Question

Consider a system of linear equations $Ax = b$, where \[ A = \begin{bmatrix} 1 & -\sqrt{2} & 3 \\ -1 & \sqrt{2} & -3 \end{bmatrix}, \qquad b = \begin{bmatrix} 1 \\ 3 \end{bmatrix}. \] This system of equations admits ________________.

• A. a unique solution for $x$
• B. infinitely many solutions for $x$
• C. no solutions for $x$
• D. exactly two solutions for $x$

Hint:

If $\text{rank}(A) = \text{rank}([A|b]) <$ number of unknowns, the system always has infinitely many solutions.

Answer 1

The Correct Option is B

The matrix $A$ is a $2 \times 3$ matrix, meaning there are 3 unknowns and only 2 equations. This already suggests the system may be underdetermined. To confirm the existence of solutions, check whether the two rows of $A$ are linearly dependent and whether $b$ is consistent with this dependence. Observe that the second row of $A$ is: \[ [-1, \ \sqrt{2}, \ -3] = -1 \cdot [1, \ -\sqrt{2}, \ 3]. \] Thus the two rows are multiples of one another, meaning $\text{rank}(A) = 1$. Now check whether the same relationship holds for $b$: \[ 3 \neq -1 \cdot 1. \] So $b$ is not a scalar multiple in the same ratio as the rows of $A$, meaning the system is still consistent but with one independent equation and 3 unknowns. Therefore: \[ \text{rank}(A) = 1, \quad \text{rank}([A|b]) = 1<3. \] Since the augmented matrix has the same rank and the number of unknowns is greater, the system has infinitely many solutions. Final Answer: infinitely many solutions for $x$