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Stitz-Zeager_College_Algebra_e-book

# In other words make sure you dont try to add apples

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Unformatted text preview: following example ﬂeshes out this method. Example 8.5.1. Use Cramer’s Rule to solve for the indicated unknowns. 1. Solve 2x1 − 3x2 = 4 for x1 and x2 5x1 + x2 = −2 2x − 3y + z = −1 x−y+z = 1 for z . 2. Solve 3x − 4z = 0 Solution. 1. Writing this system in matrix form, we ﬁnd 2 −3 5 1 A= X= x1 x2 B= 4 −2 To ﬁnd the matrix A1 , we remove the column of the coeﬃcient matrix A which holds the coeﬃcients of x1 and replace it with the corresponding entries in B . Likewise, we replace the column of A which corresponds to the coeﬃcients of x2 with the constants to form the matrix A2 . This yields A1 = 4 −3 −2 1 A2 = 2 4 5 −2 Computing determinants, we get det(A) = 17, det (A1 ) = −2 and det (A2 ) = −24, so that x1 = det (A1 ) 2 =− det(A) 17 x2 = det (A2 ) 24 =− det(A) 17 2 The reader can check that the solution to the system is − 17 , − 24 . 17 514 Systems of Equations and Matrices 2. To use Cramer’s Rule to ﬁnd z , we identify x3 as z . We have 2 −3 1 x −1 2 −3 −1 1 X...
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