If \( A=[a_{ij}]_{3\times 3} \), where \( a_{ij}=\begin{cases}0, & i\ne j\\ 2i-j, & i=j\end{cases} \), then \( A \) is
(A) Diagonal matrix
(B) Symmetric matrix
(C) Scalar matrix
(D) Unit matrix
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
On the diagonal, i = j, so \( a_{ii}=2i-i=i \). Off the diagonal every entry is 0:
\( \displaystyle A=\begin{bmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 3\end{bmatrix} \)
So A has zeros off the diagonal and the diagonal entries 1, 2, 3.
Checking the statements
(A) Diagonal matrix — Correct. All entries off the diagonal are 0.
(B) Symmetric matrix — Correct. Every diagonal matrix is symmetric.
(C) Scalar matrix — Incorrect. The diagonal entries 1, 2, 3 are not all equal.
(D) Unit matrix — Incorrect. The diagonal entries 1, 2, 3 are not all 1.