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.