Question

Let $A$ be an $n \times n$ square matrix ($n>1$) with elements \[ a_{ij} = \begin{cases} i \times j, & \text{if } i \ge j \\ 0, & \text{if } i < j \end{cases} \] The determinant of $A$ is:

• A. 0
• B. 1
• C. $n!$
• D. $(n!)^2$

Hint:

For triangular matrices, never expand—just multiply the diagonal entries!

Answer 1

The Correct Option is D

The matrix $A$ is lower triangular because $a_{ij}=0$ whenever $i<j$. For any triangular matrix (upper or lower), the determinant is equal to the product of the diagonal elements.

Step 1: Identify the diagonal elements.
For $i=j$, \[ a_{ii} = i \times i = i^2. \]

Step 2: Compute determinant.
Since the determinant of a lower triangular matrix is the product of the diagonal entries: \[ \det(A) = \prod_{i=1}^{n} i^2 \] \[ = (1^2)(2^2)(3^2)\cdots(n^2) \] \[ = (1 \cdot 2 \cdot 3 \cdots n)^2 = (n!)^2. \]

Step 3: Conclusion.
Thus, the determinant is \((n!)^2\).