Question

Let \(A=[a_{ij}]\) for \(i,j=1,2,3,\ldots,n\) be an \(n\times n\), \((n>1)\) matrix such that \[ a_{ij}= \begin{cases} i!, & \text{if } i=j, j, & \text{if } i>j, 0, & \text{if } i<j. \end{cases} \] Which one of the following is correct?

• A. \(A\) has exactly one eigenvalue
• B. \(A\) is not a triangular matrix
• C. The eigenvalues of \(A\) are \(1!,2!,3!,\ldots,n!\)
• D. \(\det(A)=0\)

Hint:

For any upper or lower triangular matrix, \[ \boxed{\text{Eigenvalues}=\text{Diagonal entries}.} \] Also, \[ \boxed{\det(A)=\prod \text{(diagonal entries)}.} \]

Answer 1

The Correct Option is C

Step 1: Identify the type of matrix. Since \[ a_{ij}=0 \quad \text{for } i<j, \] all entries above the main diagonal are zero. Hence, \[ \boxed{A \text{ is a lower triangular matrix}.} \]

Step 2: Recall the eigenvalue property of triangular matrices. For any triangular matrix, the eigenvalues are simply its diagonal entries. The diagonal entries are \[ 1!,\,2!,\,3!,\,\ldots,\,n!. \] Therefore, \[ \boxed{\lambda=1!,\,2!,\,3!,\,\ldots,\,n!} \] Hence, \[ \boxed{(C)} \] is the correct answer.