solve the system of equation for all variables using a matix
x-y-z=1
2x-3y+z=10
x+y-2z=0
form the augmented matrix
1 -1 -1 1
2 -3 1 0
1 2 -2 0
=
1 -1 -1 1
0 1 -3 -8 I did 2x#1 - #2
0 -2 1 1 I did #1 - #1
=
1 -1 -1 1
0 1 -3 -8
0 0 -5 -15 I did 2x#2 - #3
=
1 -1 -1 1
0 1 -3 -8
0 0 1 3
At this point you can work backwards since the last line says
z = 3
put that back into line 2, which says
y - 3z = -8
etc
Another way:
If you know how to find the inverse of a 3x3 matrix, find it, then multiply that inverse matrix by the column matrix
1
10
0
here is how to find the inverse of a 3x3 matrix, this is the method I use.
http://www.wikihow.com/Inverse-a-3X3-Matrix
To solve the system of equations using matrices, we can represent the system as an augmented matrix and then use matrix operations to find the solution.
Let's represent the system of equations in matrix form:
```
[ 1 -1 -1 | 1 ]
[ 2 -3 1 | 10 ]
[ 1 1 -2 | 0 ]
```
To solve the system, we can perform row operations on the matrix until we obtain a row-echelon form or a reduced row-echelon form.
Row operations include:
1. Scaling a row by a constant
2. Swapping two rows
3. Adding or subtracting a multiple of one row from another
Let's use these row operations step-by-step:
**Step 1:** Divide row 1 by 1 to get a leading coefficient of 1 in the first row:
```
[ 1 -1 -1 | 1 ]
[ 2 -3 1 | 10 ]
[ 1 1 -2 | 0 ]
```
**Step 2:** Add -2 times row 1 to row 2:
```
[ 1 -1 -1 | 1 ]
[ 0 1 3 | 8 ]
[ 1 1 -2 | 0 ]
```
**Step 3:** Add -1 times row 1 to row 3:
```
[ 1 -1 -1 | 1 ]
[ 0 1 3 | 8 ]
[ 0 2 1 | -1 ]
```
**Step 4:** Divide row 2 by 1 to get a leading coefficient of 1 in the second row:
```
[ 1 -1 -1 | 1 ]
[ 0 1 3 | 8 ]
[ 0 2 1 | -1 ]
```
**Step 5:** Add -2 times row 2 to row 3:
```
[ 1 -1 -1 | 1 ]
[ 0 1 3 | 8 ]
[ 0 0 -5 | -17 ]
```
**Step 6:** Divide row 3 by -5 to get a leading coefficient of 1 in the third row:
```
[ 1 -1 -1 | 1 ]
[ 0 1 3 | 8 ]
[ 0 0 1 | 17/5 ]
```
**Step 7:** Subtract 3 times row 3 from row 2:
```
[ 1 -1 -1 | 1 ]
[ 0 1 0 | 3/5 ]
[ 0 0 1 | 17/5 ]
```
**Step 8:** Add row 1 to row 2:
```
[ 1 -1 -1 | 1 ]
[ 1 0 -1 | 2/5 ]
[ 0 0 1 | 17/5 ]
```
**Step 9:** Add row 1 to row 3:
```
[ 1 -1 0 | 6/5 ]
[ 1 0 -1 | 2/5 ]
[ 0 0 1 | 17/5 ]
```
**Step 10:** Add 1 times row 3 to row 1:
```
[ 1 -1 0 | 6/5 ]
[ 1 0 -1 | 2/5 ]
[ 1 -1 1 | 17/5 ]
```
Now, we have reduced the augmented matrix to a form where we can easily read the solutions directly from the rightmost column:
```
x = 6/5
y = 2/5
z = 17/5
```
Therefore, the solution to the system of equations is x = 6/5, y = 2/5, and z = 17/5.