College Algebra, 2013a

8.5 Determinants and Cramer’s Rule 619
Theorem 8.8. Cramer’s Rule: Suppose AX = B is the matrix form of a system of n linear
equations in n unknowns where A is the coefficient matrix, X is the unknowns matrix, and B is
the constant matrix. If det(A) ≠ 0, then the corresponding system is consistent and independent
and the solution for unknowns x 1 , x 2 , ...x n is given by:
x j = det (A j)
det(A) ,
where A j is the matrix A whose jth column has been replaced by the constants in B.
In words, Cramer’s Rule tells us we can solve for each unknown, one at a time, by finding the ratio
of the determinant of A j to that of the determinant of the coefficient matrix. The matrix A j is
found by replacing the column in the coefficient matrix which holds the coefficients of x j with the
constants of the system. The following example fleshes out this method.
Example 8.5.1. Use Cramer’s Rule to solve for the indicated unknowns.
{ 2x1 − 3x
1. Solve
2 = 4
5x 1 + x 2 = −2 for x 1 and x 2
⎧
⎨ 2x − 3y + z = −1
2. Solve x − y + z = 1 for z.
⎩
3x − 4z = 0
Solution.
1. Writing this system in matrix form, we find
A =
[ 2 −3
5 1
]
X =
[
x1
x 2
]
[
B =
4
−2
]
To find the matrix A 1 , we remove the column of the coefficient matrix A which holds the
coefficients of x 1 and replace it with the corresponding entries in B. Likewise, we replace the
column of A which corresponds to the coefficients of x 2 with the constants to form the matrix
A 2 . This yields
[
A 1 =
4 −3
−2 1
]
A 2 =
[ 2 4
5 −2
]
Computing determinants, we get det(A) = 17, det (A 1 )=−2 and det (A 2 )=−24, so that
x 1 = det (A 1)
det(A) = − 2
17
x 2 = det (A 2)
det(A) = −24 17
The reader can check that the solution to the system is ( − 2
17 , − 24
17)
.