Question

If \(A = \begin{pmatrix} 4 & -6 -8 & 12 \end{pmatrix}\), then a nontrivial solution of \(A x = 0\) is:

• A. \(\begin{pmatrix} 1 2 \end{pmatrix}\)
• B. \(\begin{pmatrix} 2 3 \end{pmatrix}\)
• C. \(\begin{pmatrix} 3 2 \end{pmatrix}\)
• D. \(\begin{pmatrix} 2 1 \end{pmatrix}\)

Hint:

Exam Tip:
For homogeneous systems \(A x = 0\):

• Row reduce the matrix.
• Find the free variables.
• Express the solution in terms of parameters.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to solve the homogeneous system \(A x = 0\) for a nontrivial solution (i.e., not the zero vector).

Step 2: Key Formula or Approach:
The system \(A x = 0\) means: \[ \begin{pmatrix} 4 & -6 -8 & 12 \end{pmatrix} \begin{pmatrix} x_1 x_2 \end{pmatrix} = \begin{pmatrix} 0 0 \end{pmatrix} \] This gives two equations: \[ 4x_1 - 6x_2 = 0 \quad \text{(1)} \] \[ -8x_1 + 12x_2 = 0 \quad \text{(2)} \]

Step 3: Detailed Explanation:
From equation (1): \(4x_1 = 6x_2 \Rightarrow x_1 = \frac{3}{2} x_2\).
Equation (2) is just a multiple of equation (1) (multiplying (1) by -2 gives equation (2)).
So, the system reduces to \(4x_1 - 6x_2 = 0\).
Let \(x_2 = t\) (free variable), then \(x_1 = \frac{3}{2} t\).
So, the solution is \(x = \begin{pmatrix} \frac{3}{2} t t \end{pmatrix} = t \begin{pmatrix} \frac{3}{2} 1 \end{pmatrix}\).
For \(t = 2\), we get \(x = \begin{pmatrix} 3 2 \end{pmatrix}\).
This corresponds to option (C) \(\begin{pmatrix} 3 2 \end{pmatrix}\).
Wait, option (C) is \(\begin{pmatrix} 3 2 \end{pmatrix}\).
But option (B) is \(\begin{pmatrix} 2 3 \end{pmatrix}\).
Let's check: For \(x = \begin{pmatrix} 2 3 \end{pmatrix}\), \(4(2) - 6(3) = 8 - 18 = -10 \neq 0\). So, (2, 3) is not a solution.
For \(x = \begin{pmatrix} 3 2 \end{pmatrix}\), \(4(3) - 6(2) = 12 - 12 = 0\). So, (3, 2) is a solution.
Thus, option (C) is correct.
But the answer is marked as (B) in the original. Let's check again.
If the matrix was \(A = \begin{pmatrix} 4 & -6 -8 & 12 \end{pmatrix}\), then the solution is \(\begin{pmatrix} \frac{3}{2} 1 \end{pmatrix}\) times a scalar.
If we take \(t = 2\), we get \(\begin{pmatrix} 3 2 \end{pmatrix}\).
So, option (C) is the correct nontrivial solution.
There might be a typo in the original answer key.
I'll proceed with option (C).

Step 4: Final Answer:
Therefore, option (C) is correct.