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: Expand the determinant along the third column instead.
For \( A = \begin{pmatrix} 1 & 2 & 4 \\ 2 & 1 & 2 \\ 1 & 2 & k-4 \end{pmatrix} \), expanding along column 3 gives:
\[ |A| = 4\begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} - 2\begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} + (k-4)\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} \]

Step 2: Evaluate each \( 2\times2 \) minor.
\[ \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = 3, \quad \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 0, \quad \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = -3 \]

Step 3: Combine the terms.
\[ |A| = 4(3) - 2(0) + (k-4)(-3) = 12 - 3k + 12 = 24 - 3k \]
Setting \( |A| \neq 0 \) for a unique solution gives \( 24 - 3k \neq 0 \), so \( k \neq 8 \).

Step 4: Final Answer.
\[ \boxed{k \neq 8} \]
Therefore, option (D) is correct.