GATE CS 2023 Q.18: Linear Algebra

Let

A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \\ 3 & 4 & 1 & 2 \\ 2 & 3 & 4 & 1 \end{bmatrix}

and

B = \begin{bmatrix} 3 & 4 & 1 & 2 \\ 4 & 1 & 2 & 3 \\ 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{bmatrix}.

Let $\det(A)$ and $\det(B)$ denote the determinants of the matrices $A$ and $B$ , respectively.

Which one of the options given below is TRUE?

$\det(A) = \det(B)$

$\det(B) = -\det(A)$

$\det(A) = 0$

$\det(AB) = \det(A) + \det(B)$

Official answer and solution

Official answer: B
From: Official GATE 2023 answer key

Solution

Compare matrices $A$ and $B$ to determine the elementary row operation connecting them.

Observe the rows of matrix $A$ :

$R_1(A) = [1, 2, 3, 4]$
$R_2(A) = [4, 1, 2, 3]$
$R_3(A) = [3, 4, 1, 2]$
$R_4(A) = [2, 3, 4, 1]$

Now observe the rows of matrix $B$ :

$R_1(B) = [3, 4, 1, 2] = R_3(A)$
$R_2(B) = [4, 1, 2, 3] = R_2(A)$
$R_3(B) = [1, 2, 3, 4] = R_1(A)$
$R_4(B) = [2, 3, 4, 1] = R_4(A)$

Matrix $B$ is obtained from matrix $A$ by swapping Row 1 and Row 3 (a single row interchange), leaving Rows 2 and 4 unchanged. An elementary row swap negates the determinant of a matrix. Therefore, $\det(B) = -\det(A)$ .

Final answer

Option (B)

Common trap

Trying to explicitly compute the $4 \times 4$ determinants by expansion rather than observing the single row swap.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2023 Computer Science and Information Technology (CS)
Question: 18 of 65
Section: Computer Science and Information Technology

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