Question

If the homogeneous system of simultaneous equations \(AX=0\) where \(A = \begin{bmatrix} 9 & 2 & k \\ 1 & -1 & -3 \\ k-1 & 1 & 3 \end{bmatrix}\) has a nontrivial solution, then the possible values of \(k\) are

• A. 0,6
• B. 0,3
• C. 0,5
• D. 0,1
• E. 0,7

Hint:

For homogeneous systems, "nontrivial solution" is synonymous with \(\det(A) = 0\). Always simplify the inner brackets of the determinant expansion carefully to avoid errors in the resulting quadratic.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A homogeneous system of linear equations \(AX = 0\) has a non-trivial solution (solutions other than zero) if and only if the matrix \(A\) is singular, meaning its determinant is zero (\(|A| = 0\)).

Step 2: Key Formula or Approach:
Set \(\det(A) = 0\) and solve for \(k\).

Step 3: Detailed Explanation:
Calculate \(\det(A)\):
\[ |A| = \begin{vmatrix} 9 & 2 & k \\ 1 & -1 & -3 \\ k-1 & 1 & 3 \end{vmatrix} = 0 \]
Expanding along the first row:
\[ 9 [(-1)(3) - (-3)(1)] - 2 [(1)(3) - (-3)(k-1)] + k [(1)(1) - (-1)(k-1)] = 0 \]
\[ 9 [-3 + 3] - 2 [3 + 3k - 3] + k [1 + k - 1] = 0 \]
\[ 9 [0] - 2 [3k] + k [k] = 0 \]
\[ 0 - 6k + k^2 = 0 \]
\[ k^2 - 6k = 0 \]
Factoring:
\[ k(k - 6) = 0 \]
The values are \(k = 0\) and \(k = 6\).

Step 4: Final Answer:
The possible values of \(k\) are 0 and 6.