A system of linear equations and a reduced matrix for the system are given.
 

x	 − 	y	 + 	z	 = 	3
3x			 + 	2z	 = 	7
x	 − 	4y	 + 	2z	 = 	5
    

1	0	
2
3
7
3

0	1	− 
1
3
− 
2
3
0	0	0		0
(a)
Use the reduced matrix to find the general solution of the system, if one exists. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, express your answers in terms of z as in Example 3.)
(x, y, z) = 

 


The reduced matrix you provided corresponds to a system of linear equations. From the matrix, we need to interpret the solution by turning it back into a system of equations.

The given reduced matrix is:

\[
\begin{pmatrix}
1 & 0 & 2 & 7 \\
0 & 1 & -\frac{1}{3} & -\frac{2}{3} \\
0 & 0 & 0 & 0
\end{pmatrix}
\]

### Step 1: Convert the matrix into equations

The first two rows of this matrix correspond to two equations in \(x\), \(y\), and \(z\), and the last row tells us there is no information in the third row (which indicates the possibility of free variables).

From the first row:

\[
1x + 0y + 2z = 7
\]

which simplifies to:

\[
x + 2z = 7
\]

From the second row:

\[
0x + 1y - \frac{1}{3}z = -\frac{2}{3}
\]

which simplifies to:

\[
y - \frac{1}{3}z = -\frac{2}{3}
\]

Finally, the third row:

\[
0x + 0y + 0z = 0
\]

This is a true statement, indicating no contradictions.

### Step 2: Express the solution in terms of \(z\)

We now solve for \(x\) and \(y\) in terms of \(z\):

- From the equation \(x + 2z = 7\), solve for \(x\):
  \[
  x = 7 - 2z
  \]

- From the equation \(y - \frac{1}{3}z = -\frac{2}{3}\), solve for \(y\):
  \[
  y = -\frac{2}{3} + \frac{1}{3}z
  \]

### Step 3: General solution

The general solution can now be written as:

\[
(x, y, z) = (7 - 2z, -\frac{2}{3} + \frac{1}{3}z, z)
\]

where \(z\) is a free variable.

Thus, the system has infinitely many solutions, and the solution can be expressed in terms of \(z\).

### Final Answer:
\[
(x, y, z) = (7 - 2z, -\frac{2}{3} + \frac{1}{3}z, z)
\]

Would you like more details or have any questions?

Here are some related questions:
1. What is a reduced row echelon form (RREF)?
2. How can free variables help in finding general solutions?
3. What is the meaning of a matrix row with all zeros?
4. How do we know a system has infinitely many solutions?
5. Can the reduced row echelon form always be used to solve linear systems?

**Tip:** In systems with free variables, each free variable introduces a degree of freedom, often leading to infinitely many solutions.

A system of linear equations and a reduced matrix for the system are given. Use the reduced matrix to find the general solution of the system, if one exists. If there are infinitely many solutions, express your answers in terms of z.

Find the general solution to a system of linear equations using the reduced row echelon form of the matrix. Learn how to express the solution in terms of a free variable and understand the implications of free variables in infinite solutions.

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