JEE Maths›Matrices & Determinants›Visual Solution
Visual SolutionPYQ 2024 · Jan 31 Shift 1Tricky
Question
If the system of linear equations\nx−2y+z=−4x - 2y + z = -4\n2x+αy+3z=52x + \alpha y + 3z = 5\n3x−y+βz=33x - y + \beta z = 3\nhas infinitely many solutions, then 12α+13β12\alpha + 13\beta is equal to:
(A)6060
(B)6464
(C)5454
(D)5858
Solution Path
From D=D1=D2=0D = D_1 = D_2 = 0: α=13\alpha = \frac{1}{3}, 13β=5413\beta = 54. Answer: 12⋅13+54=5812 \cdot \frac{1}{3} + 54 = 58.
01Question Setup
1/4
System with infinite solutions. Find 12α+13β12\alpha + 13\beta.
12α+13β=  ?12\alpha + 13\beta = \;?
02Infinite Solutions
2/4
For infinite solutions: D=D1=D2=D3=0D = D_1 = D_2 = D_3 = 0. Set up the coefficient determinant.
D=D1=D2=D3=0D = D_1 = D_2 = D_3 = 0
03Solve ParametersKEY INSIGHT
3/4
From D1=D2=0D_1 = D_2 = 0: 13β=5413\beta = 54 and α=13\alpha = \frac{1}{3}. So 12α+13β=4+54=5812\alpha + 13\beta = 4 + 54 = 58.
4+54=584 + 54 = 58
04Final Answer
4/4
12α+13β=5812\alpha + 13\beta = 58.
58\boxed{58}
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