Let $A=\left[\begin{array}{ccc}1& 1& 1\\ -1& -1& -1\\ 0& 1& -1\end{array}\right]$ and $b=\left[\begin{array}{c}1/3\\ -1/3\\ 0\end{array}\right]$, then the system of linear equations $Ax=b$ has

Determining the Number of Solutions for Ax = b

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

$A=\left[\begin{array}{ccc}1& 1& 1\\ -1& -1& -1\\ 0& 1& -1\end{array}\right]$ and $b=\left[\begin{array}{c}1/3\\ -1/3\\ 0\end{array}\right]$.

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:

$\det (A)=1\left|\begin{array}{cc}-1& -1\\ 1& -1\end{array}\right|-1\left|\begin{array}{cc}-1& -1\\ 0& -1\end{array}\right|+1\left|\begin{array}{cc}-1& -1\\ 0& 1\end{array}\right|$ $\det (A)=1((-1)(-1)-(-1)(1))-1((-1)(-1)-(-1)(0))+1((-1)(1)-(-1)(0))$ $\det (A)=1(1+1)-1(1-0)+1(-1-0)$ $\det (A)=2-1-1=0$

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

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

$A=\left[\begin{array}{ccc}1& 1& 1\\ -1& -1& -1\\ 0& 1& -1\end{array}\right]\stackrel{R_2\leftarrow R_2+R_1}{\to }\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 0\\ 0& 1& -1\end{array}\right]\stackrel{R_2\leftrightarrow R_3}{\to }\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& -1\\ 0& 0& 0\end{array}\right]$

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 $[A|b]$:

$[A|b]=\left[\begin{array}{ccccc}1& 1& 1& |& 1/3\\ -1& -1& -1& |& -1/3\\ 0& 1& -1& |& 0\end{array}\right]$

Perform row reduction:

$[A|b]\stackrel{R_2\leftarrow R_2+R_1}{\to }\left[\begin{array}{ccccc}1& 1& 1& |& 1/3\\ 0& 0& 0& |& 0\\ 0& 1& -1& |& 0\end{array}\right]\stackrel{R_2\leftrightarrow R_3}{\to }\left[\begin{array}{ccccc}1& 1& 1& |& 1/3\\ 0& 1& -1& |& 0\\ 0& 0& 0& |& 0\end{array}\right]$

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.