then det (adj (A)) equals : [2026]

Question

Let $A = [\begin{matrix} α & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{matrix}]$ and $B = [\begin{matrix} 1 & 0 & 0 \\ 0 & - 5 α & 0 \\ 0 & 4 α & - 2 α \end{matrix}] + adj (A)$ . If det (B) = 66, then det (adj (A)) equals : [2026]

Smart Achievers · Accepted Answer

(3)

$A = [\begin{matrix} α & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{matrix}]$

$B = [\begin{matrix} 1 & 0 & 0 \\ 0 & - 5 α & 0 \\ 0 & 4 α & - 2 α \end{matrix}] + a d j (A)$

$adj (A) = {[\begin{matrix} 15 & - 10 & 8 \\ 3 & 5 α & - 4 α \\ - 6 & 4 & 3 α - 2 \end{matrix}]}^{T} = [\begin{matrix} 15 & 3 & - 6 \\ - 10 & 5 α & 4 \\ 8 & 4 α & 3 α - 2 \end{matrix}]$

$∴ B = [\begin{matrix} 1 & 0 & 0 \\ 0 & - 5 α & 0 \\ 0 & 4 α & - 2 α \end{matrix}] + [\begin{matrix} 15 & 3 & - 6 \\ - 10 & 5 α & 4 \\ 8 & 4 α & 3 α - 2 \end{matrix}]$

$\Rightarrow B = [\begin{matrix} 16 & 3 & - 6 \\ - 10 & 0 & 4 \\ 8 & 0 & α - 2 \end{matrix}]$

Now, det (B) = 66

$\Rightarrow - 3 (- 10 α + 20 - 32) = 66 \Rightarrow α = 1$

$\det (A) = | \begin{matrix} 1 & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{matrix} | = 15 - 10 + 16 = 21$

$∴ \det (adj A) = | A |^{2} = {(21)}^{2} = 441$

Solution

Q.
Let $A = [\begin{matrix} α & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{matrix}]$ and $B = [\begin{matrix} 1 & 0 & 0 \\ 0 & - 5 α & 0 \\ 0 & 4 α & - 2 α \end{matrix}] + adj (A)$ . If det (B) = 66, then det (adj (A)) equals : [2026]

1 289

2 361

3 441

4 529

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