Use Gaussian elimination to write the system of equations given below in row-echelon form.
2x+4y+z=-4
2x-4y+6z=13
4x-2y+z=6
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To write the system of equations in row-echelon form using Gaussian elimination, follow these steps:
Step 1: Write the system of equations as an augmented matrix.
The augmented matrix for the given system of equations is:
[ 2 4 1 | -4 ]
[ 2 -4 6 | 13 ]
[ 4 -2 1 | 6 ]
Step 2: Perform row operations to transform the matrix into row-echelon form.
1. Start with the first row and make the first element (2) the leading coefficient by dividing the entire row by 2:
[ 1 2 1/2 | -2 ]
2. Next, eliminate the first element of the second row (2) by subtracting the first row multiplied by 1 from the second row:
[ 1 2 1/2 | -2 ]
[ 0 -6 5 | 15 ]
3. Similarly, eliminate the first element of the third row (4) by subtracting the first row multiplied by 2 from the third row:
[ 1 2 1/2 | -2 ]
[ 0 -6 5 | 15 ]
[ 0 -2 0 | 10 ]
4. Now, we need to make the second element of the second row (6) the leading coefficient. To do this, divide the entire second row by -6:
[ 1 2 1/2 | -2 ]
[ 0 1 -5/6 | -5/2 ]
[ 0 -2 0 | 10 ]
5. Next, eliminate the second element of the third row (-2) by adding the second row multiplied by 2 to the third row:
[ 1 2 1/2 | -2 ]
[ 0 1 -5/6 | -5/2 ]
[ 0 0 5/3 | 5 ]
Step 3: The matrix is now in row-echelon form. Convert it back to a system of equations:
1x + 2y + 1/2z = -2
1y - 5/6z = -5/2
5/3z = 5
Step 4: Solve the system of equations.
- Rewrite the third equation as z = 3.
- Substitute this value of z into the second equation to find y:
1y - 5/6(3) = -5/2
1y - 5/2 = -5/2
==> y = 0
- Substitute the values of y = 0 and z = 3 into the first equation to find x:
1x + 2(0) + 1/2(3) = -2
x + 3/2 = -2
==> x = -7/2
Therefore, the solution to the system of equations is x = -7/2, y = 0, and z = 3.