In an algebraic equation such as y = kx, k = 0, we can multiply through by k1 = 1/k in
Chapter 12, Problem 12.5.13(choose chapter or problem)
In an algebraic equation such as y = kx, k = 0, we can multiply through by k1 = 1/k in order to solve for x, giving x = k1y. We can do a similar thing with matrices: if v = Au , it is sometimes possible to find an inverse of A, written A1, such that u = A1 v . (a) Let A = 2 1 3 2 . It can be shown that A1 = 2 1 3 2 . Check this fact using the vector u = (3, 5), first by multiplying u by A to get v , and then by multiplying v by A1 to show that we get u back. (b) Rework part (a), this time for u = (1, 7). (c) Rework part (a), this time for u = (a, b). 14. In Pro
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