Question: Use Gauss Method to solve the following system:
$-5x + 2y + 6z = 1$
$7x + -\frac{12}{5} y + -\frac{2}{5} z = -5$
$23x + -\frac{38}{5} y + \frac{22}{5} z = -19$
Step 1: Use $-\frac{7}{-5} R_1 + R_2$; $-\frac{23}{-5} R_1 + R_3$ to eliminate $x$ in $R_2$ and $R_3$:
$-5x + 2y + 6z = 1$
$0x + \frac{2}{5} y + 8 z = -\frac{18}{5}$
$0x + ___ y + ___ z = ___$
Step 2: Use $0R_1 + -4R_2 + 1R_3$ to eliminate $y$ in $R_3$:
$-5x + 2y + 6z = 1$
$0x + \frac{2}{5} y + 8 z = -\frac{18}{5}$
$0x + 0y + 0z = 0$
The last equation indicates there are infinite solutions for the given system.
1. **State the problem:** Solve the system using Gauss Method:
$$
\begin{cases}
-5x + 2y + 6z = 1 \\
7x - \frac{12}{5}y - \frac{2}{5}z = -5 \\
23x - \frac{38}{5}y + \frac{22}{5}z = -19
\end{cases}
$$
2. **Step 1: Eliminate $x$ from $R_2$ and $R_3$ using $R_1$:**
Calculate multipliers:
$$m_2 = -\frac{7}{-5} = \frac{7}{5}, \quad m_3 = -\frac{23}{-5} = \frac{23}{5}$$
Apply row operations:
$$R_2 \to R_2 + m_2 R_1 = 7x - \frac{12}{5}y - \frac{2}{5}z + \frac{7}{5}(-5x + 2y + 6z)$$
Simplify $R_2$:
$$7x - \frac{12}{5}y - \frac{2}{5}z + \frac{7}{5}(-5x) + \frac{7}{5}(2y) + \frac{7}{5}(6z) = 7x - \frac{12}{5}y - \frac{2}{5}z - 7x + \frac{14}{5}y + \frac{42}{5}z$$
Combine like terms:
$$0x + \left(-\frac{12}{5} + \frac{14}{5}\right)y + \left(-\frac{2}{5} + \frac{42}{5}\right)z = 0x + \frac{2}{5}y + 8z$$
Calculate right side:
$$-5 + \frac{7}{5} \times 1 = -5 + \frac{7}{5} = -\frac{25}{5} + \frac{7}{5} = -\frac{18}{5}$$
So,
$$R_2: 0x + \frac{2}{5}y + 8z = -\frac{18}{5}$$
Similarly for $R_3$:
$$R_3 \to R_3 + m_3 R_1 = 23x - \frac{38}{5}y + \frac{22}{5}z + \frac{23}{5}(-5x + 2y + 6z)$$
Simplify $R_3$:
$$23x - \frac{38}{5}y + \frac{22}{5}z - 23x + \frac{46}{5}y + \frac{138}{5}z = 0x + \left(-\frac{38}{5} + \frac{46}{5}\right)y + \left(\frac{22}{5} + \frac{138}{5}\right)z$$
Combine like terms:
$$0x + \frac{8}{5}y + 32z$$
Calculate right side:
$$-19 + \frac{23}{5} \times 1 = -\frac{95}{5} + \frac{23}{5} = -\frac{72}{5}$$
So,
$$R_3: 0x + \frac{8}{5}y + 32z = -\frac{72}{5}$$
3. **Step 2: Eliminate $y$ from $R_3$ using $R_2$:**
Use operation:
$$R_3 \to R_3 - 4 R_2$$
Calculate:
$$\left(0x + \frac{8}{5}y + 32z\right) - 4\left(0x + \frac{2}{5}y + 8z\right) = 0x + \left(\frac{8}{5} - \frac{8}{5}\right)y + (32 - 32)z = 0x + 0y + 0z$$
Right side:
$$-\frac{72}{5} - 4\left(-\frac{18}{5}\right) = -\frac{72}{5} + \frac{72}{5} = 0$$
So,
$$R_3: 0x + 0y + 0z = 0$$
4. **Interpretation:**
The last equation is $0=0$, which is always true, indicating the system has infinitely many solutions.
**Final system:**
$$
\begin{cases}
-5x + 2y + 6z = 1 \\
0x + \frac{2}{5}y + 8z = -\frac{18}{5} \\
0x + 0y + 0z = 0
\end{cases}
$$
This confirms infinite solutions exist.
**Answer:** The last equation indicates there are infinite solutions for the given system.