Evaluating matrices with negative integers?

geko · 04-15-2008

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 · 04-16-2008

Glad you figured it out!

CraigD · 04-18-2008

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 $A$ means a matrix $B$ such that $A \cdot B = I$ , where $I$ is a unit matrix. For the example given in post one, the matrix, its inverse, their 2x2 unit matrix product are:
$\begin{bmatrix}3 & -2 \\ 5 & 4\end{bmatrix} \cdot \begin{bmatrix}\frac{4}{22} & \frac{2}{22} \\ \frac{-5}{22} & \frac{3}{22}\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$

It’s neater to extract a constant from B and write it
$\frac{1}{22} \begin{bmatrix}4 & 2 \\ -5 & 3 \end{bmatrix}$

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
$\begin{bmatrix}a & b \\ c & d\end{bmatrix}$
is
$\frac{1}{ad - bc} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$
From this, we can get the quick-and-easy rule “swap the diagonals, negative the other diagonals, and divide by the determinant.

Evaluating matrices with negative integers?

Hypography?

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