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Given matrix:

[tex]\[
\begin{aligned}
\left[\begin{array}{rrr|r}
1 & 2 & -3 & 3 \\
2 & -1 & -2 & 14 \\
3 & -1 & 5 & -5
\end{array}\right]
& \rightarrow
\left[\begin{array}{rrr|r}
1 & 2 & -3 & 3 \\
0 & -5 & 4 & 8 \\
0 & -7 & 14 & -14
\end{array}\right]
\\
& \rightarrow
\left[\begin{array}{rrr|r}
1 & 2 & -3 & 3 \\
0 & 1 & -\frac{4}{5} & -\frac{8}{5} \\
0 & -7 & 14 & -14
\end{array}\right]
\end{aligned}
\][/tex]

Sagot :

GinnyAnswer
To solve the system of linear equations using Gaussian elimination, follow these steps carefully. We'll transform the augmented matrix step-by-step to reach its row-echelon form:

Given the augmented matrix:
[tex]\[ \begin{bmatrix} 1 & 2 & -3 & 3 \\ 2 & -1 & -2 & 14 \\ 3 & -1 & 5 & -5 \end{bmatrix} \][/tex]

### Step 1: Eliminate the first element of the second and third rows

1. Make the first element of the second row zero:

[tex]\[ R2 \rightarrow R2 - 2 \cdot R1 \][/tex]
[tex]\[ R2 = [2, -1, -2, 14] - 2 \cdot [1, 2, -3, 3] \][/tex]
[tex]\[ R2 = [2 - 2 \cdot 1, -1 - 2 \cdot 2, -2 + 2 \cdot 3, 14 - 2 \cdot 3] \][/tex]
[tex]\[ R2 = [0, -5, 4, 8] \][/tex]

Our matrix now looks like this:
[tex]\[ \begin{bmatrix} 1 & 2 & -3 & 3 \\ 0 & -5 & 4 & 8 \\ 3 & -1 & 5 & -5 \end{bmatrix} \][/tex]

2. Make the first element of the third row zero:

[tex]\[ R3 \rightarrow R3 - 3 \cdot R1 \][/tex]
[tex]\[ R3 = [3, -1, 5, -5] - 3 \cdot [1, 2, -3, 3] \][/tex]
[tex]\[ R3 = [3 - 3 \cdot 1, -1 - 3 \cdot 2, 5 + 3 \cdot 3, -5 - 3 \cdot 3] \][/tex]
[tex]\[ R3 = [0, -7, 14, -14] \][/tex]

Our matrix now looks like this:
[tex]\[ \begin{bmatrix} 1 & 2 & -3 & 3 \\ 0 & -5 & 4 & 8 \\ 0 & -7 & 14 & -14 \end{bmatrix} \][/tex]

### Step 2: Make the second element of the second row one

To make the second element of the second row one:
[tex]\[ R2 \rightarrow \frac{R2}{-5} \][/tex]
[tex]\[ R2 = \frac{1}{-5} \cdot [0, -5, 4, 8] \][/tex]
[tex]\[ R2 = [0, 1, -0.8, -1.6] \][/tex]

Our matrix now looks like this:
[tex]\[ \begin{bmatrix} 1 & 2 & -3 & 3 \\ 0 & 1 & -0.8 & -1.6 \\ 0 & -7 & 14 & -14 \end{bmatrix} \][/tex]

### Step 3: Eliminate the second element of the third row

To make the second element of the third row zero:
[tex]\[ R3 \rightarrow R3 + 7 \cdot R2 \][/tex]
[tex]\[ R3 = [0, -7, 14, -14] + 7 \cdot [0, 1, -0.8, -1.6] \][/tex]
[tex]\[ R3 = [0 + 7 \cdot 0, -7 + 7 \cdot 1, 14 + 7 \cdot (-0.8), -14 + 7 \cdot (-1.6)] \][/tex]
[tex]\[ R3 = [0, 0, 14 - 5.6, -14 - 11.2] \][/tex]
[tex]\[ R3 = [0, 0, 8.4, -25.2] \][/tex]

Our final row-echelon form of the augmented matrix is:
[tex]\[ \begin{bmatrix} 1 & 2 & -3 & 3 \\ 0 & 1 & -0.8 & -1.6 \\ 0 & 0 & 8.4 & -25.2 \end{bmatrix} \][/tex]
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