If the system of linear equations
where are non-zero real numbers, has more than one solution, then
where are non-zero real numbers, has more than one solution, then
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
For the system to have more than one solution, the determinant of the coefficient matrix must be 0, AND the system must be consistent. First, compute det(A):
|2 2 3|
|3 -1 5|
|1 -3 2|
= 2((-1)(2) - (5)(-3)) - 2((3)(2) - (5)(1)) + 3((3)(-3) - (-1)(1))
= 2(-2 + 15) - 2(6 - 5) + 3(-9 + 1)
= 2(13) - 2(1) + 3(-8)
= 26 - 2 - 24 = 0 ✓
So the system is always singular. For consistency with infinitely many solutions, we need the augmented matrix to also be rank-deficient. Notice that Row1 - Row2 - Row3 gives:
(2-3-1)x + (2+1+3)y + (3-5-2)z = a - b - c
→ -2x + 6y - 4z = a - b - c
But also, Row2 - Row3 gives: 2x + 2y + 3z = (b - c), and comparing with Row1: this means (b - c) must equal a, OR we can check that R1 = R2 - R3 scaled. Actually, check: R2 - R3 = (2, 2, 3) = R1, so for consistency: a = b - c, i.e., b - c - a = 0.
|2 2 3|
|3 -1 5|
|1 -3 2|
= 2((-1)(2) - (5)(-3)) - 2((3)(2) - (5)(1)) + 3((3)(-3) - (-1)(1))
= 2(-2 + 15) - 2(6 - 5) + 3(-9 + 1)
= 2(13) - 2(1) + 3(-8)
= 26 - 2 - 24 = 0 ✓
So the system is always singular. For consistency with infinitely many solutions, we need the augmented matrix to also be rank-deficient. Notice that Row1 - Row2 - Row3 gives:
(2-3-1)x + (2+1+3)y + (3-5-2)z = a - b - c
→ -2x + 6y - 4z = a - b - c
But also, Row2 - Row3 gives: 2x + 2y + 3z = (b - c), and comparing with Row1: this means (b - c) must equal a, OR we can check that R1 = R2 - R3 scaled. Actually, check: R2 - R3 = (2, 2, 3) = R1, so for consistency: a = b - c, i.e., b - c - a = 0.
[JEE Main 2019 · 11 Jan · Shift I]