geko Posted April 15, 2008 Report Share Posted April 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 :) Quote Link to comment Share on other sites More sharing options...

Tormod Posted April 16, 2008 Report Share Posted April 16, 2008 Glad you figured it out! :xparty: Quote Link to comment Share on other sites More sharing options...

CraigD Posted April 18, 2008 Report Share Posted April 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 [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. Quote Link to comment Share on other sites More sharing options...

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