Inverse and Adjugate MatricesDate: 09/21/98 at 13:39:14 From: Khamisi Easton Subject: Linear algebra I know that you have already been asked this question but I didn't really understand the answer that you gave. I am trying to solve a 3x3 matrix. I know how to solve for a 2x2 matrix. (2 equations 2 unknowns). I don't know how to solve for three equations three unknowns. Here is what I have so far. I would appreciate it if you could help me out. Can you please solve any 3x3 matrix using this format? A^(-1) * A = I [3 -1 2] A = [1 4 -3] [2 5 3] So: [c11 c12 c13] [3 -1 2] [1 0 0] [c21 c22 c23] * [1 4 -3] = [0 1 0] [c31 c32 c33] [2 5 3] [0 0 1] 3c11 + 1c12 + 2c13 = 1 -1c21 + 4c22 + 5c23 = 0 2c31 + -3c32 + 3c33 =0 ...and so on. Thank you, Khamisi Date: 09/21/98 at 17:43:03 From: Doctor Anthony Subject: Re: Linear algebra Hello Khamisi, You certainly don't want to solve 9 simultaneous equations, which your method suggests. The two ways that are normally used are either row operations or finding the adjugate matrix. Below is an example of the latter method. Use the inverse matrix method to solve the following set of equations. x + 3y - z = -3 3x - y + 2z = 1 2x -y + z = -1 You require the inverse of the matrix formed by the coefficients of the equations. A1, A2, B1, C2, etc are the cofactors of the elements a1, a2, b1, c2 and the way they are formed is described below. If M = [1 3 -1| then: A1 = 1 -A2 = 2 A3 = 5 |3 -1 2| -B1 = -1 B2 = 3 -B3 = 5 |2 -1 1] C1 = -1 -C2 = -7 C3 = -10 labeling the positions of the matrix as: [a1 b1 c1| |a2 b2 c2| |a3 b3 c3] Note that A1 is the cofactor of a1, A2 is the cofactor of a2, and so on. To get the cofactors, cross-out the row and column and find the determinant of the remaining matrix. For example, to find the cofactor of A1, cross out its corresponding row and column to get the matrix: [-1 2| |-1 1] which has determinant 1. We write the cofactors in the transposed positions as shown, and use the rule of signs with alternating + and - signs. The adjugate matrix is then: [A1 A2 A3| = [1 -2 5| |B1 B2 B3| |1 3 -5| |C1 C2 C3] |-1 7 -10] Finally the inverse matrix is: M^(-1) = ADJ(M)/det(M) so we must get det(M) to complete the calculation. det(M) = -1+12+3-2+2-9 = 5 and so: M^(-1) = (1/5)[1 -2 5| |1 3 -5| |-1 7 -10] We then have: [x| = (1/5)[1 -2 5| [-3| |y| |1 3 -5| |1 | |z] |-1 7 -10] |-1] The solutions are then found by multiplying out the matrices on the right: [x| [-2| |y| = | 1| |z] | 4] - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/