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

Answer: (D)

$k \neq 0$

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:

$k \neq 0.$