Question

For What value of \(k\), the system of equations \(\begin{pmatrix} 1 & 2 & 4 \\ 2 & 1 & 2 \\ 1 & 2 & k-4 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \\ k \end{pmatrix}\) have a unique solution?

• A. For every real number \(k\)
• B. \(k = 8\)
• C. \(k \ne 6\)
• D. \(k \ne 8\)

Hint:

To determine if a system of linear equations has a unique solution, always calculate the determinant of the coefficient matrix first and set it to non-zero.
The right-hand side constant vector \(\vec{b}\) does not affect the uniqueness condition of the solution, only whether the non-unique cases are consistent or inconsistent.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the condition on the parameter \(k\) such that the given system of three linear equations in three variables (\(x, y, z\)) has a unique solution.

Step 2: Key Formula or Approach:
A system of linear equations of the form \(M\vec{x} = \vec{b}\) has a unique solution if and only if the coefficient matrix \(M\) is non-singular, which means its determinant is non-zero:
\[ \det(M) \ne 0 \] If the determinant is zero, the system will either have infinitely many solutions or no solution, depending on the vector \(\vec{b}\).

Step 3: Detailed Explanation:
Let us extract the coefficient matrix \(M\):
\[ M = \begin{pmatrix} 1 & 2 & 4 \\ 2 & 1 & 2 \\ 1 & 2 & k-4 \end{pmatrix} \] We compute the determinant of \(M\) by expanding along the first row:
\[ \det(M) = 1 \cdot \begin{vmatrix} 1 & 2 \\ 2 & k-4 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 2 \\ 1 & k-4 \end{vmatrix} + 4 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} \] Now, calculate each \(2 \times 2\) minor determinant:
1. First minor:
\[ \begin{vmatrix} 1 & 2 \\ 2 & k-4 \end{vmatrix} = 1(k-4) - 2(2) = k - 4 - 4 = k - 8 \] 2. Second minor:
\[ \begin{vmatrix} 2 & 2 \\ 1 & k-4 \end{vmatrix} = 2(k-4) - 2(1) = 2k - 8 - 2 = 2k - 10 \] 3. Third minor:
\[ \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = 2(2) - 1(1) = 4 - 1 = 3 \] Substitute these values back into the determinant expansion:
\[ \det(M) = 1(k - 8) - 2(2k - 10) + 4(3) \] Expand the terms:
\[ \det(M) = k - 8 - 4k + 20 + 12 \] Combine the like terms:
\[ \det(M) = (k - 4k) + (-8 + 20 + 12) = -3k + 24 \] For the system to have a unique solution, we set \(\det(M) \ne 0\):
\[ -3k + 24 \ne 0 \] \[ 3k \ne 24 \implies k \ne 8 \] Thus, a unique solution exists for any real number \(k\) except \(8\).

Step 4: Final Answer
The system has a unique solution when \(k \ne 8\), which corresponds to option (D).