Answer :
To solve the given system of equations using Gauss-Jordan elimination, we'll follow these steps to transform the matrix into reduced row-echelon form (RREF):
The given system of equations is:
1. [tex]\( 2x_1 - 4x_2 + 10x_3 = -12 \)[/tex]
2. [tex]\( 3x_1 + 19x_2 - 46x_3 = -4 \)[/tex]
3. [tex]\( x_1 + 3x_2 - 7x_3 = -3 \)[/tex]
First, we'll write the augmented matrix for this system:
[tex]\[
\begin{bmatrix}
2 & -4 & 10 & | & -12 \\
3 & 19 & -46 & | & -4 \\
1 & 3 & -7 & | & -3 \\
\end{bmatrix}
\][/tex]
Step 1: Make the first element of the first row a pivot (1). We can divide the entire first row by 2:
[tex]\[
\begin{bmatrix}
1 & -2 & 5 & | & -6 \\
3 & 19 & -46 & | & -4 \\
1 & 3 & -7 & | & -3 \\
\end{bmatrix}
\][/tex]
Step 2: Eliminate the [tex]\(x_1\)[/tex] term from the second and third row by adjusting them with the first row.
For the second row, subtract 3 times the first row from the second row:
[tex]\[
R2 = R2 - 3 \times R1
\][/tex]
[tex]\[
\begin{bmatrix}
1 & -2 & 5 & | & -6 \\
0 & 25 & -61 & | & 14 \\
1 & 3 & -7 & | & -3 \\
\end{bmatrix}
\][/tex]
For the third row, subtract the first row from the third row:
[tex]\[
R3 = R3 - R1
\][/tex]
[tex]\[
\begin{bmatrix}
1 & -2 & 5 & | & -6 \\
0 & 25 & -61 & | & 14 \\
0 & 5 & -12 & | & 3 \\
\end{bmatrix}
\][/tex]
Step 3: Make the second element in the second row a pivot (1) by dividing the entire second row by 25:
[tex]\[
\begin{bmatrix}
1 & -2 & 5 & | & -6 \\
0 & 1 & -\frac{61}{25} & | & \frac{14}{25} \\
0 & 5 & -12 & | & 3 \\
\end{bmatrix}
\][/tex]
Step 4: Eliminate the [tex]\(x_2\)[/tex] term from the first and third rows using the second row.
For the first row, add 2 times the second row to the first row:
[tex]\[
R1 = R1 + 2 \times R2
\][/tex]
[tex]\[
\begin{bmatrix}
1 & 0 & \frac{10}{25} & | & -\frac{92}{25} \\
0 & 1 & -\frac{61}{25} & | & \frac{14}{25} \\
0 & 5 & -12 & | & 3 \\
\end{bmatrix}
\][/tex]
For the third row, subtract 5 times the second row from the third row:
[tex]\[
R3 = R3 - 5 \times R2
\][/tex]
[tex]\[
\begin{bmatrix}
1 & 0 & \frac{10}{25} & | & -\frac{92}{25} \\
0 & 1 & -\frac{61}{25} & | & \frac{14}{25} \\
0 & 0 & 1 & | & 1 \\
\end{bmatrix}
\][/tex]
Step 5: Finally, simplify the third column in the first row by replacing it with the corresponding operation from the third row to achieve the final reduced row-echelon form:
Subtract [tex]\(\frac{10}{25}\)[/tex] times the third row from the first row:
[tex]\[
R1 = R1 - \frac{10}{25} \times R3
\][/tex]
[tex]\[
\begin{bmatrix}
1 & 0 & 0 & | & -5 \\
0 & 1 & 0 & | & 3 \\
0 & 0 & 1 & | & 1 \\
\end{bmatrix}
\][/tex]
Now that the matrix is in reduced row-echelon form, we have the solutions:
- [tex]\(x_1 = -5\)[/tex]
- [tex]\(x_2 = 3\)[/tex]
- [tex]\(x_3 = 1\)[/tex]
So, the solution to the system of equations is [tex]\((-5, 3, 1)\)[/tex].
The given system of equations is:
1. [tex]\( 2x_1 - 4x_2 + 10x_3 = -12 \)[/tex]
2. [tex]\( 3x_1 + 19x_2 - 46x_3 = -4 \)[/tex]
3. [tex]\( x_1 + 3x_2 - 7x_3 = -3 \)[/tex]
First, we'll write the augmented matrix for this system:
[tex]\[
\begin{bmatrix}
2 & -4 & 10 & | & -12 \\
3 & 19 & -46 & | & -4 \\
1 & 3 & -7 & | & -3 \\
\end{bmatrix}
\][/tex]
Step 1: Make the first element of the first row a pivot (1). We can divide the entire first row by 2:
[tex]\[
\begin{bmatrix}
1 & -2 & 5 & | & -6 \\
3 & 19 & -46 & | & -4 \\
1 & 3 & -7 & | & -3 \\
\end{bmatrix}
\][/tex]
Step 2: Eliminate the [tex]\(x_1\)[/tex] term from the second and third row by adjusting them with the first row.
For the second row, subtract 3 times the first row from the second row:
[tex]\[
R2 = R2 - 3 \times R1
\][/tex]
[tex]\[
\begin{bmatrix}
1 & -2 & 5 & | & -6 \\
0 & 25 & -61 & | & 14 \\
1 & 3 & -7 & | & -3 \\
\end{bmatrix}
\][/tex]
For the third row, subtract the first row from the third row:
[tex]\[
R3 = R3 - R1
\][/tex]
[tex]\[
\begin{bmatrix}
1 & -2 & 5 & | & -6 \\
0 & 25 & -61 & | & 14 \\
0 & 5 & -12 & | & 3 \\
\end{bmatrix}
\][/tex]
Step 3: Make the second element in the second row a pivot (1) by dividing the entire second row by 25:
[tex]\[
\begin{bmatrix}
1 & -2 & 5 & | & -6 \\
0 & 1 & -\frac{61}{25} & | & \frac{14}{25} \\
0 & 5 & -12 & | & 3 \\
\end{bmatrix}
\][/tex]
Step 4: Eliminate the [tex]\(x_2\)[/tex] term from the first and third rows using the second row.
For the first row, add 2 times the second row to the first row:
[tex]\[
R1 = R1 + 2 \times R2
\][/tex]
[tex]\[
\begin{bmatrix}
1 & 0 & \frac{10}{25} & | & -\frac{92}{25} \\
0 & 1 & -\frac{61}{25} & | & \frac{14}{25} \\
0 & 5 & -12 & | & 3 \\
\end{bmatrix}
\][/tex]
For the third row, subtract 5 times the second row from the third row:
[tex]\[
R3 = R3 - 5 \times R2
\][/tex]
[tex]\[
\begin{bmatrix}
1 & 0 & \frac{10}{25} & | & -\frac{92}{25} \\
0 & 1 & -\frac{61}{25} & | & \frac{14}{25} \\
0 & 0 & 1 & | & 1 \\
\end{bmatrix}
\][/tex]
Step 5: Finally, simplify the third column in the first row by replacing it with the corresponding operation from the third row to achieve the final reduced row-echelon form:
Subtract [tex]\(\frac{10}{25}\)[/tex] times the third row from the first row:
[tex]\[
R1 = R1 - \frac{10}{25} \times R3
\][/tex]
[tex]\[
\begin{bmatrix}
1 & 0 & 0 & | & -5 \\
0 & 1 & 0 & | & 3 \\
0 & 0 & 1 & | & 1 \\
\end{bmatrix}
\][/tex]
Now that the matrix is in reduced row-echelon form, we have the solutions:
- [tex]\(x_1 = -5\)[/tex]
- [tex]\(x_2 = 3\)[/tex]
- [tex]\(x_3 = 1\)[/tex]
So, the solution to the system of equations is [tex]\((-5, 3, 1)\)[/tex].
