College Algebra 9th txtbk

SECTION 7–4 Solving Systems of Linear Equations Using Matrix Inverse Methods 479
and multiply it by the constant matrix B on the left. In Example 1, we found that the inverse
of matrix A is
So the equation X A 1 B is
3 3 1
A 1 £ 2 2 1§
4 5 2
X A 1 B
x 1 3 3 1 1 5
£ x 2 § £ 2 2 1§£
1 § £ 3 §
x 3 4 5 2 1 7
and we can conclude that x 1 5, x 2 3, and x 3 7. Check this result in system (4).
MATCHED PROBLEM 5
Use matrix inverse methods to solve the system:
3x 1 x 2 x 3 1
x 1 x 2 3
x 1 x 3 2
[Note: The inverse of the coefficient matrix was found in Matched Problem 1.]
Z USING INVERSE METHODS TO SOLVE SYSTEMS OF EQUATIONS
If the number of equations in a system equals the number of variables and the coefficient
matrix has an inverse, then the system will always have a unique solution
that can be found by using the inverse of the coefficient matrix to solve the corresponding
matrix equation.
Matrix equation
AX B
Solution
X A 1 B
At first, matrix inverse methods don’t seem any better than Gauss–Jordan elimination—
both require applying row operations to an augmented matrix. The advantage of the inverse
method becomes apparent when solving a number of systems with a common coefficient
matrix, as in Example 6.
EXAMPLE 6 Using Inverses to Solve Systems of Equations
Use matrix inverse methods to solve each of the following systems:
(A) x 1 x 2 x 3 3 (B) x 1 x 2 x 3 5
2x 2 x 3 1
2x 2 x 3 2
2x 1 3x 2 4 2x 1 3x 2 3
SOLUTIONS
Notice that both systems have the same coefficient matrix A as system (4) in Example 5.
Only the constant terms have been changed. So we can use A 1 to solve these systems just
as we did in Example 5.