Question

The determinant of a matrix $A = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ is}

• A. 1
• B. -1
• C. 2
• D. -2

Hint:

For any $n \times n$ anti-identity matrix, the determinant is determined by the term \( \frac{n(n-1)}{2} \):
- If \( \frac{n(n-1)}{2} \) is even, the determinant is +1.
- If \( \frac{n(n-1)}{2} \) is odd, the determinant is -1.
For \( n = 6 \), the exponent is 15 (odd), so the determinant is -1.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires calculating the determinant of a special square matrix.
The given matrix $A$ is a $6 \times 6$ permutation matrix with $1$s along its anti-diagonal (from top-right to bottom-left) and $0$s elsewhere. This is known as the anti-identity matrix, denoted as \( J_6 \).
Key Formula or Approach:
The determinant of the anti-identity matrix \( J_n \) of order $n$ is given by the formula:
\[ \det(J_n) = (-1)^{\frac{n(n-1)}{2}} \]
Alternatively, we can find the determinant by swapping rows to transform the matrix into the identity matrix \( I_6 \), tracking how many swaps are performed.

Step 2: Detailed Explanation:
Let us use the row-swap approach to convert $A$ into the identity matrix \( I_6 \):
To align the $1$s along the main diagonal (from top-left to bottom-right):
1. Swap Row 1 and Row 6.
2. Swap Row 2 and Row 5.
3. Swap Row 3 and Row 4.
These 3 row swaps transform the matrix into the standard identity matrix \( I_6 \):
\[ \det(A) = (-1)^{\text{number of swaps}} \cdot \det(I_6) \]
Since \( \det(I_6) = 1 \) and we performed 3 swaps:
\[ \det(A) = (-1)^3 \cdot 1 = -1 \cdot 1 = -1 \]
Let us verify this result using our general formula for \( n = 6 \):
\[ \det(J_6) = (-1)^{\frac{6(6-1)}{2}} = (-1)^{\frac{30}{2}} = (-1)^{15} = -1 \]
Both methods confirm the determinant is $-1$.

Step 3: Final Answer:
The determinant of the matrix is -1.