Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find a 3x3 nondiagonal matrix whose eigenvalues are $-2,-2,$ and $3$, and associated eignenvectors are $\begin{pmatrix} 1 \\ 0\\1 \end{pmatrix}$ , $\begin{pmatrix} 0 \\ 1 \\1\end{pmatrix}$, and $\begin{pmatrix} 1 \\ 1 \\1\end{pmatrix}$, respectively.

Answer: $\begin{pmatrix} 3&5&-5 \\ 5&3&-5 \\ 5&5&-7\end{pmatrix}$

I keep getting $\begin{pmatrix} 1&3&-3 \\ 5&3&-5 \\ 3&3&-5\end{pmatrix}$, so I am only getting the second row correct. I know that you're supposed to use the formula pA = PD$P^{-1}$. I had $\begin{pmatrix} 1&0&1 \\ 0&1&1 \\ 1&1&1\end{pmatrix}$ as $P$, $\begin{pmatrix} -2&0&0 \\ 0&-2&0 \\ 0&0&3\end{pmatrix}$ as $D$, and $\begin{pmatrix} 1&0&0 \\ -1&0&1 \\ 1&1&-1\end{pmatrix}$ as $P^{-1}$, and found the product to be $\begin{pmatrix} 1&3&-3 \\ 5&3&-5 \\ 3&3&-5\end{pmatrix}$ which is not right. Am I doing something wrong?

share|improve this question
    
Your inverse is incorrect...you can check this yourself. –  fretty Jun 24 '12 at 16:29
    
I checked it more than three times, and the elementary operations for both the inverse and the original matrix appears to be right for me... –  Ashley Jun 24 '12 at 16:34
    
But clearly those two matrices do NOT multiply to give the identity matrix. For a start the top left entry would be $2$. –  fretty Jun 24 '12 at 16:35
    
Yeah, that's the problem. I don't get how the inverse is wrong though. I thought I derived it right –  Ashley Jun 24 '12 at 16:36
    
Oh, I think I got it. Thanks fretty! –  Ashley Jun 24 '12 at 16:39

1 Answer 1

I think that $$P^{-1}=\begin{pmatrix}0&-1&1\\-1&0&1\\1&1&-1 \end{pmatrix}$$

Here is what maple says about this :

> with(linalg):P:=matrix(3,3,[1,0,1,0,1,1,1,1,1]);

                           [1    0    1]
                           [           ]
                      P := [0    1    1]
                           [           ]
                           [1    1    1]

PP:=evalm(inverse(P));

                          [ 0    -1     1]
                          [              ]
                    PP := [-1     0     1]
                          [              ]
                          [ 1     1    -1]

DD:=matrix(3,3,[-2,0,0,0,-2,0,0,0,3]);

                          [-2     0    0]
                          [             ]
                    DD := [ 0    -2    0]
                          [             ]
                          [ 0     0    3]

evalm((P &* DD) &* PP);

                        [3    5    -5]
                        [            ]
                        [5    3    -5]
                        [            ]
                        [5    5    -7]
share|improve this answer
    
Thanks a lot Mohamed! –  Ashley Jun 24 '12 at 17:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.