For some , let A = [ 2 1 2 ] a n d B = [ 1 1 1 ] be such that A 2 - 4 A + 2 I = B 2 - 3 B + I = O . Then det ( adj ( A 3 - B 3 ) ) 2 is equal to [2026]

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Solution

Q.
For some $α, β \in ℝ$ , let $A = [\begin{matrix} α & 2 \\ 1 & 2 \end{matrix}] and B = [\begin{matrix} 1 & 1 \\ 1 & β \end{matrix}]$ be such that $A^{2} - 4 A + 2 I = B^{2} - 3 B + I = O .$ Then $(\det (adj {(A^{3} - B^{3})))}^{2} is equal to$ ________ [2026]

Ans.
(225)

$Tr (A) = 4 \Rightarrow α + 2 = 4 \Rightarrow α = 2$

$Tr (B) = 3 \Rightarrow β + 1 = 3 \Rightarrow β = 2$

$A^{2} - 4 A + 2 I = 0$

$A^{3} = 4 A^{2} - 2 A = 16 A - 8 I - 2 A = 14 A - 8 I$

$= [\begin{matrix} 28 & 28 \\ 14 & 28 \end{matrix}] + [\begin{matrix} - 8 & 0 \\ 0 & - 8 \end{matrix}]$

$= A^{3} = [\begin{matrix} 20 & 28 \\ 14 & 20 \end{matrix}]$

$B^{2} - 3 B + I = 0$

$B^{2} = 3 B - I$

$\Rightarrow B^{3} = 3 B^{2} - B = 3 (3 B - I) - B = 8 B - 3 I$

$B^{3} = [\begin{matrix} 8 & 8 \\ 8 & 16 \end{matrix}] + [\begin{matrix} - 3 & 0 \\ 0 & - 3 \end{matrix}] = [\begin{matrix} 5 & 8 \\ 8 & 13 \end{matrix}]$

$A^{3} - B^{3} = [\begin{matrix} 20 & 28 \\ 14 & 20 \end{matrix}] - [\begin{matrix} 5 & 8 \\ 8 & 13 \end{matrix}] = [\begin{matrix} 15 & 20 \\ 6 & 7 \end{matrix}]$

$\Rightarrow$ $| A^{3} - B^{3} |$ $= 105 - 120 = - 15$

$\Rightarrow$ $| adj (A^{3} - B^{3}) |$ $=$ $| A^{3} - B^{3} |$ $= - 15$

$\Rightarrow | adj (A^{3} - B^{3}) |^{2} = 225$

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