GATE CS 2014 Set 2 Q.14 (CS Q.4): Linear Algebra

If the matrix $A$ is such that

A = \begin{bmatrix} 2 \\ -4 \\ 7 \end{bmatrix} \begin{bmatrix} 1 & 9 & 5 \end{bmatrix}

then the determinant of $A$ is equal to _____.

Your answer

Official answer and solution

Solution

Recognize that the outer product of two vectors produces a rank-1 matrix, whose determinant for dimension >= 2 is zero.

The matrix $A$ is given by the outer product of a $3 \times 1$ column vector $u$ and a $1 \times 3$ row vector $v^T$ :
$u = \begin{bmatrix} 2 \\ -4 \\ 7 \end{bmatrix}, \quad v^T = \begin{bmatrix} 1 & 9 & 5 \end{bmatrix}.$
The outer product $A = u v^T$ is a $3 \times 3$ matrix:
$A = \begin{bmatrix} 2(1) & 2(9) & 2(5) \\ -4(1) & -4(9) & -4(5) \\ 7(1) & 7(9) & 7(5) \end{bmatrix} = \begin{bmatrix} 2 & 18 & 10 \\ -4 & -36 & -20 \\ 7 & 63 & 35 \end{bmatrix}.$
Rank and Determinant Analysis:
- Each row of $A$ is a scalar multiple of the row vector $[1, 9, 5]$ : Row 2 is $-2 \times$ Row 1, and Row 3 is $\frac{7}{2} \times$ Row 1.
- Therefore, the rows are linearly dependent, and $\text{rank}(A) = 1$ .
- Because the rank of a $3 \times 3$ matrix of rank 1 is strictly less than 3, the matrix is singular.
Consequently, $\det(A) = 0$ .

Final answer

Common trap

Multiplying out the full 3x3 determinant expansion unnecessarily when rank properties immediately establish det(A) = 0.

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