Stitz-Zeager_College_Algebra_e-book

# We see immediately that x 0 is a zero of multiplicity

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Unformatted text preview: s of Equations and Matrices also know that the process by which we ﬁnd A−1 is determined completely by A, and not by the constants in B . This answers the question as to why we would bother doing row operations on a super-sized augmented matrix to ﬁnd A−1 instead of an ordinary augmented matrix to solve a system; by ﬁnding A−1 we have done all of the row operations we ever need to do, once and for all, since we can quickly solve any equation AX = B using one multiplication, A−1 B . We illustrate this in our next example. 3 1 2 5 Example 8.4.1. Let A = 0 −1 2 1 4 1. Use row operations to ﬁnd A−1 . Check your answer by ﬁnding A−1 A and AA−1 . 2. Use A−1 to solve the 3x + y + 2 z −y + 5z (a) 2x + y + 4 z following systems of equations = 26 3x + y + 2 z = 4 = 39 −y + 5z = 2 (b) = 117 2x + y + 4 z = 5 3x + y + 2 z = 1 −y + 5z = 0 (c) 2x + y + 4 z = 0 Solution. 1. We begin with a super-sized augmented matrix and proceed with Gauss-Jordan elimination. 3 1 2 0 −1 5 2 1 4 2 1 1 3 3 0 −1 5 2 1 4 1 2 1 3 3 0 −1 5 8 1 0 3 3 1 2 1 3 3 0 1 −5...
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