If the system of equations x+2 y-3 z=2 2 x+λ y+5 z=5 14 x+3 y+μ z=33 has infinitely many solutions, then λ+μ is equal to:

Mathematics · Matrices

JEE Main 2025 — 24 January, Evening Shift — Question 9

If the system of equations

$x+2 y-3 z=2$

$2 x+\lambda y+5 z=5$

$14 x+3 y+\mu z=33$

has infinitely many solutions, then $\lambda+\mu$ is equal to:

For infinitely many solutions the three rows of the coefficient matrix are dependent.

$\text{Try \; } ; \text{row}_3 = a,\text{row}_1 + b,\text{row}_2.$

$14 = a + 2b \quad\Rightarrow\quad a=14-2b.$

$33 = 2a + 5b \quad\Rightarrow\quad 33 = 2(14-2b)+5b = 28 + b ;\Rightarrow; b=5.$

$\Rightarrow a=14-2\cdot5=4.$

$3 = 2a + b\lambda \quad\Rightarrow\quad 3=8+5\lambda ;\Rightarrow; \lambda=-1.$

$\mu = -3a + 5b = -12+25=13.$

$\therefore \lambda+\mu = -1+13 = 12.$

$\boxed{12}$

Answer key and solution verified before publishing.

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