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Stitz-Zeager_College_Algebra_e-book

# This means that the numerator of the fraction with x

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Unformatted text preview: the matrix A itself. (We will come back to this observation in a moment.) 2 3 Much like Carl’s quest to ﬁnd Sasquatch. Since matrix multiplication isn’t necessarily commutative, at this stage, these are two diﬀerent equations. 8.4 Systems of Linear Equations: Matrix Inverses 495 2x1 − 3x3 = 1 3x1 + 4 x3 = 0 −− − − − − − − − − − − −→ Encode into a matrix 2 −3 1 3 40 2x2 − 3x4 = 0 3x2 + 4 x4 = 1 −− − − − − − − − − − − −→ Encode into a matrix 2 −3 0 3 41 To solve these two systems, we use Gauss-Jordan Elimination to put the augmented matrices into reduced row echelon form. (We leave the details to the reader.) For the ﬁrst system, we get 2 −3 1 3 40 4 17 3 − 17 10 Gauss Jordan Elimination −− − − − − − −→ −−−−−−−− 01 4 3 which gives x1 = 17 and x3 = − 17 . To solve the second system, we use the exact same row operations, in the same order, to put its augmented matrix into reduced row echelon form (Think about why that works.) and we obtain 2 −3 0 3 41 which means x2 = 3 17 and x4 = 2 17 . 10 Gauss Jordan Elimination −− − − − − − −→ −−−−−−−− 01 3 17 2 17 Hence, A−1 = x1 x2 x3 x4 = 4 17 3 − 17 3 17 2 17 We can check to see that A−1 behaves as it should by computing AA−1 AA−1 = 4 17 3 − 17 2 −3 3 4 3 17 2 17 = 10 01 = I2 2 −3 3 4 = 10 01 = I2 As an added bonus, −1 A A= 4 17 3 − 17 3 17 2 17 We can now return to t...
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