Solve the system of equations using Row Operations
x + y - z = 6
3x - 2y + z = -5
x + 3y - 2z = 14
let rows be a,b,c
a+b>d
2b+a>e
d) 4x-y=1
e)7x-y=-4
d-e>f
f: -3x=5
check those.
To solve the system of equations using row operations, we can represent the system as an augmented matrix and perform row operations to transform it into row-echelon form.
Step 1: Represent the system as an augmented matrix
The augmented matrix can be formed by writing down the coefficients of the variables along with the constants on the right-hand side of each equation. The system of equations can be represented as follows:
1 1 -1 | 6
3 -2 1 | -5
1 3 -2 | 14
Step 2: Perform row operations to transform the augmented matrix into row-echelon form
To perform row operations, we can use the following three operations:
1. Swapping two rows.
2. Multiplying a row by a non-zero constant.
3. Adding or subtracting multiples of one row to/from another row.
We'll use these operations to transform the augmented matrix into row-echelon form.
First, let's create a zero underneath the 1 in the first column:
R2 - 3R1 -> R2
R3 - R1 -> R3
This operation means subtracting three times the first row from the second row and subtracting the first row from the third row.
The augmented matrix becomes:
1 1 -1 | 6
0 -5 4 | -23
0 2 -3 | 8
Next, let's create a zero underneath the 2 in the second column:
R2 + 2R3 -> R2
This operation means adding two times the third row to the second row.
The augmented matrix becomes:
1 1 -1 | 6
0 0 -2 | -7
0 2 -3 | 8
Finally, let's create a zero underneath the -3 in the third column:
2R3 + 3R2 -> R3
This operation means adding three times the second row to two times the third row.
The augmented matrix becomes:
1 1 -1 | 6
0 0 -2 | -7
0 0 -9 | -6
Step 3: Solve for the variables
Now that the augmented matrix is in row-echelon form, we can back-substitute starting from the bottom row. We can work our way up to find the values of the variables.
From the third row in the row-echelon form, we have:
-9z = -6
Solving for z, we find z = 2/3.
Using this value of z, we can substitute it back into the second row of the row-echelon form:
-2y - 2 = -7
Solving for y, we find y = 5/2.
Finally, using the values of y and z, we can substitute them back into the first row of the row-echelon form:
x + 5/2 - 2/3 = 6
Simplifying the equation, we find x = 47/6.
Therefore, the solution to the system of equations is x = 47/6, y = 5/2, and z = 2/3.