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**Unformatted text preview: **llows that
A−1 10.5 3 −9
3
−3
9 −3
1
1
1
adj(A) = −
=
−9
9 −9 =
9 −9
9 . 2
det(A)
18
18
5 −3 −1
−5
3
1 Solving Linear Systems of Equations Finally we show how we can use inverse matrices to solve linear systems of equations. We will explain the idea ﬁrst and then we apply it in two examples.
Let us start by looking at an example of a linear system of 2 equations in 2 unknowns.
1 x + 2 y = 5,
3 x + 4 y = 6. (10.15) 288 10. Matrices We can rewrite this linear system in the form
Ax = b
with the 2 × 2 matrix
A= 12
34 and the vectors
x= x
y and b= 5
6 . That is, (10.15) is equivalent to the linear system in matrix form
12
34 x
y = 5
6 . After this motivation we will think from now on of square linear systems with n
equations and n unknowns as an expression of the form
A x = b, (10.16) where A is an n × n matrix, and b is a given n × 1 matrix (an n-vector in Rn ) and x
is an n × 1 matrix (an n-vector in Rn ) that contains the unknowns. Without further
information about A and b, we cannot know whether an x exists that satisﬁes
(10.16) and if such an x exists whether it is unique.
However, if det(A) = 0, then there exists a uniquely determined n-vector x
that satisﬁes (10.16). ‘Uniquely determined’ means that x is the only vector that
satisﬁes (10.16). Now we explain how we can use the inverse matrix A−1 of A to
solve (10.16) if det(A) = 0.
If det(A) = 0, then A is invertible and has an inverse matrix A−1 (see Theorem
10.43). We multiply (10.16) from the left with the inverse matrix A−1 and obtain
A−1 (A x) = A−1 b. (10.17) Using the associative law of matrix multiplication (see Theorem 10.20) and A−1 A =
In and In x = x (from (10.1)), we then transform the left-hand side
A−1 (A x) = (A−1 A) x = In x = x. (10.18) Combing (10.17) and (10.18) yields that the solution x of (10.16) is given by
x = A−1 b. (10.19) 10. Matrices 289 If it is not too demanding to determine the inverse matrix A−1 , then (10.19) provides
an easy way for computing the solution x of (10.16).
We illustrate this for two examples.
Example 10.50 (solve linear system of equations)
Solve the linear system of equations
3 x + 2 y = 4,
5 x + 3 y = 7.
Solution: We start by writing the linear system of equations in matrix form as
x
y 32
53 4
7 = . Hence A x = b, with
A= 32
53 , x
y x= and 4
7 b= . Since det(A) = 3 × 3 − 2 × 5 = 9 − 10 = −1 = 0, the matrix A is invertible. Its
inverse...

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