How Do I Proceed with Variable Row Reduction?

[stoffer]

Can anyone help me with this. I am asked to solve the following matrix with Row Red. Algorithm,

[ a 3 (1+a) | -a ]
[ 1 a 3 | -2a]
[ 1 1 1 | 0 ]

I get to something like

[ 1 1 1 | 0 ]
[ 0 a-1 2 | -2a]
[ 0 3-a 1 | -a]

but what do I do now?

 

[matt grime]

if a isn't 1 then you can divide row two by a-1. (if a=1...)

 

[M98Ranger]

Sure, I can definitely help you with this problem! It looks like you're on the right track with your row reduction so far. The next step would be to use row operations to get the matrix into reduced row echelon form, where all leading coefficients are 1 and all other entries in the same column are 0. Here's how you can do that:

1. Start with the first row and use row operations to make the first entry (a) equal to 1. You can do this by dividing the first row by a.

2. Next, use row operations to make the second entry in the second row (a-1) equal to 0. You can do this by adding the first row to the second row.

3. Similarly, use row operations to make the second entry in the third row (3-a) equal to 0. You can do this by adding the first row to the third row.

4. Now, focus on the second row and use row operations to make the second entry (2) equal to 1. You can do this by dividing the second row by (a-1).

5. Finally, use row operations to make the third entry in the third row (1) equal to 0. You can do this by subtracting 2 times the second row from the third row.

Your final matrix should look like this:

[ 1 0 0 | 0 ]
[ 0 1 0 | 0 ]
[ 0 0 1 | a ]

This means that the solution to this system of equations is a=0. You can plug this value back into the original equations to verify that it works. I hope this helps! Let me know if you have any other questions.