Question

Let \( \mathbf{A} = \begin{bmatrix} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 0 & 1 & -1 \end{bmatrix} \), and \( \mathbf{b} = \begin{bmatrix} \frac{1}{3} \\ -\frac{1}{3} \\ 0 \end{bmatrix} \). Then, the system of linear equations \( \mathbf{A} \mathbf{x} = \mathbf{b} \) has

• A. a unique solution.
• B. infinitely many solutions.
• C. a finite number of solutions.
• D. no solution.

Hint:

If the rank of the coefficient matrix equals the rank of the augmented matrix and is {less than} the number of variables, the system has {infinitely many solutions}.

Answer 1

The Correct Option is B

We are given a system of linear equations of the form \( \mathbf{A} \mathbf{x} = \mathbf{b} \), where:
\[ \mathbf{A} = \begin{bmatrix} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 0 & 1 & -1 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} \frac{1}{3} \\ -\frac{1}{3} \\ 0 \end{bmatrix} \]
First, observe that Row 2 is the negative of Row 1, which means the rank of \( \mathbf{A} \) is at most 2.

Let's write the augmented matrix and row reduce:

\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & \frac{1}{3} \\ -1 & -1 & -1 & -\frac{1}{3} \\ 0 & 1 & -1 & 0 \end{array} \right] \]

Add Row 1 to Row 2:
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & \frac{1}{3} \\ 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \end{array} \right] \]

Now subtract \( 1 \times \) Row 2 from Row 1:
\[ \left[ \begin{array}{ccc|c} 1 & 0 & 2 & \frac{1}{3} \\ 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \end{array} \right] \]

This system has two non-zero rows, and three variables, so the rank of \( \mathbf{A} \) = rank of augmented matrix = 2 < number of variables = 3.

\[ \Rightarrow \text{Infinitely many solutions} \]