# Application - Solving System of Equations Using Matrices

## Application - Solving System of Equations Using Matrices - How it Works - Video

### First Steps

First Steps:

Our first steps is with the system as a coefficient matrix. Then our next step is add our constant terms as another column to form an augmented matrix. Now we have to do some row operations to make our coefficient part of the augmented matrix into an identity matrix.

### Example 1 - Step 1

Example 1 - Step 1:

Here we had 2 * R1 - R2 to replace R2 and the reason is that we want all the numbers beneath a1,1 or 1 in this case to be 0.

The main number that we are focusing on here is the 2 underneath the 1. We multiplied the first row by 2 so that we can get 2, -4, 6, 8. So we get 0, -5, 10, 5 for our new R2. Our main focus is 2 - 2 = 0 for the element a2,1.

If we get any other 1s or 0s, then that is great. If not, more arithmetic.

### Example 1 - Step 2

Example 1 - Step 2:

Here we had 3 * R1 + R3 to replace R3 and the reason is that we want all the numbers beneath a1,1 or 1 in this case to be 0.

The main number that we are focusing on here is the -3 underneath the 1. We multiplied the first row by 3 so that we can get 3, -6, 9, 12. So we get 0, -2, 10, 14 for our new R3. Our main focus is 3 + (-3) = 0 for the element a3,1.

If we get any other 1s or 0s, then that is great. If not, more arithmetic.

### Example 1 - Step 3

Example 1 - Step 3:

Here we had (-1/5) * R2 to replace R2 and the reason is that we want the first number to be 1 or a2,2.

The main number that we are focusing on here is the -5 underneath the -2. We multiplied the second row by (-1/5) so that we can get 0, 1, -2, -1, which is our new R2. Our main focus is (-1/5) * -5 = 1 for the element a2,2.

Since all of the elements in this row are multiples of 5, we don't get any fractions. Otherwise I would wait until all the numbers not in the principal diagonal are 0. Then multiply by a scalar to get 1.

If we get any other 1s or 0s, then that is great. If not, more arithmetic.

### Example 1 - Step 4

Example 1 - Step 4:

Here we had (-1/2) * R3 to replace R3 and the reason is that we want the first number to be 1 or a3,2.

The main number that we are focusing on here is the -2 underneath the 1. We multiplied the third row by (-1/2) so that we can get 0, 1, -5, -7, which is our new R3. Our main focus is (-1/2) * -2 = 1 for the element a3,2.

Since all of the elements in this row are multiples of 2, we don't get any fractions. Otherwise I would wait until all the numbers not in the principal diagonal are 0. Then multiply by a scalar to get 1.

If we get any other 1s or 0s, then that is great. If not, more arithmetic.

### Example 1 - Step 5

Example 1 - Step 5:

Here we had R2 - R3 to replace R3 and the reason is that we want all the numbers beneath a2,2 or 1 in this case to be 0.

The main number that we are focusing on here is the 1 underneath the 1. We subtract the third row from the second row so that we can get 0, 0, 3, 6 which is our new R3. Our main focus is 1 - 1 = 0 for the element a3,2.

If we get any other 1s or 0s, then that is great. If not, more arithmetic.

### Example 1 - Step 6

Example 1 - Step 6:

Here we had (1/3) * R3 to replace R3 and the reason is that we want the first number to be 1 or a3,3.

The main number that we are focusing on here is the 3 underneath the -2. We multiplied the third row by (1/3) so that we can get 0, 0, 1, 2, which is our new R3. Our main focus is (1/3) * 3 = 1 for the element a3,3.

Since all of the elements in this row are multiples of 2, we don't get any fractions. Otherwise I would wait until all the numbers not in the principal diagonal are 0. Then multiply by a scalar to get 1.

If we get any other 1s or 0s, then that is great. If not, more arithmetic.

### Example 1 - Answer with Substitution

Example 1 - Answer with Substitution:

Since we row echelon here, we can use substitution to solve for the variables. The last row is z = 2. Once we substitute that back into for the second row, we get y = 3. Finally, we substitute y and z to find x and we get x = 4.

Weirdly the answers are the same as the constant that rarely happens, but on occasion math is cool like that.

### Example 1 - Step 7

Example 1 - Step 7:

Here we had R1 + 2 * R2 to replace R1 and the reason is that we want all the numbers above a2,2 or 1 in this case to be 0.

The main number that we are focusing on here is the -2 above the 1. We multiply the second row to get 1, -2, 3, 4. Now we add the first row and 2 times the second row. So we get 1, 0, -1, 2 for our new R1. Our main focus is (-2) + 2 = 0 for the element a1,2.

If we get any other 1s or 0s, then that is great. If not, more arithmetic.

### Example 1 - Step 8

Example 1 - Step 8:

Here we had R1 + R2 to replace R1 and the reason is that we want all the numbers above a3,3 or 1 in this case to be 0.

The main number that we are focusing on here is the -1 above the -2. We add the first row and the third row so that we can get 1, 0, 0, 4 which is our new R1. Our main focus is (-1) + 1 = 0 for the element a1,3.

If we get any other 1s or 0s, then that is great. If not, more arithmetic.

### Example 1 - Step 9

Example 1 - Step 9:

Here we had R2 + 2 * R3 to replace R2 and the reason is that we want all the numbers above a3,3 or 1 in this case to be 0.

The main number that we are focusing on here is the -2 above the 1. We multiply the third row by 2 to get 0, 0, 2, 4. Now we add the second row and two times the third row so that we can get 1, 0, 0, 4 which is our new R1. Our main focus is (-2) + 2 = 0 for the element a2,3.

If we get any other 1s or 0s, then that is great. If not, more arithmetic.

### Example 1 - Answer with Gauss-Jordan

Example 1 - Answer with Gauss-Jordan:

After all of that, we found the answers for each variable; x = 2; y = 3; and z = 2.

Weirdly the answers are the same as the constant that rarely happens, but on occasion math is cool like that.