# Matrix operations

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1. Aug 19, 2016

### Mark53

1. The problem statement, all variables and given/known data

\begin{array}{cc}1 & 1&1\\ 1&1-s&1-s\\-s&1-s&s^2-1\end{array}

a)For which values of s does the inverse exist, and why? You should be using row operations and ideally head for reduced row echelon form

b) In the process of calculating part a), you will come across a row operation that will not work for all values of s, and in fact these values of s make A a non-invertible matrix. Write down the elementary matrix for this row operation.

3. The attempt at a solution

From doing row operations I get to

\begin{array}{cc}1 & 1&1\\ 0&-s&-s\\0&0&s^2-1\end{array}

I am unsure how to show for what values of s the inverse exists

2. Aug 19, 2016

### Staff: Mentor

Will the original matrix be invertible if you end up with a RREF matrix with a row of zeros?

Note that I didn't check your work.

3. Aug 19, 2016

### Mark53

does that mean that when s is less than -1 and greater than 1 the inverse will exist

for part b would that just mean that s cannot equal -1,0,1

4. Aug 19, 2016

### SammyS

Staff Emeritus
I do not get that for a row echelon form.

5. Aug 19, 2016

### Ray Vickson

I agree with SammyS that your echelon form is not correct.

However, more to the point: you need to show the steps you took to get the final form, because for some values of $s$ you may not even be allowed to proceed as far as you did. In other words, for some values of $s$ you may need to stop before reaching your echelon form.!

6. Aug 19, 2016

### Mark53

This is how I came to the answer

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7. Aug 20, 2016

### SammyS

Staff Emeritus

That doesn't quite match what you gave initially.

$\left(\begin{array}{cc}1 & 1&1\\ 1&1-s&1-s\\-s&1-s&s^2-1\end{array} \right)$

Look at entry 3, 3 .

8. Aug 20, 2016

### Mark53

Sorry I made a mistake earlier entry 3,3 should be s^2-s

9. Aug 20, 2016

### Staff: Mentor

In one of your steps you are dividing row 2 by s, and adding it to the 3rd row. If s happens to be zero, this step is invalid.

10. Aug 20, 2016

### Mark53

Does this mean the elementary matrix for this row operation is:

$\left(\begin{array}{cc}1 & 0&0\\ 0&1&0\\0&1/s&1\end{array} \right)$

11. Aug 20, 2016

### Staff: Mentor

Not if s = 0.
Notice that if s = 0, the determinant of your original matrix is 0. Since for this value of s, the determinant is zero, the matrix is not invertible for that value. There are also two other values of s for which the matrix is not invertible. One of the values you listed is incorrect.

12. Aug 20, 2016

### Mark53

when s = -1 or 1 the matrix would also not be invertible

part b says In the process of calculating part a), you will come across a row operation that will not work for all values of s, and in fact these values of s make A a non-invertible matrix. Write down the elementary matrix for this row operation.

How would i write down the elementary matrix for the row operation?

13. Aug 20, 2016

### Staff: Mentor

I believe you for one of these values, but not the other.
The problem comes from replacing a row by 1/s times itself. What does the elementary matrix for this row operation look like?

14. Aug 20, 2016

### Mark53

\begin{array}{cc}1 & 0&0\\ 0&1&0\\0&0&1/s\end{array}

Is this correct?

which row should I be doing the operation to?

15. Aug 20, 2016

### Ray Vickson

You cannot divide by s if s = 0, end of story. So, when s = 0 you are forced to stop at your third matrix above, which is
$$\pmatrix{1&1&1\\0&0&0\\0&1&0}$$
when s = 0. This form is already sufficient to allow you to answer the original question about the existence or non-existence of an inverse. In fact, when s = 0 you do not even need to perform any row operations at all to answer the original question.

Last edited: Aug 20, 2016
16. Aug 20, 2016

### SammyS

Staff Emeritus

I don't get this with either initial matrix.

17. Aug 20, 2016

### Mark53

$$\pmatrix{1&1&1\\0&0&0\\0&1&0}$$ How is this an elementary matrix I thought that only one row could be changed from the identity matrix to form an elementary matrix?

18. Aug 21, 2016

### Ray Vickson

Yes. but that is completely irrelevant here. The matrix above is NOT supposed to be a "transformation matrix" (like an elementary matrix or whatever); it is a matrix that results from the application of two elementary matrices acting on the original matrix to form a new matrix. If you had written out the original problem as three equations in three unknowns, the matrix above would correspond to the new system of equations resulting from the operations of Gaussian elimination.

Look at your own work displayed in post #7. Look at the first line, and the second of your modified matrices---the one at the far right on the first line. That is a matrix that you got as a result of some row operations. Now put s = 0 in that matrix.

19. Aug 21, 2016

### Mark53

thanks I got it now