0
$\begingroup$

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?

  • 0
    Oh, I think I got it. Thanks fretty!2012-06-24

1 Answers 1

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]