By Kenneth Kuttler

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1 2 0 2 1 1  1 2 0  0 3 1  −2 1 1 The first thing you need to check before doing anything else is whether it is possible to do the multiplication. The first matrix is a 2 × 3 and the second matrix is a 3 × 3. Therefore, is it possible to multiply these matrices. According to the above discussion it should be a 2 × 3 matrix of the form   Second column Third column First column        1 2 0   1 2 1 1 2 1 1 2 1   0 ,  3 ,  1    0 2 1 0 2 1  0 2 1  −2 1 1   You know how to multiply a matrix times a three columns.

1 1 −1 Form the augmented matrix,  1  1 1 0 −1 1  1 1 0 0 1 0 1 0 . −1 0 0 1 Now do row operations untill the n × n matrix on the left becomes the identity matrix. This yields after some computations,   1 1 1 0 0 0 2 2  0 1 0 1 −1 0  0 0 1 1 − 21 − 12 and so the inverse of A is the matrix on the right,   1 1 0 2 2  1 −1 0 . 1 1 − 2 − 21 Checking the answer is easy. Just multiply the matrices and see if     1 1 1 0 1 0 1 0 2 2  1 −1 1   1 −1 0 = 0 1 1 1 −1 0 0 1 − 12 − 12 Always check your answer because if mistake.

Before reading this, review the definition of matrix multiplication in terms of entries of the matrices. (A (BC))ij Aik (BC)kj = k = Aik k = Bkl Clj l (AB)il Clj l = ((AB) C)ij . 15. Another important operation on matrices is that of taking the transpose. The following example shows what is meant by this operation, denoted by placing a T as an exponent on the matrix.  T 1 1 + 2i 1 3 2  3  = 1 1 + 2i 1 6 2 6 What happened? The first column became the first row and the second column became the second row.

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