To solve the given system of equations using an augmented matrix and row echelon form, we start by writing the system of equations as an augmented matrix. The system of equations is:
- \( x + y + z = 2 \)
- \( 6x - 4y + 5z = 31 \)
- \( 5x + 2y + 2z = 13 \)
The augmented matrix for this system is:
\[ \begin{pmatrix} 1 & 1 & 1 & | & 2 \ 6 & -4 & 5 & | & 31 \ 5 & 2 & 2 & | & 13 \end{pmatrix} \]
Next, we perform row operations to bring the matrix into row echelon form.
Step 1: Eliminate \(x\) from the second and third rows.
We can eliminate \( x \) from the second row by replacing it with \( R_2 - 6R_1 \):
\[ R_2 = R_2 - 6R_1 \rightarrow (6 - 6 \cdot 1, -4 - 6 \cdot 1, 5 - 6 \cdot 1, 31 - 6 \cdot 2) \] \[ \Rightarrow (0, -10, -1, 19) \]
Next, we eliminate \( x \) from the third row by replacing it with \( R_3 - 5R_1 \):
\[ R_3 = R_3 - 5R_1 \rightarrow (5 - 5 \cdot 1, 2 - 5 \cdot 1, 2 - 5 \cdot 1, 13 - 5 \cdot 2) \] \[ \Rightarrow (0, -3, -3, 3) \]
The matrix now looks like this:
\[ \begin{pmatrix} 1 & 1 & 1 & | & 2 \ 0 & -10 & -1 & | & 19 \ 0 & -3 & -3 & | & 3 \end{pmatrix} \]
Step 2: Normalize and eliminate the second column entry.
Next, we simplify the second row by dividing it by \(-10\):
\[ R_2 = \frac{1}{-10} R_2 \rightarrow (0, 1, \frac{1}{10}, -\frac{19}{10}) \]
Now we eliminate the \(y\) from the third row using \(R_3 + 3R_2\):
\[ R_3 = R_3 + 3R_2 \rightarrow (0, 0, -3 + 3 \cdot \frac{1}{10}, 3 - 3 \cdot \frac{19}{10}) \] Calculating the third row gives:
\[ (0, 0, -3 + \frac{3}{10}, 3 - \frac{57}{10}) \rightarrow (0, 0, -\frac{27}{10}, -\frac{27}{10}) \]
The matrix is now:
\[ \begin{pmatrix} 1 & 1 & 1 & | & 2 \ 0 & 1 & \frac{1}{10} & | & -\frac{19}{10} \ 0 & 0 & -\frac{27}{10} & | & -\frac{27}{10} \end{pmatrix} \]
Step 3: Back substitution to solve for values.
We can simplify the third row further by multiplying by \(-\frac{10}{27}\):
\[ R_3 = \frac{10}{-27}R_3 \rightarrow (0, 0, 1 | 1) \]
The system in row echelon form now looks like this:
\[ \begin{pmatrix} 1 & 1 & 1 & | & 2 \ 0 & 1 & \frac{1}{10} & | & -\frac{19}{10} \ 0 & 0 & 1 & | & 1 \end{pmatrix} \]
From the final \( R_3 \):
- \( z = 1 \)
Substituting \( z = 1 \) into \( R_2 \):
\[ 0y + \frac{1}{10}(1) = -\frac{19}{10} \implies y + \frac{1}{10} = -\frac{19}{10} \implies y = -\frac{20}{10} = -2 \]
Substituting \( y = -2 \) and \( z = 1 \) into \( R_1 \):
\[ x - 2 + 1 = 2 \implies x - 1 = 2 \implies x = 3 \]
Thus, the value of \( x \) is
\[ \boxed{3} \]
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