Question

Under which one of the following conditions does 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 \(\neq\) 6
• D. k \(\neq\) 8

Hint:

Notice that the first and third rows of the matrix \( A \) are almost identical:
- Row 1: \( (1, 2, 4) \)
- Row 3: \( (1, 2, k-4) \)
If \( k - 4 = 4 \implies k = 8 \), the two rows become identical, making the determinant zero and violating the unique solution condition. This visual shortcut yields the answer instantly.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This question asks for the condition on parameter \( k \) such that the given system of linear equations has a unique solution.

Step 2: Key Formula or Approach:
A system of linear equations \( A X = B \) has a unique solution if and only if the determinant of the coefficient matrix \( A \) is non-zero:
\[ |A| \neq 0 \]
If \( |A| = 0 \), the system will either have no solution or infinitely many solutions.

Step 3: Detailed Explanation:

• Calculate the Determinant of Coefficient Matrix \( A \):
- The coefficient matrix is:
\[ A = \begin{pmatrix} 1 & 2 & 4 \\ 2 & 1 & 2 \\ 1 & 2 & k-4 \end{pmatrix} \]
- Expand the determinant \( |A| \) along the first row:
\[ |A| = 1 \cdot \left| \begin{matrix} 1 & 2 \\ 2 & k-4 \end{matrix} \right| - 2 \cdot \left| \begin{matrix} 2 & 2 \\ 1 & k-4 \end{matrix} \right| + 4 \cdot \left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| \]
- Calculate each \( 2 \times 2 \) determinant:
\[ \left| \begin{matrix} 1 & 2 \\ 2 & k-4 \end{matrix} \right| = 1(k-4) - (2)(2) = k - 4 - 4 = k - 8 \]
\[ \left| \begin{matrix} 2 & 2 \\ 1 & k-4 \end{matrix} \right| = 2(k-4) - (2)(1) = 2k - 8 - 2 = 2k - 10 \]
\[ \left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| = (2)(2) - (1)(1) = 4 - 1 = 3 \]
- Substitute these back into the determinant expression:
\[ |A| = 1(k - 8) - 2(2k - 10) + 4(3) \]
\[ |A| = k - 8 - 4k + 20 + 12 \]
\[ |A| = -3k + 24 \]

• Apply the Uniqueness Condition:
- For a unique solution, set \( |A| \neq 0 \):
\[ -3k + 24 \neq 0 \]
\[ 3k \neq 24 \implies k \neq 8 \]


Step 4: Final Answer:
The system has a unique solution under the condition \( k \neq 8 \).
Therefore, the correct choice is option (D).