Evaluating matrices with negative integers?

[geko]

Im struggling with how to evaluate matrices with negative integers and am looking for help. As far as i can understand the negative integer causes the matrix to be inverted, so does this mean that i multiply the matrix by 1/A and also invert the vectors?

 

For example, consider a 2x2 matrix (a11, b12, a21, b22) with 3, -2, 4, -5 respectively. Does this become (b22, -b12, -a21, a11), -5, 2, -4, 3?

 

Also, will the inverse be 1/(ad-bc)? Such that it becomes 1/(3*-5) - (2*-4) leading to a multiple of -0.142857142..

 

This seems bizarre and i think im doing something wrong, can anyone tell me if im doing it wrong or im on the right track please?

 

Edit: Nevermind, im on the right track. The multiple can be left as a fraction making it seem less bizarre :)

[Tormod]

Glad you figured it out! :xparty:

[CraigD]

Although geko notes he’s answered his own question, for anyone puzzled by this thread, let me offer my take on the question and its answer.

 

I believe the question is “how do you get the inverse of a matrix?”

 

As I’ve heard the term most commonly used, the inverse of a matrix [math]A[/math] means a matrix [math]B[/math] such that [math]A \cdot B = I[/math], where [math]I[/math] is a unit matrix. For the example given in post one, the matrix, its inverse, their 2x2 unit matrix product are:

[math]\begin{bmatrix}3 & -2 \\ 5 & 4\end{bmatrix} \cdot \begin{bmatrix}\frac{4}{22} & \frac{2}{22} \\ \frac{-5}{22} & \frac{3}{22}\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}[/math]

 

It’s neater to extract a constant from B and write it

[math] \frac{1}{22} \begin{bmatrix}4 & 2 \\ -5 & 3 \end{bmatrix}[/math]

 

Inverse of this kind are only meaningful for square (same number of columns and rows) matrixes.

 

Calculating the inverse of a matrix is pretty easy:

Start with a matrix A and a unit matrix B of the same size;

Multiply the rows of A by scalars and add them together as needed to change it into a unit matrix;

For every operation performed on a row of A, perform the same operation on B;

When complete, B is the inverse of A. If it’s impossible to change A into a unit matrix, it has no inverse.

 

If you do this calculation symbolically (with variables), you’ll find that the inverse of

[math]\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/math]

is

[math]\frac{1}{ad - bc} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}[/math]

From this, we can get the quick-and-easy rule “swap the diagonals, negative the other diagonals, and divide by the determinant.