Let A and B be diagonal matrices of order 3. Consider the following statements:
(A) \(AB = BA\)
(B) \(A + B\) is a diagonal matrix.
(C) A is a symmetric matrix.
(D) A is always invertible.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (C) only
Let \(A = \begin{bmatrix}a_1&0&0\\0&a_2&0\\0&0&a_3\end{bmatrix}\) and \(B = \begin{bmatrix}b_1&0&0\\0&b_2&0\\0&0&b_3\end{bmatrix}\). Then
\( \displaystyle AB = \begin{bmatrix}a_1b_1&0&0\\0&a_2b_2&0\\0&0&a_3b_3\end{bmatrix} = BA \)
\( \displaystyle A + B = \begin{bmatrix}a_1 + b_1&0&0\\0&a_2 + b_2&0\\0&0&a_3 + b_3\end{bmatrix} \)
So diagonal matrices of the same order commute, their sum is diagonal, every diagonal matrix equals its transpose, and a zero diagonal element is allowed.
Checking the statements
(A) \(AB = BA\) — Correct. Both products have diagonal elements \(a_1b_1, a_2b_2, a_3b_3\).
(B) \(A + B\) is a diagonal matrix. — Correct. All its non-diagonal elements are \(0 + 0 = 0\).
(C) A is a symmetric matrix. — Correct. Every non-diagonal element is 0, so \(a_{ij} = a_{ji}\) and \(A' = A\).
(D) A is always invertible. — Incorrect. \(A = \begin{bmatrix}1&0&0\\0&0&0\\0&0&2\end{bmatrix}\) is diagonal, but every product AX has second row zero, so \(AX \ne I\).