anyone help me on this.
the eigenvectors are 2,-1,3.
This is B,
1 1 1
2 1 2
3 2 4
This is B-1,
0 2 -1
2 -1 0
-1 -1 1
I cant seem to workout how you manage to get from B to B-1.
1. Use Cramer's rule
2. Augment $\displaystyle B$ with $\displaystyle I_{3 \times 3}$ then reduce the first three columns to
$\displaystyle I_{3 \times 3}$ using row operations, and then what is left in the
augmentation part of the matrix will be $\displaystyle B^{-1}$
RonL
Naisy, what are you doing in this material? You simply MUST be able to find the inverse of a simple 3x3 matrix.
Cpt. Black used some terms with which you simply must be familiar. If you are not, you must look them up and get familiar. It will not serve you at all to remain absent this information.
Here's a way to find an inverse, but it's rather cumbersome. May not be taught much these days. I don't know. It's comes from the adjoint.
$\displaystyle \begin{bmatrix}1&1&1\\2&1&2\\3&2&4\end{bmatrix}$
Using the cofactors:
$\displaystyle C_{11}=det(\begin{bmatrix}1&2\\2&4\end{bmatrix})=0$
$\displaystyle C_{12}= -det(\begin{bmatrix}2&2\\3&4\end{bmatrix})=-2$
$\displaystyle C_{13}=det(\begin{bmatrix}2&1\\3&2\end{bmatrix})=1$
$\displaystyle C_{21}= -det(\begin{bmatrix}1&1\\2&4\end{bmatrix})=-2$
$\displaystyle C_{22}= det(\begin{bmatrix}1&1\\3&4\end{bmatrix})=1$
$\displaystyle C_{23}= -det(\begin{bmatrix}1&1\\3&2\end{bmatrix})=1$
$\displaystyle C_{31} = det(\begin{bmatrix}1&1\\1&2\end{bmatrix})=1$
$\displaystyle C_{32} = -det(\begin{bmatrix}1&1\\2&2\end{bmatrix})=0$
$\displaystyle C_{33} = det(\begin{bmatrix}1&1\\2&1\end{bmatrix}) = -1$
From these we build a matrix:
$\displaystyle \begin{bmatrix}0&-2&1\\-2&1&1\\1&0&-1\end{bmatrix}$
Take the transpose and we get:
$\displaystyle \begin{bmatrix}0&-2&1\\-2&1&0\\1&1&-1\end{bmatrix}$
Now, finally, $\displaystyle B^{-1}=\frac{1}{det(B)}\cdot{adj(B)}$
$\displaystyle det(B)=-1$
So, we have $\displaystyle B^{-1}=\begin{bmatrix}0&2&-1\\2&-1&0\\-1&-1&1\end{bmatrix}$