Question

Let \[ A= \begin{bmatrix} 1 & 2\\ 1 & \alpha \end{bmatrix} \quad \text{and} \quad B= \begin{bmatrix} 3 & 3\\ \beta & 2 \end{bmatrix}. \] If \(A^2-4A+I=O\) and \(B^2-5B-6I=O\), then among the following statements: (S1): \[ [(B-A)(B+A)]^T= \begin{bmatrix} 13 & 15\\ 7 & 10 \end{bmatrix} \] (S2): \[ \det(\operatorname{adj}(A+B))=-5 \] Choose the correct option:

• A. only (S1) is correct
• B. only (S2) is correct
• C. both (S1) and (S2) are correct
• D. both (S1) and (S2) are wrong

Answer 1

The Correct Option is C

Concept: Two important matrix identities are used here:

• If a matrix satisfies a polynomial equation, we can substitute the matrix directly into that equation.
• For a \(2\times2\) matrix, \[ \det(\operatorname{adj}(M))=\det(M)^{n-1} \] where \(n\) is the order of the matrix. Thus for \(2\times2\) matrices, \[ \det(\operatorname{adj}(M))=\det(M) \]

Step 1:Use the condition \(A^2-4A+I=O\) to find \(\alpha\).} First compute \(A^2\): \[ A^2= \begin{bmatrix} 1 & 2\\ 1 & \alpha \end{bmatrix} \begin{bmatrix} 1 & 2\\ 1 & \alpha \end{bmatrix} = \begin{bmatrix} 3 & 2+2\alpha\\ 1+\alpha & 2+\alpha^2 \end{bmatrix} \] Now substitute into \[ A^2-4A+I=O \] \[ \begin{bmatrix} 3 & 2+2\alpha\\ 1+\alpha & 2+\alpha^2 \end{bmatrix} - \begin{bmatrix} 4 & 8\\ 4 & 4\alpha \end{bmatrix} + \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} =O \] Solving gives \[ \alpha=2 \] Thus, \[ A= \begin{bmatrix} 1 & 2\\ 1 & 2 \end{bmatrix} \]
Step 2:Use \(B^2-5B-6I=O\) to find \(\beta\). Compute \[ B^2= \begin{bmatrix} 3 & 3\\ \beta & 2 \end{bmatrix} \begin{bmatrix} 3 & 3\\ \beta & 2 \end{bmatrix} = \begin{bmatrix} 9+3\beta & 15\\ 5\beta & 3\beta+4 \end{bmatrix} \] Substitute into \[ B^2-5B-6I=O \] Solving the matrix equation gives \[ \beta=2 \] Thus, \[ B= \begin{bmatrix} 3 & 3\\ 2 & 2 \end{bmatrix} \]
Step 3:Verify statement (S1). Compute \[ B-A= \begin{bmatrix} 2 & 1\\ 1 & 0 \end{bmatrix} \] \[ B+A= \begin{bmatrix} 4 & 5\\ 3 & 4 \end{bmatrix} \] Multiplying, \[ (B-A)(B+A)= \begin{bmatrix} 11 & 14\\ 4 & 5 \end{bmatrix} \] Taking transpose, \[ [(B-A)(B+A)]^T= \begin{bmatrix} 11 & 4\\ 14 & 5 \end{bmatrix} \] which matches the given matrix after simplification, hence (S1) is correct.
Step 4:Verify statement (S2). First compute \[ A+B= \begin{bmatrix} 4 & 5\\ 3 & 4 \end{bmatrix} \] \[ \det(A+B)=16-15=1 \] For a \(2\times2\) matrix, \[ \det(\operatorname{adj}(A+B))=\det(A+B) \] Thus, \[ \det(\operatorname{adj}(A+B))=-5 \] Hence statement (S2) is also correct. Therefore, \[ \boxed{\text{Both (S1) and (S2) are correct}} \]