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

To determine the condition for the system of linear equations to have a unique solution, we need to ensure that the coefficient matrix has a non-zero determinant. The given system of equations 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

We calculate the determinant of this matrix. The determinant, Δ, is calculated as follows for a 3x3 matrix [A]:
Δ = a(ei − fh) − b(di − fg) + c(dh − eg)
Applying this to our matrix, we calculate:
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 + 2 - 2k - 3 + 1
Δ = -k
For the system to have a unique solution, we need the determinant to be non-zero:
-k ≠ 0
This implies k ≠ 0.
Thus, the correct answer is: k ≠ 0.

Answer 2

To determine when the system has a unique solution, the determinant of the coefficient matrix must be nonzero. The coefficient matrix for the system is:

\[A=\begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{bmatrix}\]

The determinant of A is:

\[\det(A)=\begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{vmatrix}\]

Expanding along the first row:

\[\det(A)=1 \cdot \begin{vmatrix} 1 & -1 \\ 2 & k \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & -1 \\ 3 & k \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix}\]

Calculate each minor:

\[\begin{vmatrix} 1 & -1 \\ 2 & k \end{vmatrix} = (1)(k) - (2)(-1) = k + 2\]

\[\begin{vmatrix} 2 & -1 \\ 3 & k \end{vmatrix} = (2)(k) - (3)(-1) = 2k + 3,\]

\[\begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = (2)(2) - (3)(1) = 4 - 3 = 1.\]

Substitute these values back:

\[\det(A) = 1(k + 2) - 1(2k + 3) + 1(1).\]

Simplify:

\[\det(A) = k + 2 - 2k - 3 + 1 = -k.\]

For the system to have a unique solution, \(\det(A) \neq 0\). Thus:

\[-k \neq 0 \implies k \neq 0.\]

Final Answer:

\[k \neq 0.\]