Diagonalisation of a matrix

**naisy18** · December 24th 2007, 05:22 AM

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.

**CaptainBlack** · December 24th 2007, 05:38 AM

Originally Posted by naisy18

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 $B$ with $I_{3 \times 3}$ then reduce the first three columns to
$I_{3 \times 3}$ using row operations, and then what is left in the
augmentation part of the matrix will be $B^{-1}$

RonL

**naisy18** · December 24th 2007, 05:50 AM

Originally Posted by CaptainBlack

1. Use Cramer's rule

2. Augment $B$ with $I_{3 \times 3}$ then reduce the first three columns to
$I_{3 \times 3}$ using row operations, and then what is left in the
augmentation part of the matrix will be $B^{-1}$

RonL

Im still not too sure what your saying to be honest.

If you or any other poster has time could you show me where you would start and how you would start it, as im still pretty stuck. cheers

**TKHunny** · December 24th 2007, 08:06 AM

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.

**Isomorphism** · December 24th 2007, 10:23 AM

Can't we use the given eigenvalues to make the computation a tad easier

**galactus** · December 24th 2007, 04:29 PM

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.

$\begin{bmatrix}1&1&1\\2&1&2\\3&2&4\end{bmatrix}$

Using the cofactors:

$C_{11}=det(\begin{bmatrix}1&2\\2&4\end{bmatrix})=0$

$C_{12}= -det(\begin{bmatrix}2&2\\3&4\end{bmatrix})=-2$

$C_{13}=det(\begin{bmatrix}2&1\\3&2\end{bmatrix})=1$

$C_{21}= -det(\begin{bmatrix}1&1\\2&4\end{bmatrix})=-2$

$C_{22}= det(\begin{bmatrix}1&1\\3&4\end{bmatrix})=1$

$C_{23}= -det(\begin{bmatrix}1&1\\3&2\end{bmatrix})=1$

$C_{31} = det(\begin{bmatrix}1&1\\1&2\end{bmatrix})=1$

$C_{32} = -det(\begin{bmatrix}1&1\\2&2\end{bmatrix})=0$

$C_{33} = det(\begin{bmatrix}1&1\\2&1\end{bmatrix}) = -1$

From these we build a matrix:

$\begin{bmatrix}0&-2&1\\-2&1&1\\1&0&-1\end{bmatrix}$

Take the transpose and we get:

$\begin{bmatrix}0&-2&1\\-2&1&0\\1&1&-1\end{bmatrix}$

Now, finally, $B^{-1}=\frac{1}{det(B)}\cdot{adj(B)}$

$det(B)=-1$

So, we have $B^{-1}=\begin{bmatrix}0&2&-1\\2&-1&0\\-1&-1&1\end{bmatrix}$

**Isomorphism** · December 25th 2007, 06:52 AM

Originally Posted by galactus

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.

Actually this is how I originally learned it (while learning engineering).

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