JEE Maths›Matrices & Determinants›Visual Solution
Visual SolutionPYQ 2024 · Jan (1 Feb) Shift 2Standard
Question
Let the system of equations x+2y+3z=5x + 2y + 3z = 5, 2x+3y+z=92x + 3y + z = 9, 4x+3y+λz=μ4x + 3y + \lambda z = \mu have infinite number of solutions. Then λ+2μ\lambda + 2\mu is equal to:
(A)2828
(B)1717
(C)2222
(D)1515
Solution Path
D=0D = 0 gives λ=−13\lambda = -13. D1=0D_1 = 0 gives μ=15\mu = 15. λ+2μ=17\lambda + 2\mu = 17.
01Question Setup
1/4
Find λ+2μ\lambda + 2\mu for a system of equations with determinant conditions.
λ+2μ=  ?\lambda + 2\mu = \;?
02Determinant D = 0
2/4
Expand the 3×33 \times 3 determinant. Setting D=0D = 0 gives λ=−13\lambda = -13.
D=0⇒λ=−13D = 0 \Rightarrow \lambda = -13
03Find muKEY INSIGHT
3/4
D1=0D_1 = 0 gives μ=15\mu = 15. Therefore λ+2μ=−13+30=17\lambda + 2\mu = -13 + 30 = 17.
λ+2μ=−13+30=17\lambda + 2\mu = -13 + 30 = 17
04Final Answer
4/4
λ=−13\lambda = -13 and μ=15\mu = 15 satisfy both determinant conditions.
17\boxed{17}
Concepts from this question1 concepts unlocked
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Determinant Product Rule

EASY

det(AB) = det(A) × det(B). Also det(Aᵀ) = det(A) and det(kA) = kⁿ det(A) for n×n matrix.

∣AB∣=∣A∣⋅∣B∣,∣AT∣=∣A∣,∣kA∣=kn∣A∣|AB| = |A| \cdot |B|, \quad |A^T| = |A|, \quad |kA| = k^n|A|

JEE loves chained matrix products like det(ABA^T). This rule avoids explicit multiplication entirely.

Product of matricesTranspose determinantScalar multiplication
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