Answer :
To solve the given system of equations using Gaussian elimination, we need to represent the system in matrix form and then apply the Gaussian elimination method to find the solution. The system of equations given are:
1.
[tex]\[ \begin{array}{l} x - 2y - 3z = 0 \\ -x + y + 2z = 3 \\ 12y + z = -8 \end{array} \][/tex]
### Step 1: Express the system as an augmented matrix
We need to write the system of equations in augmented matrix form, which is:
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ -1 & 1 & 2 & 3 \\ 0 & 12 & 1 & -8 \end{pmatrix} \][/tex]
### Step 2: Apply Gaussian elimination to obtain the matrix in Row Echelon Form (REF)
We start by transforming the augmented matrix into its Row Echelon Form through Gaussian elimination.
First, let's transform the augmented matrix into a simpler form, aiming to create zeros below the leading coefficients (the pivots) in each row.
#### Step 2.1: Augmented matrix at the start
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ -1 & 1 & 2 & 3 \\ 0 & 12 & 1 & -8 \end{pmatrix} \][/tex]
We see that the pivot in the first row is already 1. The next step is to make the first element in the second row to be 0.
#### Step 2.2: Eliminate the first element in the second row
We add the first row to the second row:
[tex]\[ R2 = R2 + R1 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ 0 & -1 & -1 & 3 \\ 0 & 12 & 1 & -8 \end{pmatrix} \][/tex]
Next, we make the second element in the third row to be 0:
#### Step 2.3: Eliminate the second element in the third row
Divide the second row by -1 to make the pivot equal to 1:
[tex]\[ R2 = \frac{R2}{-1} \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ 0 & 1 & 1 & -3 \\ 0 & 12 & 1 & -8 \end{pmatrix} \][/tex]
Now, we subtract 12 times the second row from the third row to create a zero element below the pivot:
[tex]\[ R3 = R3 - 12R2 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & -11 & 28 \end{pmatrix} \][/tex]
Next, we divide the third row by -11 to make the pivot equal to 1:
[tex]\[ R3 = \frac{R3}{-11} \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & 1 & -\frac{28}{11} \end{pmatrix} \][/tex]
#### Step 2.4: Back-substitution to form the identity matrix on the left side
- For the second row, subtract the third row to create a zero at position (2,3):
[tex]\[ R2 = R2 - R3 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ 0 & 1 & 0 & -\frac{5}{11} \\ 0 & 0 & 1 & -\frac{28}{11} \end{pmatrix} \][/tex]
- For the first row, add 3 times the third row:
[tex]\[ R1 = R1 + 3R3 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & 0 & -\frac{84}{11} \\ 0 & 1 & 0 & -\frac{5}{11} \\ 0 & 0 & 1 & -\frac{28}{11} \end{pmatrix} \][/tex]
Finally, we simplify the first row:
- For the first row, add 2 times the second row:
[tex]\[ R1 = R1 + 2R2 \][/tex]
[tex]\[ \begin{pmatrix} 1 & 0 & 0 & -\frac{94}{11} \\ 0 & 1 & 0 & -\frac{5}{11} \\ 0 & 0 & 1 & -\frac{28}{11} \end{pmatrix} \][/tex]
### Step 3: Interpret the result
From the row-reduced form of the augmented matrix, we can read off the solutions directly:
[tex]\[ x = -\frac{94}{11}, \quad y = -\frac{5}{11}, \quad z = -\frac{28}{11} \][/tex]
These fractions can be left as they are or converted to decimals if preferred:
- [tex]\( x \approx -8.545 \)[/tex]
- [tex]\( y \approx -0.455 \)[/tex]
- [tex]\( z \approx -2.545 \)[/tex]
Thus, the solution to the system of equations is:
[tex]\[ (x, y, z) = \left( -\frac{94}{11}, -\frac{5}{11}, -\frac{28}{11} \right) \][/tex]
1.
[tex]\[ \begin{array}{l} x - 2y - 3z = 0 \\ -x + y + 2z = 3 \\ 12y + z = -8 \end{array} \][/tex]
### Step 1: Express the system as an augmented matrix
We need to write the system of equations in augmented matrix form, which is:
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ -1 & 1 & 2 & 3 \\ 0 & 12 & 1 & -8 \end{pmatrix} \][/tex]
### Step 2: Apply Gaussian elimination to obtain the matrix in Row Echelon Form (REF)
We start by transforming the augmented matrix into its Row Echelon Form through Gaussian elimination.
First, let's transform the augmented matrix into a simpler form, aiming to create zeros below the leading coefficients (the pivots) in each row.
#### Step 2.1: Augmented matrix at the start
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ -1 & 1 & 2 & 3 \\ 0 & 12 & 1 & -8 \end{pmatrix} \][/tex]
We see that the pivot in the first row is already 1. The next step is to make the first element in the second row to be 0.
#### Step 2.2: Eliminate the first element in the second row
We add the first row to the second row:
[tex]\[ R2 = R2 + R1 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ 0 & -1 & -1 & 3 \\ 0 & 12 & 1 & -8 \end{pmatrix} \][/tex]
Next, we make the second element in the third row to be 0:
#### Step 2.3: Eliminate the second element in the third row
Divide the second row by -1 to make the pivot equal to 1:
[tex]\[ R2 = \frac{R2}{-1} \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ 0 & 1 & 1 & -3 \\ 0 & 12 & 1 & -8 \end{pmatrix} \][/tex]
Now, we subtract 12 times the second row from the third row to create a zero element below the pivot:
[tex]\[ R3 = R3 - 12R2 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & -11 & 28 \end{pmatrix} \][/tex]
Next, we divide the third row by -11 to make the pivot equal to 1:
[tex]\[ R3 = \frac{R3}{-11} \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & 1 & -\frac{28}{11} \end{pmatrix} \][/tex]
#### Step 2.4: Back-substitution to form the identity matrix on the left side
- For the second row, subtract the third row to create a zero at position (2,3):
[tex]\[ R2 = R2 - R3 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & -3 & 0 \\ 0 & 1 & 0 & -\frac{5}{11} \\ 0 & 0 & 1 & -\frac{28}{11} \end{pmatrix} \][/tex]
- For the first row, add 3 times the third row:
[tex]\[ R1 = R1 + 3R3 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -2 & 0 & -\frac{84}{11} \\ 0 & 1 & 0 & -\frac{5}{11} \\ 0 & 0 & 1 & -\frac{28}{11} \end{pmatrix} \][/tex]
Finally, we simplify the first row:
- For the first row, add 2 times the second row:
[tex]\[ R1 = R1 + 2R2 \][/tex]
[tex]\[ \begin{pmatrix} 1 & 0 & 0 & -\frac{94}{11} \\ 0 & 1 & 0 & -\frac{5}{11} \\ 0 & 0 & 1 & -\frac{28}{11} \end{pmatrix} \][/tex]
### Step 3: Interpret the result
From the row-reduced form of the augmented matrix, we can read off the solutions directly:
[tex]\[ x = -\frac{94}{11}, \quad y = -\frac{5}{11}, \quad z = -\frac{28}{11} \][/tex]
These fractions can be left as they are or converted to decimals if preferred:
- [tex]\( x \approx -8.545 \)[/tex]
- [tex]\( y \approx -0.455 \)[/tex]
- [tex]\( z \approx -2.545 \)[/tex]
Thus, the solution to the system of equations is:
[tex]\[ (x, y, z) = \left( -\frac{94}{11}, -\frac{5}{11}, -\frac{28}{11} \right) \][/tex]
