Solve the following system of linear equations:
2x2+4x3+2x4 = −5
5x2+10x3+5x4 = −15
x1+2x2+2x3−3x4 =3
If the system has no solution, demonstrate this by giving a row-echelon form of the augmented matrix for the system.
First we'll form an augmented matrix for the system:
$\begin{bmatrix} 0 & 2 & 4 & 2 & | & -5 \\ 0 & 5 & 10 & 5 & | & -15 \\ 1 & 2 & 2 & -3 & | & 3 \end{bmatrix}$
Swap the first and third rows to move the 1 to the upper left corner:
$\begin{bmatrix} 1 & 2 & 2 & -3 & | & 3 \\ 0 & 5 & 10 & 5 & | & -15 \\ 0 & 2 & 4 & 2 & | & -5 \end{bmatrix}$
We'll use the first row to eliminate the 2 in the second row by subtracting twice the second row from the third row, and then scale down the second row by 1/5:
$\begin{bmatrix} 1 & 2 & 2 & -3 & | & 3 \\ 0 & 1 & 2 & 1 & | & -3 \\ 0 & 0 & 0 & 0 & | & 1 \end{bmatrix}$
Since we have a row with 0's on the variable side and a non-zero constant on the other side, and thus an inconsistency, the system has no solution.
To solve the system of linear equations, we can use the Gaussian elimination method. Let's represent the augmented matrix for the system:
[A|B] =
| 0 2 4 2 | -5 |
| 0 5 10 5 | -15 |
| 1 1 2 -3 | 3 |
To simplify the matrix, we will perform row operations:
R2 = R2 - 2R1
R3 = R3 - R1
The updated matrix is:
[A|B] =
| 0 2 4 2 | -5 |
| 0 1 2 1 | -5 |
| 1 1 2 -3 | 3 |
Next, we'll normalize R2 by dividing it by 1:
R2 = R2/1, which simplifies to:
R2 = R2
The updated matrix is:
[A|B] =
| 0 2 4 2 | -5 |
| 0 1 2 1 | -5 |
| 1 1 2 -3 | 3 |
Now, we'll perform the following row operations:
R1 = R1 - 2R2
R3 = R3 - R2
The updated matrix is:
[A|B] =
| 0 0 0 0 | 5 |
| 0 1 2 1 | -5 |
| 1 0 0 -4 | 8 |
Now, we can see that the first row consists of all zeros on the left-hand side. This implies that the system is inconsistent and has no solution.
Therefore, the row-echelon form of the augmented matrix for the system is:
[A|B] =
| 0 0 0 0 | 5 |
| 0 1 2 1 | -5 |
| 1 0 0 -4 | 8 |
To solve the system of linear equations, we can use the method of Gaussian elimination to get the row-echelon form of the augmented matrix. Let's begin:
Step 1: Create the augmented matrix for the system:
[ 2 4 2 | -5 ]
[ 5 10 5 | -15 ]
[ 1 2 2 | 3 ]
Step 2: Apply row operations to get zeros below the leading coefficient in the first column:
R2 = R2 - (5/2)R1
R3 = R3 - (1/2)R1
The updated matrix becomes:
[ 2 4 2 | -5 ]
[ 0 2 1 | 0 ]
[ 0 1 1 | 4 ]
Step 3: Apply row operations to get zeros below the leading coefficient in the second column:
R3 = R3 - (1/2)R2
The updated matrix becomes:
[ 2 4 2 | -5 ]
[ 0 2 1 | 0 ]
[ 0 0 1/2 | 4 ]
Step 4: Multiply equation 3 by 2 to clear the fraction:
R3 = 2R3
The updated matrix becomes:
[ 2 4 2 | -5 ]
[ 0 2 1 | 0 ]
[ 0 0 1 | 8 ]
Step 5: Apply row operations to get zeros above the leading coefficient in the third column:
R1 = R1 - 2R3
R2 = R2 - R3
The updated matrix becomes:
[ 2 4 0 | -21 ]
[ 0 2 0 | -8 ]
[ 0 0 1 | 8 ]
Step 6: Apply row operations to get the reduced row-echelon form:
R1 = R1 - 2R2
The updated matrix becomes:
[ 2 0 0 | 5 ]
[ 0 2 0 | -8 ]
[ 0 0 1 | 8 ]
Now we have obtained the row-echelon form of the augmented matrix.
Looking at the augmented matrix, we can see that there are no rows of the form [0 0 0 | c], where c is a non-zero constant, in the matrix. This means that we do not have an inconsistent system, and thus, there is no row of all zeros to the left of the vertical line. Therefore, the system of equations has a unique solution.
To find the values of x1, x2, x3, and x4:
From the row-echelon form, we can write the system of equations:
2x1 = 5 => x1 = 5/2
2x2 = -8 => x2 = -4
x3 = 8
(there is no equation for x4, so it can take any value)
Therefore, the solution to the system of linear equations is:
x1 = 5/2, x2 = -4, x3 = 8, and x4 can take any value.