Determining the Number of Solutions for Ax = b

We are given the system of linear equations represented by $Ax = b$, where:

$A = \begin{bmatrix} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 0 & 1 & -1 \end{bmatrix}$ and $b = \begin{bmatrix} 1/3 \\ -1/3 \\ 0 \end{bmatrix}$.

To find the number of solutions, we analyze the ranks of the coefficient matrix $A$ and the augmented matrix $[A|b]$.

Calculating the Rank of Matrix A

First, find the determinant of $A$ to check for a unique solution:

Since $\det(A) = 0$, the system does not have a unique solution.

Now, let's find the rank of $A$ using row reduction:

The row echelon form has 2 non-zero rows. Thus, rank(A) = 2.

Calculating the Rank of the Augmented Matrix [A|b]

Form the augmented matrix $[A|b]$:

Perform row reduction:

The row echelon form has 2 non-zero rows. Thus, rank([A|b]) = 2.

Conclusion on the Number of Solutions

Compare the ranks:

• rank(A) = 2
• rank([A|b]) = 2

The number of variables ($n$) in the system is 3.

Since rank(A) = rank([A|b]) < n (specifically, 2 < 3), the system has infinitely many solutions.