Question

Let $A=[a_{ij}]$ be a square matrix of order $3\times3$, where \[ a_{ij}= \begin{cases} i-2j, & i=j \\ 0, & i>j \\ 1, & i<j \end{cases} \] Then the value of $|A^T|$ is:

• A. $1$
• B. $-6$
• C. $-11$
• D. $-5$

Hint:

Whenever you see a triangular matrix, never expand the determinant manually. Just multiply the diagonal elements directly.

Answer 1

The Correct Option is B

Concept: This problem is based on properties of determinants and triangular matrices. Important facts:
• The determinant of a matrix and its transpose are equal: \[ |A^T|=|A| \]
• The determinant of a triangular matrix is simply the product of its diagonal elements.

Step 1: Construct the matrix using the given definition.
The matrix is of order $3\times3$. We calculate each element carefully. For diagonal elements $(i=j)$: \[ a_{11}=1-2(1)=-1 \] \[ a_{22}=2-2(2)=-2 \] \[ a_{33}=3-2(3)=-3 \] For elements below the diagonal $(i>j)$: \[ a_{21}=0,\qquad a_{31}=0,\qquad a_{32}=0 \] For elements above the diagonal $(i<j)$: \[ a_{12}=1,\qquad a_{13}=1,\qquad a_{23}=1 \] Thus, \[ A= \begin{bmatrix} -1 & 1 & 1 \\ 0 & -2 & 1 \\ 0 & 0 & -3 \end{bmatrix} \]

Step 2: Identify the type of matrix.
Observe that all entries below the principal diagonal are zero. Hence, $A$ is an upper triangular matrix. For any triangular matrix: \[ |A|=\text{product of diagonal entries} \] Therefore, \[ |A|=(-1)(-2)(-3) \] \[ |A|=2(-3) \] \[ |A|=-6 \]

Step 3: Use determinant property of transpose.
We know: \[ |A^T|=|A| \] Hence, \[ |A^T|=-6 \] Therefore, \[ \boxed{-6} \]