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02-14-03

# 02-14-03 - Lecture 7 Andrei Antonenko 1 Matrix equations...

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Lecture 7 Andrei Antonenko February 14, 2003 1 Matrix equations and the inverse 1.1 Algorithm of solving Suppose we have 2 given matrices A and B , and we’d like to find a matrix X such that AX = B. (1) Note, that B should not be a square matrix, for example in case of a linear system B is a matrix with only 1 column, and X is a matrix with 1 column as well. Suppose matrix A is invertible, i.e. there exists an inverse A - 1 . So, we can multiply both parts of the equation (1) by A - 1 from the left side. We’ll get: A - 1 · AX = A - 1 B = ( A - 1 A ) X = A - 1 B = X = A - 1 B . So, the solution of this equation in case when A is invertible is X = A - 1 B. (2) Now let’s suppose we need to find the inverse of any square matrix A . If matrix X is an inverse for A then it satisfies the following equation: AX = I , where I is an identity matrix. It is a special case of the equation (1). So we see, that finding an inverse is a special case of solving the matrix equation (1). We’ll provide an algorithm of solving the general case of matrix equation (1). Algorithm of solving the equation AX = B . Step 1 Construct the augmented matrix from A and B writing the matrix B to the right of matrix A , i.e. if A = a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . . . . . . . a n 1 a n 2 . . . a nn , and B = b 11 b 12 . . . b 1 m b 21 b 22 . . . b 2 m . . . . . . . . . . . . . . . . . . b n 1 b n 2 . . . b nm , 1

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then augmented matrix is ( A | B ) = a 11 a 12 . . . a 1 n fl fl fl b 11 b 12 . . . b 1 m a 21 a 22 . . . a 2 n fl fl fl b 21 b 22 . . . b 2 m . . . . . . . . . . . . . . . . . . fl fl fl . . . . . . . . . . . . . . . . . . a n 1 a n 2 . . . a nn fl fl fl b n 1 b n 2 . . . b nm (3) Step 2 Perform elementary row operations to reduce the matrix A in the left side of the augmented matrix to its RREF. (But perform elementary row operations with whole rows of ( A | B )!) If we get a zero-row in A -half of the matrix ( A | B ), then the matrix A is not invertible, and we should stop our algorithm. Otherwise, the RREF of the matrix
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