$$ \left[ \begin{array} { c c c | c } { 3 } & { - 2 } & { 4 } & { - 16 } \ { 2 } & { 3 } & { - 2 } & { - 5 } \ { 5 } & { - 1 } & { - 1 } & { - 3 } \end{array} ight] $$

Question

$$ \left[ \begin{array} { c c c | c } { 3 } & { - 2 } & { 4 } & { - 16 } \ { 2 } & { 3 } & { - 2 } & { - 5 } \ { 5 } & { - 1 } & { - 1 } & { - 3 } \end{array} ight] $$

UpStudy AI · Accepted Answer

To solve the system of linear equations represented by the augmented matrix:

$$
\left[ \begin{array} { c c c | c } 
3 & -2 & 4 & -16 \ 
2 & 3 & -2 & -5 \ 
5 & -1 & -1 & -3 
\end{array} \right]
$$

we can use **Gaussian Elimination** to reduce the matrix to row-echelon form and then solve for the variables $$ x $$, $$ y $$, and $$ z $$.

### Step 1: Write the System of Equations

The augmented matrix corresponds to the following system:

$$
\begin{cases}
3x - 2y + 4z = -16 \quad \text{(Equation 1)} \
2x + 3y - 2z = -5 \quad \text{(Equation 2)} \
5x - y - z = -3 \quad \text{(Equation 3)}
\end{cases}
$$

### Step 2: Eliminate $$ x $$ from Equations 2 and 3

**a. Eliminate $$ x $$ from Equation 2**

To eliminate $$ x $$ from Equation 2 using Equation 1:

- Multiply Equation 1 by $$ \frac{2}{3} $$:

$$
2x - \frac{4}{3}y + \frac{8}{3}z = -\frac{32}{3}
$$

- Subtract this result from Equation 2:

$$
\begin{align*}
(2x + 3y - 2z) - \left(2x - \frac{4}{3}y + \frac{8}{3}z\right) &= -5 - \left(-\frac{32}{3}\right) \
0x + \left(3 + \frac{4}{3}\right)y + \left(-2 - \frac{8}{3}\right)z &= \frac{17}{3}
\end{align*}
$$

Simplifying:

$$
\frac{13}{3}y - \frac{14}{3}z = \frac{17}{3} \quad \text{(Equation 2')}
$$

Multiply through by 3 to eliminate fractions:

$$
13y - 14z = 17 \quad \text{(Equation 2')}
$$

**b. Eliminate $$ x $$ from Equation 3**

To eliminate $$ x $$ from Equation 3 using Equation 1:

- Multiply Equation 1 by $$ \frac{5}{3} $$:

$$
5x - \frac{10}{3}y + \frac{20}{3}z = -\frac{80}{3}
$$

- Subtract this result from Equation 3:

$$
\begin{align*}
(5x - y - z) - \left(5x - \frac{10}{3}y + \frac{20}{3}z\right) &= -3 - \left(-\frac{80}{3}\right) \
0x + \left(-1 + \frac{10}{3}\right)y + \left(-1 - \frac{20}{3}\right)z &= \frac{71}{3}
\end{align*}
$$

Simplifying:

$$
\frac{7}{3}y - \frac{23}{3}z = \frac{71}{3} \quad \text{(Equation 3')}
$$

Multiply through by 3 to eliminate fractions:

$$
7y - 23z = 71 \quad \text{(Equation 3')}
$$

### Step 3: Solve for $$ y $$ and $$ z $$

Now, we have two equations with two variables:

$$
\begin{cases}
13y - 14z = 17 \quad \text{(Equation 2')} \
7y - 23z = 71 \quad \text{(Equation 3')}
\end{cases}
$$

**a. Eliminate $$ y $$**

- Multiply Equation 2' by 7:

$$
91y - 98z = 119 \quad \text{(Equation 2'')}
$$

- Multiply Equation 3' by 13:

$$
91y - 299z = 923 \quad \text{(Equation 3'')}
$$

- Subtract Equation 2'' from Equation 3'':

$$
(91y - 299z) - (91y - 98z) = 923 - 119 \
-201z = 804 \
z = -4
$$

**b. Solve for $$ y $$**

Plug $$ z = -4 $$ into Equation 2':

$$
13y - 14(-4) = 17 \
13y + 56 = 17 \
13y = -39 \
y = -3
$$

### Step 4: Solve for $$ x $$

Use Equation 1 to solve for $$ x $$:

$$
3x - 2y + 4z = -16 \
3x - 2(-3) + 4(-4) = -16 \
3x + 6 - 16 = -16 \
3x - 10 = -16 \
3x = -6 \
x = -2
$$

### Step 5: Summary of Solutions

$$
\begin{cases}
x = -2 \
y = -3 \
z = -4
\end{cases}
$$

**Therefore, the solution to the system of equations is:**

$$
x = -2,\quad y = -3,\quad z = -4
$$
The solution to the system of equations is $$ x = -2 $$, $$ y = -3 $$, and $$ z = -4 $$.

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