Let and . If and , then among the two statements : (S1): and (S2): det (adj (A + B)) = –5 [2026]

Question

Let $A = [\begin{matrix} 1 & 2 \\ 1 & α \end{matrix}]$ and $B = [\begin{matrix} 3 & 3 \\ β & 2 \end{matrix}]$ . If $A^{2} - 4 A + I = O$ and $B^{2} - 5 B - 6 I = O$ , then among the two statements :

(S1): ${[(B - A) (B + A)]}^{T} = [\begin{matrix} 13 & 15 \\ 7 & 10 \end{matrix}]$ and

(S2): det (adj (A + B)) = –5 [2026]

Smart Achievers · Accepted Answer

(2)

Given: $A = [\begin{matrix} 1 & 2 \\ 1 & α \end{matrix}]$ & $B = [\begin{matrix} 3 & 3 \\ β & 2 \end{matrix}]$

Now, $A^{2} - 4 A + I = O$

$\Rightarrow [\begin{matrix} 1 & 2 \\ 1 & α \end{matrix}] [\begin{matrix} 1 & 2 \\ 1 & α \end{matrix}] - [\begin{matrix} 4 & 8 \\ 4 & 4 α \end{matrix}] + [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

$\Rightarrow [\begin{matrix} 3 & 2 + 2 α \\ 1 + α & 2 + α^{2} \end{matrix}] - [\begin{matrix} 4 & 8 \\ 4 & 4 α \end{matrix}] + [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

$\Rightarrow 2 + 2 α - 8 = 0 \Rightarrow 1 + α - 4 = 0$

$\Rightarrow α = 3$

Also, $B^{2} - 5 B - 6 I = O$

$\Rightarrow [\begin{matrix} 3 & 3 \\ β & 2 \end{matrix}] [\begin{matrix} 3 & 3 \\ β & 2 \end{matrix}] - [\begin{matrix} 15 & 15 \\ 5 β & 10 \end{matrix}] + [\begin{matrix} 6 & 0 \\ 0 & 6 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

$\Rightarrow [\begin{matrix} 9 + 3 β & 15 \\ 5 β & 3 β + 4 \end{matrix}] - [\begin{matrix} 15 & 15 \\ 5 β & 10 \end{matrix}] + [\begin{matrix} 6 & 0 \\ 0 & 6 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

$\Rightarrow 9 + 3 β - 15 - 6 = 0 \Rightarrow β = 4$

Thus, $A = [\begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix}]$ & $B = [\begin{matrix} 3 & 3 \\ 4 & 2 \end{matrix}]$

(S1): L.H.S. = ${[(B - A) (B + A)]}^{T}$

$[[\begin{matrix} 2 & 1 \\ 3 & - 1 \end{matrix}] {[\begin{matrix} 4 & 5 \\ 5 & 5 \end{matrix}]]}^{T} = [\begin{matrix} 13 & 7 \\ 15 & 10 \end{matrix}]$

R.H.S. = $[\begin{matrix} 13 & 15 \\ 7 & 10 \end{matrix}]$

$∴$ L.H.S. $\neq$ R.H.S.

Hence, (S1) is wrong.

(S2): $A + B = [\begin{matrix} 4 & 5 \\ 5 & 5 \end{matrix}]$

Adj(A + B) = $[\begin{matrix} 5 & - 5 \\ - 5 & 4 \end{matrix}]$

|adj . (A + B)| = 20 – 25 = –5 = R.H.S.

Hence, (S2) is correct.

Solution

1 Only (S1) is correct

2 Only (S2) is correct

3 Both (S1) and (S2) are correct

4 Both (S1) and (S2) are wrong

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