Consider the system of equations:
x1 + x2 + x3 = 6,
−x1 − 2x2 + 3x3 = 1,
3x1 − 4x2 + 4x3 = 5.
(a) Write down the augmented matrix for this system
(b) Use elementary row operations to reduced the augmented matrix to reduced row-ecehelon form
FOR A: Ive got:
1 1 1 | 6
-1 -2 3 | 1
3 -4 4 | 5
I need help with B thanks.
1 0 0 41 /27
0 1 0 59/27
0 0 1 62 /27
How did you get that though
1 1 1 | 6
-1 -2 3 | 1
3 -4 4 | 5
1 1 1 | 6
0 -1 4 | 7 from: add #1 and #2
0 7 -1 | 13 from #3 x 1 - #3
1 1 1 | 6
0 -1 4 | 7
0 0 27| 62 from #2 x 7 + #3
1 1 1 6
0 1 -4 -7 switch signs
0 0 1 62/27 divide by 27
so x3 = 62/27
now back substitute
x2 - 4(62/27) = -7
x2 = 248/27 -7 = 59/27
x1 + 59/27 + 62/27 = 6
x1 = 41/27
x1=41/27 , x2=59/27, x3=62/27
To solve the system of equations, we can perform elementary row operations on the augmented matrix to bring it to reduced row-echelon form. Here are the steps to do that:
1. Start with the given augmented matrix:
```
1 1 1 | 6
-1 -2 3 | 1
3 -4 4 | 5
```
2. We want to make the leading entry (the leftmost entry) of the second row to be equal to 1. To do this, we can multiply the second row by -1:
```
1 1 1 | 6
1 2 -3 | -1
3 -4 4 | 5
```
3. Now, we want to eliminate the 1 below the leading entry of the second row. To do this, we can add the first row to the second row:
```
1 1 1 | 6
0 3 -2 | 5
3 -4 4 | 5
```
4. Next, we want to make the leading entry of the third row equal to 1. We can achieve this by dividing the third row by 3:
```
1 1 1 | 6
0 3 -2 | 5
1 -4 4/3 | 5/3
```
5. Now, we want to eliminate the 1 below the leading entry of the third row. We can do this by subtracting the first row from the third row:
```
1 1 1 | 6
0 3 -2 | 5
0 -5 1/3 | -1/3
```
6. Lastly, we want to eliminate the -5 below the leading entry of the third row. We can accomplish this by adding 5 times the second row to the third row:
```
1 1 1 | 6
0 3 -2 | 5
0 0 1 | 0
```
This is the reduced row-echelon form of the augmented matrix.
To interpret this form, we can now write down the corresponding system of equations:
```
x1 + x2 + x3 = 6
3x2 - 2x3 = 5
x3 = 0
```
From the third equation, we know that x3 = 0. Substituting this into the second equation, we have:
3x2 - 2(0) = 5
3x2 = 5
x2 = 5/3
Finally, substituting the values of x2 and x3 into the first equation, we have:
x1 + 5/3 + 0 = 6
x1 = 6 - 5/3
x1 = 13/3
So the solution to the system of equations is x1 = 13/3, x2 = 5/3, and x3 = 0.