NDA 2 2025 Maths, question 14

Topic: Matrices and determinants

Q14

In obtaining the solution of the system of equations $x+y+z=7$ , $x+2y+3z=16$ and $x+3y+4z=22$ by Cramer’s rule, the value of $y$ is obtained by dividing $D$ by $D_{2}$ , where

$D=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{matrix} \right|$

What is the value of the determinant $D_{2}$ ?

(a) $-13$
(b) $-3$ Correct answer
(c) $3$
(d) $13$

Solution.

Answer: (b) $-3$

$D_{2}$ is $D$ with its second column replaced by the constants $(7, 16, 22)$ .
$D_{2} = 1(64 - 66) - 7(4 - 3) + 1(22 - 16) = -2 - 7 + 6 = -3$ .
Check: $D = -1$ , so $y = D_{2}/D = 3$ , and $x = 1$ , $z = 3$ satisfy all three equations.

Trap: The question says "dividing $D$ by $D_{2}$ "; Cramer's rule is $y = D_{2}/D$ , but the value asked for, $D_{2}$ , is $-3$ either way.