If the system of equations
has infinitely many solutions, then the ordered pair is equal to :
has infinitely many solutions, then the ordered pair is equal to :
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
For infinitely many solutions, the determinant of the coefficient matrix must be 0, and the system must be consistent.
Finding λ using det(A) = 0:
|1 2 3 |
|4 3 -4 |
|8 4 -λ |
= 1(3·(-λ) - (-4)·4) - 2(4·(-λ) - (-4)·8) + 3(4·4 - 3·8)
= 1(-3λ + 16) - 2(-4λ + 32) + 3(16 - 24)
= -3λ + 16 + 8λ - 64 + 3(-8)
= 5λ - 72 = 0
So λ = 72/5.
Finding μ using consistency:
With λ = 72/5, use augmented matrix. From equations 1 and 2, eliminate x: multiply eq1 by 4 and subtract from eq2:
4x+3y-4z - (4x+8y+12z) = 4-12 → -5y - 16z = -8 → 5y + 16z = 8.
Multiply eq1 by 8 and subtract from eq3:
8x+4y-(72/5)z - (8x+16y+24z) = 9+μ-24 → -12y - (72/5+24)z = μ-15
→ -12y - (192/5)z = μ-15.
From 5y + 16z = 8: y = (8
Finding λ using det(A) = 0:
|1 2 3 |
|4 3 -4 |
|8 4 -λ |
= 1(3·(-λ) - (-4)·4) - 2(4·(-λ) - (-4)·8) + 3(4·4 - 3·8)
= 1(-3λ + 16) - 2(-4λ + 32) + 3(16 - 24)
= -3λ + 16 + 8λ - 64 + 3(-8)
= 5λ - 72 = 0
So λ = 72/5.
Finding μ using consistency:
With λ = 72/5, use augmented matrix. From equations 1 and 2, eliminate x: multiply eq1 by 4 and subtract from eq2:
4x+3y-4z - (4x+8y+12z) = 4-12 → -5y - 16z = -8 → 5y + 16z = 8.
Multiply eq1 by 8 and subtract from eq3:
8x+4y-(72/5)z - (8x+16y+24z) = 9+μ-24 → -12y - (72/5+24)z = μ-15
→ -12y - (192/5)z = μ-15.
From 5y + 16z = 8: y = (8
[JEE Main 2023 · 24 Jan · Shift 2]