Matrix Operations: Inverse Existence & Row Op.

[Mark53]

Homework Statement

[/B]


\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.

The Attempt at a Solution

[/B]
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

 

[Mark44]

Homework Statement

[/B]


\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.

The Attempt at a Solution

[/B]
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

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.

 

[Mark53]

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.

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

 

[SammyS]

The Attempt at a Solution

[/B]
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

I do not get that for a row echelon form.

 

[Ray Vickson]

Homework Statement

[/B]


\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.

The Attempt at a Solution

[/B]
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

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.!

 

[Mark53]

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.!

I do not get that for a row echelon form.

This is how I came to the answer

 

[SammyS]

This is how I came to the answer

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 .

 

[Mark53]

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 .

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

 

[Mark44]

This is how I came to the answer

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.

 

[Mark53]

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.

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

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

 

[Mark44]

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

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

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.

 

[Mark53]

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.

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?

 

[Mark44]

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

I believe you for one of these values, but not the other.

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?

The problem comes from replacing a row by 1/s times itself. What does the elementary matrix for this row operation look like?

 

[Mark53]

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?

\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?

 

[Ray Vickson]

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 .

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.

 

[SammyS]

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

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

What matrix did you start with to get this?

I don't get this with either initial matrix.

Again, please show steps.

 

[Mark53]

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.

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

 

[Ray Vickson]

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

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.

 

[Mark53]

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.

thanks I got it now