If the system of linear equations : x+y+2 z=6 , 2 x+3 y+a z=a+1 , -x-3 y+b z=2 b , where a , b in R , has infinitely many solutions, then 7 a+3 b is equal to :

Mathematics · Matrices

JEE Main 2025 — 22 January, Evening Shift — Question 14

If the system of linear equations :

$x+y+2 z=6$ ,

$2 x+3 y+a z=a+1$ ,

$-x-3 y+b z=2 b$ ,

where $\mathrm{a}, \mathrm{b} \in \mathbf{R}$ , has infinitely many solutions, then $7 a+3 b$ is equal to $:$

\Delta = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{vmatrix} = 0

\Rightarrow 2a + b - 6 = 0 \qquad \qquad \qquad (1)

\Delta_1 = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 3 & a+1 \\ -1 & -3 & 2b \end{vmatrix} = 0

\Rightarrow a + b - 8 = 0 \qquad \qquad \qquad (2)

\text{Solving (1) + (2)}

a = -2, \qquad b = 10

\Rightarrow 7a + 3b = 16

Answer key and solution verified before publishing.

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