Question

If the system of linear equations
x + y + z = 2, 2x + y − z = 3, 3x + 2y + kz = 4
has a unique solution, then:

• A. $k = 0$
• B. $-1 < k < 1$
• C. $-3 < k < 3$
• D. $k \neq 0$

Answer 1

The Correct Option is D

For a system of linear equations to possess a unique solution, its coefficient matrix must have a non-zero determinant. The given system is:
x + y + z = 2
2x + y - z = 3
3x + 2y + kz = 4
The corresponding coefficient matrix is:

1	1	1
2	1	-1
3	2	k

The determinant of this matrix, denoted as Δ, is calculated using the formula for a 3x3 matrix [A]:
Δ = a(ei − fh) − b(di − fg) + c(dh − eg)
Applying this formula to the given matrix yields:
Δ = 1((1)(k) - (-1)(2)) - 1((2)(k) - (-1)(3)) + 1((2)(2) - (1)(3))
Δ = 1(k + 2) - 1(2k + 3) + 1(4 - 3)
Δ = k + 2 - 2k - 3 + 1
Δ = -k
A unique solution exists if the determinant is non-zero:
-k ≠ 0
Therefore, k ≠ 0.
The condition for a unique solution is: k ≠ 0.