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Let $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 1 & \alpha\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}3 & 3 \\ \beta & 2\end{array}\right]$. If $\mathrm{A}^{2}-4 \mathrm{~A}+\mathrm{I}=0$ and $\mathrm{B}^{2}-5 \mathrm{~B}-6 \mathrm{I}=0$, then among the two statements
$(\mathrm{S} 1):[(\mathrm{B}-\mathrm{A})(\mathrm{B}+\mathrm{A})]^{\mathrm{T}}=\left[\begin{array}{cc}13 & 15 \\ 7 & 10\end{array}\right]$ and
$(\mathrm{S} 2): \operatorname{det}(\operatorname{adj}(\mathrm{A}+\mathrm{B}))=-5$,
[JEE MAIN 02 April 2026 S I]
(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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Solution

For 2 × 2 order matrix M

\[ \begin{aligned} & M^{2}-\operatorname{tr}(M) M+\operatorname{det}(M) I=0 \\ & A=\left[\begin{array}{ll} 1 & 2 \\ 1 & \alpha \end{array}\right] \Rightarrow A^{2}-4 A+I=0 ; \text { trace of matrix } \alpha+1=4 \Rightarrow \alpha=3 \\ & B=\left[\begin{array}{ll} 3 & 3 \\ \beta & 2 \end{array}\right] \Rightarrow B^{2}-5 B-6 I=0 ; \operatorname{det} .(B)=6-3 \beta=-6 \Rightarrow 3 \beta=12 \Rightarrow \beta=4 \\ & B-A=\left[\begin{array}{ll} 2 & 1 \\ 3 & -1 \end{array}\right] ; B+A=\left[\begin{array}{ll} 4 & 5 \\ 5 & 5 \end{array}\right] \Rightarrow \operatorname{adj}(A+B)=\left[\begin{array}{cc} 5 & -5 \\ -5 & 4 \end{array}\right] \\ & (B-A)(B+A)=\left[\begin{array}{cc} 13 & 15 \\ 7 & 10 \end{array}\right] \end{aligned} \]

[CHAPTER M05 PROGRESSION TOPIC T1 ]

[shahkb4]

Answer

(2)

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