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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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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