Finding the inverse of a matrix; I disagree with professor's answer

[ksdhart2]

Hi all, I have a question from my Linear Algebra course. I successfully solved the problem, but the answer I got is not the same as my professor's. I believe my answer is correct and his is wrong. The problem is as follows:

4) Find the inverse of F (or explain why it does not exist):

\(\displaystyle F=\begin{pmatrix}1-c&0&c\\ -c&1-c&2c\\ -c&-c&2c+1\end{pmatrix}\)

Hint: F is a solution of p(x) = 0 for \(\displaystyle p(x)=x^3-3cx^2+3c^2x-c^3\)

I interpreted the hint as meaning that p(F) = 0, so that's the direction I started working in. I first calculated F³ and F² as follows:

\(\displaystyle F^2=\begin{pmatrix}1-2c&-c^2&c^2+2c\\ -2c&-c^2-2c+1&c^2+4c\\ -2c&-c^2-2c&c^2+4c+1\end{pmatrix}\)

\(\displaystyle F^3=\begin{pmatrix}1-3c&-3c^2&3c^2+3c\\ -3c&-3c^2-3c+1&3c^2+6c\\ -3c&-3c^2-3c&3c^2+6c+1\end{pmatrix}\)

From these, I then calculated 3cF^2 and 3c^2F:

\(\displaystyle 3cF^2=\begin{pmatrix}3c-6c^2&-3c^3&3c^3+6c^2\\ \:-6c^2&-3c^3-6c^2+3c&3c^3+12c^2\\ \:-6c^2&-3c^3-6c^2&3c^3+12c^2+3c\end{pmatrix}\)

\(\displaystyle 3c^2F=\begin{pmatrix}3c^2-3c^3&0&3c^3\\ \:\:\:\:\:\:\:-3c^3&3c^2-3c^3&6c^3\\ \:\:\:\:\:\:\:-3c^3&-3c^3&6c^3+3c^2\end{pmatrix}\)

I then crafted the polynomial:

\(\displaystyle F^3-3cF^2+3c^2F-c^3I=0\)

\(\displaystyle F^3-3cF^2+3c^2F+c^3I=\begin{pmatrix}1-6c+9c^2-4c^3&3c^3-3c^2&3c-3c^2\\ \:\:6c^2-3c^3-3c&6c^2-6c+1-c^3&3c^3+6c-9c^2\\ \:\:6c^2-3c^3-3c&3c^2-3c&2c^3-6c^2+3c+1\end{pmatrix}\)

Factoring each of these expressions and setting equal to the 3x3 zero matrix gives:

\(\displaystyle \begin{pmatrix}-\left(c-1\right)^2\left(4c-1\right)&3c^2\left(c-1\right)&-3c\left(c-1\right)\\ \:\:-3c\left(c-1\right)^2&-\left(c-1\right)\left(c^2-5c+1\right)&3c\left(c-1\right)\left(c-2\right)\\ \:\:-3c\left(c-1\right)^2&3c^2\left(c-1\right)&\left(c-1\right)\left(2c^2-4c-1\right)\end{pmatrix}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\)

That gives me nine equations, all of which are equal to 0 and share a common factor of (c-1). Therefore, I concluded that c = 1. Plugging that in to the original F matrix gives:

\(\displaystyle \begin{pmatrix}0&0&1\\ -1&0&2\\ -1&-1&3\end{pmatrix}\)

From there, it was a simple matter of calculating the inverse, giving me the final answer of:

\(\displaystyle \begin{pmatrix}0&0&1\\ \:-1&0&2\\ \:-1&-1&3\end{pmatrix}^{-1}=\begin{pmatrix}2&-1&0\\ 1&1&-1\\ 1&0&0\end{pmatrix}\)

-----

What my professor did is to rewrite the polynomial p(x):

\(\displaystyle F^3-3cF^2+3c^2F-c^3I=0\)

\(\displaystyle F^3-3cF^2+3c^2F=c^3I\)

\(\displaystyle F(F^2-3cF+3c^2I)=c^3I\)

\(\displaystyle F \left( \dfrac{1}{c^3}(F^2-3cF+3c^2I) \right)=I\)

Then, because \(\displaystyle FF^{-1}=I\), he concludes that \(\displaystyle F^{-1}=\dfrac{1}{c^3}(F^2-3cF+3c^2I)\)

...For brevity's sake, I'll omit the remaining calculations...

His final answer is thus:

\(\displaystyle F^{-1}=\begin{pmatrix}1-5c+6c^2&-c^2&2c-2c^2\\ -2c+3c^2&1-5c+5c^2&4c-5c^2\\ -2c+3c^2&-2c+2c^2&1+c-2c^2\end{pmatrix}\)

I note that if I let c=1, then our solutions match... we used different processes but we arrived at the same solution - all I did was realize that since I was given p(F)=0, c must be 1. When I got back the homework assignment, the part where I factor the expressions and arrive at c = 1 was marked with a large ? and the entire problem was marked wrong. Am I missing something? Or am I right that the professor made a mistake?

 

[Steven G]

Hi all, I have a question from my Linear Algebra course. I successfully solved the problem, but the answer I got is not the same as my professor's. I believe my answer is correct and his is wrong. The problem is as follows:



I interpreted the hint as meaning that p(F) = 0, so that's the direction I started working in. I first calculated F³ and F² as follows:

\(\displaystyle F^2=\begin{pmatrix}1-2c&-c^2&c^2+2c\\ -2c&-c^2-2c+1&c^2+4c\\ -2c&-c^2-2c&c^2+4c+1\end{pmatrix}\)

\(\displaystyle F^3=\begin{pmatrix}1-3c&-3c^2&3c^2+3c\\ -3c&-3c^2-3c+1&3c^2+6c\\ -3c&-3c^2-3c&3c^2+6c+1\end{pmatrix}\)

From these, I then calculated 3cF^2 and 3c^2F:

\(\displaystyle 3cF^2=\begin{pmatrix}3c-6c^2&-3c^3&3c^3+6c^2\\ \:-6c^2&-3c^3-6c^2+3c&3c^3+12c^2\\ \:-6c^2&-3c^3-6c^2&3c^3+12c^2+3c\end{pmatrix}\)

\(\displaystyle 3c^2F=\begin{pmatrix}3c^2-3c^3&0&3c^3\\ \:\:\:\:\:\:\:-3c^3&3c^2-3c^3&6c^3\\ \:\:\:\:\:\:\:-3c^3&-3c^3&6c^3+3c^2\end{pmatrix}\)

I then crafted the polynomial:

\(\displaystyle F^3-3cF^2+3c^2F-c^3I=0\)

\(\displaystyle F^3-3cF^2+3c^2F+c^3I=\begin{pmatrix}1-6c+9c^2-4c^3&3c^3-3c^2&3c-3c^2\\ \:\:6c^2-3c^3-3c&6c^2-6c+1-c^3&3c^3+6c-9c^2\\ \:\:6c^2-3c^3-3c&3c^2-3c&2c^3-6c^2+3c+1\end{pmatrix}\)

Factoring each of these expressions and setting equal to the 3x3 zero matrix gives:

\(\displaystyle \begin{pmatrix}-\left(c-1\right)^2\left(4c-1\right)&3c^2\left(c-1\right)&-3c\left(c-1\right)\\ \:\:-3c\left(c-1\right)^2&-\left(c-1\right)\left(c^2-5c+1\right)&3c\left(c-1\right)\left(c-2\right)\\ \:\:-3c\left(c-1\right)^2&3c^2\left(c-1\right)&\left(c-1\right)\left(2c^2-4c-1\right)\end{pmatrix}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\)

That gives me nine equations, all of which are equal to 0 and share a common factor of (c-1). Therefore, I concluded that c = 1. Plugging that in to the original F matrix gives:

\(\displaystyle \begin{pmatrix}0&0&1\\ -1&0&2\\ -1&-1&3\end{pmatrix}\)

From there, it was a simple matter of calculating the inverse, giving me the final answer of:

\(\displaystyle \begin{pmatrix}0&0&1\\ \:-1&0&2\\ \:-1&-1&3\end{pmatrix}^{-1}=\begin{pmatrix}2&-1&0\\ 1&1&-1\\ 1&0&0\end{pmatrix}\)

-----

What my professor did is to rewrite the polynomial p(x):

\(\displaystyle F^3-3cF^2+3c^2F-c^3I=0\)

\(\displaystyle F^3-3cF^2+3c^2F=c^3I\)

\(\displaystyle F(F^2-3cF+3c^2I)=c^3I\)

\(\displaystyle F \left( \dfrac{1}{c^3}(F^2-3cF+3c^2I) \right)=I\)

Then, because \(\displaystyle FF^{-1}=I\), he concludes that \(\displaystyle F^{-1}=\dfrac{1}{c^3}(F^2-3cF+3c^2I)\)

...For brevity's sake, I'll omit the remaining calculations...

His final answer is thus:

\(\displaystyle F^{-1}=\begin{pmatrix}1-5c+6c^2&-c^2&2c-2c^2\\ -2c+3c^2&1-5c+5c^2&4c-5c^2\\ -2c+3c^2&-2c+2c^2&1+c-2c^2\end{pmatrix}\)

I note that if I let c=1, then our solutions match... we used different processes but we arrived at the same solution - all I did was realize that since I was given p(F)=0, c must be 1. When I got back the homework assignment, the part where I factor the expressions and arrive at c = 1 was marked with a large ? and the entire problem was marked wrong. Am I missing something? Or am I right that the professor made a mistake?

Someone has a small typo. Your teachers result for F^(-1) should be multiplied by 1/c^3.

If you let c=2 then the product is not I, so you are correct and the teacher is wrong! Now where did the teacher go wrong?

 

[Steven G]

Hi all, I have a question from my Linear Algebra course. I successfully solved the problem, but the answer I got is not the same as my professor's. I believe my answer is correct and his is wrong. The problem is as follows:



I interpreted the hint as meaning that p(F) = 0, so that's the direction I started working in. I first calculated F³ and F² as follows:

\(\displaystyle F^2=\begin{pmatrix}1-2c&-c^2&c^2+2c\\ -2c&-c^2-2c+1&c^2+4c\\ -2c&-c^2-2c&c^2+4c+1\end{pmatrix}\)

\(\displaystyle F^3=\begin{pmatrix}1-3c&-3c^2&3c^2+3c\\ -3c&-3c^2-3c+1&3c^2+6c\\ -3c&-3c^2-3c&3c^2+6c+1\end{pmatrix}\)

From these, I then calculated 3cF^2 and 3c^2F:

\(\displaystyle 3cF^2=\begin{pmatrix}3c-6c^2&-3c^3&3c^3+6c^2\\ \:-6c^2&-3c^3-6c^2+3c&3c^3+12c^2\\ \:-6c^2&-3c^3-6c^2&3c^3+12c^2+3c\end{pmatrix}\)

\(\displaystyle 3c^2F=\begin{pmatrix}3c^2-3c^3&0&3c^3\\ \:\:\:\:\:\:\:-3c^3&3c^2-3c^3&6c^3\\ \:\:\:\:\:\:\:-3c^3&-3c^3&6c^3+3c^2\end{pmatrix}\)

I then crafted the polynomial:

\(\displaystyle F^3-3cF^2+3c^2F-c^3I=0\)

\(\displaystyle F^3-3cF^2+3c^2F+c^3I=\begin{pmatrix}1-6c+9c^2-4c^3&3c^3-3c^2&3c-3c^2\\ \:\:6c^2-3c^3-3c&6c^2-6c+1-c^3&3c^3+6c-9c^2\\ \:\:6c^2-3c^3-3c&3c^2-3c&2c^3-6c^2+3c+1\end{pmatrix}\)

Factoring each of these expressions and setting equal to the 3x3 zero matrix gives:

\(\displaystyle \begin{pmatrix}-\left(c-1\right)^2\left(4c-1\right)&3c^2\left(c-1\right)&-3c\left(c-1\right)\\ \:\:-3c\left(c-1\right)^2&-\left(c-1\right)\left(c^2-5c+1\right)&3c\left(c-1\right)\left(c-2\right)\\ \:\:-3c\left(c-1\right)^2&3c^2\left(c-1\right)&\left(c-1\right)\left(2c^2-4c-1\right)\end{pmatrix}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\)

That gives me nine equations, all of which are equal to 0 and share a common factor of (c-1). Therefore, I concluded that c = 1. Plugging that in to the original F matrix gives:

\(\displaystyle \begin{pmatrix}0&0&1\\ -1&0&2\\ -1&-1&3\end{pmatrix}\)

From there, it was a simple matter of calculating the inverse, giving me the final answer of:

\(\displaystyle \begin{pmatrix}0&0&1\\ \:-1&0&2\\ \:-1&-1&3\end{pmatrix}^{-1}=\begin{pmatrix}2&-1&0\\ 1&1&-1\\ 1&0&0\end{pmatrix}\)

-----

What my professor did is to rewrite the polynomial p(x):

\(\displaystyle F^3-3cF^2+3c^2F-c^3I=0\)

\(\displaystyle F^3-3cF^2+3c^2F=c^3I\)

\(\displaystyle F(F^2-3cF+3c^2I)=c^3I\)

\(\displaystyle F \left( \dfrac{1}{c^3}(F^2-3cF+3c^2I) \right)=I\)

Then, because \(\displaystyle FF^{-1}=I\), he concludes that \(\displaystyle F^{-1}=\dfrac{1}{c^3}(F^2-3cF+3c^2I)\)

...For brevity's sake, I'll omit the remaining calculations...

His final answer is thus:

\(\displaystyle F^{-1}=\begin{pmatrix}1-5c+6c^2&-c^2&2c-2c^2\\ -2c+3c^2&1-5c+5c^2&4c-5c^2\\ -2c+3c^2&-2c+2c^2&1+c-2c^2\end{pmatrix}\)

I note that if I let c=1, then our solutions match... we used different processes but we arrived at the same solution - all I did was realize that since I was given p(F)=0, c must be 1. When I got back the homework assignment, the part where I factor the expressions and arrive at c = 1 was marked with a large ? and the entire problem was marked wrong. Am I missing something? Or am I right that the professor made a mistake?

I finally got a handle on this problem. The statement that P(F) = 0 should be an identity but it is a conditional statement.
Now if it was an identity statement your instructor, in my opinion, did it the best way but your method works quite well.

You can easily convince your instructor that there is a problem by saying that it is not an identity which should get him/her interested in what you're saying. Simply showing your instructor that the product of F and it's 'inverse' is not I for c=2 will get you your test points back. The reason your instructor marked you wrong was obviously because the teacher did not realize the equality was conditionally.

**** you rattle my brains for awhile but in the end I got it!!

Good job!

 

[ksdhart2]

Hm. Interesting. So, the "given" equation simply wasn't true, except for one specific value of c? I guess that could explain why I found that c has to be 1. I'll confer with my professor later today and see what he has to say about this.

 

[pka]

Hi all, I have a question from my Linear Algebra course. I successfully solved the problem, but the answer I got is not the same as my professor's. I believe my answer is correct and his is wrong. The problem is as follows:I interpreted the hint as meaning that p(F) = 0, so that's the direction I started working in. I first calculated F³ and F² as follows:

\(\displaystyle F^2=\begin{pmatrix}1-2c&-c^2&c^2+2c\\ -2c&-c^2-2c+1&c^2+4c\\ -2c&-c^2-2c&c^2+4c+1\end{pmatrix}\)

\(\displaystyle F^3=\begin{pmatrix}1-3c&-3c^2&3c^2+3c\\ -3c&-3c^2-3c+1&3c^2+6c\\ -3c&-3c^2-3c&3c^2+6c+1\end{pmatrix}\)

From these, I then calculated 3cF^2 and 3c^2F:

\(\displaystyle 3cF^2=\begin{pmatrix}3c-6c^2&-3c^3&3c^3+6c^2\\ \:-6c^2&-3c^3-6c^2+3c&3c^3+12c^2\\ \:-6c^2&-3c^3-6c^2&3c^3+12c^2+3c\end{pmatrix}\)

\(\displaystyle 3c^2F=\begin{pmatrix}3c^2-3c^3&0&3c^3\\ \:\:\:\:\:\:\:-3c^3&3c^2-3c^3&6c^3\\ \:\:\:\:\:\:\:-3c^3&-3c^3&6c^3+3c^2\end{pmatrix}\)

I then crafted the polynomial:

\(\displaystyle F^3-3cF^2+3c^2F-c^3I=0\)

\(\displaystyle F^3-3cF^2+3c^2F+c^3I=\begin{pmatrix}1-6c+9c^2-4c^3&3c^3-3c^2&3c-3c^2\\ \:\:6c^2-3c^3-3c&6c^2-6c+1-c^3&3c^3+6c-9c^2\\ \:\:6c^2-3c^3-3c&3c^2-3c&2c^3-6c^2+3c+1\end{pmatrix}\)

Factoring each of these expressions and setting equal to the 3x3 zero matrix gives:

\(\displaystyle \begin{pmatrix}-\left(c-1\right)^2\left(4c-1\right)&3c^2\left(c-1\right)&-3c\left(c-1\right)\\ \:\:-3c\left(c-1\right)^2&-\left(c-1\right)\left(c^2-5c+1\right)&3c\left(c-1\right)\left(c-2\right)\\ \:\:-3c\left(c-1\right)^2&3c^2\left(c-1\right)&\left(c-1\right)\left(2c^2-4c-1\right)\end{pmatrix}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\)

That gives me nine equations, all of which are equal to 0 and share a common factor of (c-1). Therefore, I concluded that c = 1. Plugging that in to the original F matrix gives:

\(\displaystyle \begin{pmatrix}0&0&1\\ -1&0&2\\ -1&-1&3\end{pmatrix}\)

From there, it was a simple matter of calculating the inverse, giving me the final answer of:

\(\displaystyle \begin{pmatrix}0&0&1\\ \:-1&0&2\\ \:-1&-1&3\end{pmatrix}^{-1}=\begin{pmatrix}2&-1&0\\ 1&1&-1\\ 1&0&0\end{pmatrix}\)

-----

What my professor did is to rewrite the polynomial p(x):

\(\displaystyle F^3-3cF^2+3c^2F-c^3I=0\)

\(\displaystyle F^3-3cF^2+3c^2F=c^3I\)

\(\displaystyle F(F^2-3cF+3c^2I)=c^3I\)

\(\displaystyle F \left( \dfrac{1}{c^3}(F^2-3cF+3c^2I) \right)=I\)

Then, because \(\displaystyle FF^{-1}=I\), he concludes that \(\displaystyle F^{-1}=\dfrac{1}{c^3}(F^2-3cF+3c^2I)\)

...For brevity's sake, I'll omit the remaining calculations...

His final answer is thus:

\(\displaystyle F^{-1}=\begin{pmatrix}1-5c+6c^2&-c^2&2c-2c^2\\ -2c+3c^2&1-5c+5c^2&4c-5c^2\\ -2c+3c^2&-2c+2c^2&1+c-2c^2\end{pmatrix}\)

I note that if I let c=1, then our solutions match... we used different processes but we arrived at the same solution - all I did was realize that since I was given p(F)=0, c must be 1. When I got back the homework assignment, the part where I factor the expressions and arrive at c = 1 was marked with a large ? and the entire problem was marked wrong. Am I missing something? Or am I right that the professor made a mistake?

Have a look at this.

 

[Steven G]

Hi all, I have a question from my Linear Algebra course. I successfully solved the problem, but the answer I got is not the same as my professor's. I believe my answer is correct and his is wrong. The problem is as follows:



I interpreted the hint as meaning that p(F) = 0, so that's the direction I started working in. I first calculated F³ and F² as follows:

\(\displaystyle F^2=\begin{pmatrix}1-2c&-c^2&c^2+2c\\ -2c&-c^2-2c+1&c^2+4c\\ -2c&-c^2-2c&c^2+4c+1\end{pmatrix}\)

\(\displaystyle F^3=\begin{pmatrix}1-3c&-3c^2&3c^2+3c\\ -3c&-3c^2-3c+1&3c^2+6c\\ -3c&-3c^2-3c&3c^2+6c+1\end{pmatrix}\)

From these, I then calculated 3cF^2 and 3c^2F:

\(\displaystyle 3cF^2=\begin{pmatrix}3c-6c^2&-3c^3&3c^3+6c^2\\ \:-6c^2&-3c^3-6c^2+3c&3c^3+12c^2\\ \:-6c^2&-3c^3-6c^2&3c^3+12c^2+3c\end{pmatrix}\)

\(\displaystyle 3c^2F=\begin{pmatrix}3c^2-3c^3&0&3c^3\\ \:\:\:\:\:\:\:-3c^3&3c^2-3c^3&6c^3\\ \:\:\:\:\:\:\:-3c^3&-3c^3&6c^3+3c^2\end{pmatrix}\)

I then crafted the polynomial:

\(\displaystyle F^3-3cF^2+3c^2F-c^3I=0\)

\(\displaystyle F^3-3cF^2+3c^2F+c^3I=\begin{pmatrix}1-6c+9c^2-4c^3&3c^3-3c^2&3c-3c^2\\ \:\:6c^2-3c^3-3c&6c^2-6c+1-c^3&3c^3+6c-9c^2\\ \:\:6c^2-3c^3-3c&3c^2-3c&2c^3-6c^2+3c+1\end{pmatrix}\)

Factoring each of these expressions and setting equal to the 3x3 zero matrix gives:

\(\displaystyle \begin{pmatrix}-\left(c-1\right)^2\left(4c-1\right)&3c^2\left(c-1\right)&-3c\left(c-1\right)\\ \:\:-3c\left(c-1\right)^2&-\left(c-1\right)\left(c^2-5c+1\right)&3c\left(c-1\right)\left(c-2\right)\\ \:\:-3c\left(c-1\right)^2&3c^2\left(c-1\right)&\left(c-1\right)\left(2c^2-4c-1\right)\end{pmatrix}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\)

That gives me nine equations, all of which are equal to 0 and share a common factor of (c-1). Therefore, I concluded that c = 1. Plugging that in to the original F matrix gives:

\(\displaystyle \begin{pmatrix}0&0&1\\ -1&0&2\\ -1&-1&3\end{pmatrix}\)

From there, it was a simple matter of calculating the inverse, giving me the final answer of:

\(\displaystyle \begin{pmatrix}0&0&1\\ \:-1&0&2\\ \:-1&-1&3\end{pmatrix}^{-1}=\begin{pmatrix}2&-1&0\\ 1&1&-1\\ 1&0&0\end{pmatrix}\)

-----

What my professor did is to rewrite the polynomial p(x):

\(\displaystyle F^3-3cF^2+3c^2F-c^3I=0\)

\(\displaystyle F^3-3cF^2+3c^2F=c^3I\)

\(\displaystyle F(F^2-3cF+3c^2I)=c^3I\)

\(\displaystyle F \left( \dfrac{1}{c^3}(F^2-3cF+3c^2I) \right)=I\)

Then, because \(\displaystyle FF^{-1}=I\), he concludes that \(\displaystyle F^{-1}=\dfrac{1}{c^3}(F^2-3cF+3c^2I)\)

...For brevity's sake, I'll omit the remaining calculations...

His final answer is thus:

\(\displaystyle F^{-1}=\begin{pmatrix}1-5c+6c^2&-c^2&2c-2c^2\\ -2c+3c^2&1-5c+5c^2&4c-5c^2\\ -2c+3c^2&-2c+2c^2&1+c-2c^2\end{pmatrix}\)

I note that if I let c=1, then our solutions match... we used different processes but we arrived at the same solution - all I did was realize that since I was given p(F)=0, c must be 1. When I got back the homework assignment, the part where I factor the expressions and arrive at c = 1 was marked with a large ? and the entire problem was marked wrong. Am I missing something? Or am I right that the professor made a mistake?

Here I go again. I assumed that your calculations were correct. If they were I would stand by comment about the problem being conditional. However, your calculations are flawed. I bothered to check when I was presented with the inverse. Your F^2 is not correct so I can assume that F^3 is not correct.

Your teacher's method, less calculations errors, is the way to proceed. Now go and fix your errors.

What will happen when you try to solve those 9 eqs for c?????? Look into that!! What does it mean to get a solution c=k for some real number k? I need to stop doing these problems at 2am as it makes me look bad. Especially when Denis thinks I am a research physicist.

 

[Steven G]

Have a look at this.

OK, so the inverse works. Hence the teacher made at least one silly error in getting the inverse. Thanks!

 

[ksdhart2]

Alright, so apparently I'm not the only one who was confused by this problem. An addendum to the homework solutions was passed out. It says:

The given equation for p(x) was actually incorrect. I made an error. F is actually a solution of p(x) = 0 for \(\displaystyle p(x)=x^3-3x^2+3x-1=(x-1)^3\). If I had not made the error, here is what F inverse would look like:

\(\displaystyle \begin{pmatrix}1-c&-c^2&-c+c^2\\ c&1+c-c^2&-2c+c^2\\ c&c-c^2&1-2c+c^2\end{pmatrix}\)

So, it seems that the entire problem was one big error, as the "given" was, in fact, not true. Now all I'm left with is a question regarding one of Jomo's comments:

Here I go again. I assumed that your calculations were correct. If they were I would stand by comment about the problem being conditional. However, your calculations are flawed. I bothered to check when I was presented with the inverse. Your F^2 is not correct so I can assume that F^3 is not correct.

Erm... hold on. Sorry, but how can the result for F² be wrong? I double checked with Wolfram Alpha when doing the homework, and I just now triple checked. What I wrote matches WolframAlpha. Same deal for F³ too.

 

[Steven G]

Alright, so apparently I'm not the only one who was confused by this problem. An addendum to the homework solutions was passed out. It says:



So, it seems that the entire problem was one big error, as the "given" was, in fact, not true. Now all I'm left with is a question regarding one of Jomo's comments: But wait a minute, then how did PKA get the inverse. I need to look at this problem and see if F^3 -3cF^2 +3c^2F-c^3I= 0



Erm... hold on. Sorry, but how can the result for F² be wrong? I double checked with Wolfram Alpha when doing the homework, and I just now triple checked. What I wrote matches WolframAlpha. Same deal for F³ too.

I checked three times doing it by hand and did not get your result for F^2. Besides even your own result did not show that F^3 -3cF^2 +3c^2F-c^3I equals the 3X3 matrix consisting of all Os. Also you did NOT get the answer that PKA got for the inverse of F which I checked and is correct. If I remember correctly I found two errors in your F^2 and I think one of them was the result for the 2,2 entry. Please try and see for yourself.