JEE Main 2026 Apr 5 Shift 2, Mathematics Q4: System Of Linear Equations

JEE Main2026Apr 5, Shift 2Mathematics

Let $M$ be a $3 \times 3$ matrix such that

$M 100 = 123, M 010 = 012$ and $M 001 = - 1 11$ .

If $M x y z = 1711$ , then $x + y + z$ equals:

A
4
B
5
C
7
D
11

Solution

The three given products give the columns of $M$ directly:

$M = 10 - 1 211321$ .

The system $M v = (1, 7, 11)^{T}$ reads:

$x - z = 1, 2 x + y + z = 7, 3 x + 2 y + z = 11$ .

Solving: from the first $z = x - 1$ . Substituting into the second: $3 x + y = 8$ . Into the third: $4 x + 2 y = 12 \Rightarrow 2 x + y = 6$ . Subtracting: $x = 2$ , hence $y = 2$ , $z = 1$ .

$∴ x + y + z = 5$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q4: System Of Linear Equations

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