线性代数英文课件4.3

1 6 −1 2 0 0
Байду номын сангаас
First, we find det A. We can zero out column 1 and then expand along that column: 1 1 −1 1 1 −1 −2 2 1 = det 0 −2 2 = 1 det det A = det 1 −1 −0+0 −1 −1 1 0 −2 0 −1 −1 = (−2)(−1) − (−1)(2) = 2 − (−2) = 4
158
Unit 12
a system are the quotients of the determinants of these new matrices over the determinant of the coefficient matrix. Neat trick, eh? We won’t attempt to prove or even explain why this works. We’ll just take Mr. Cramer’s word for it. (And that means that you can just ignore pages 170 and 171 in the text.)
Now, how do we use these matrices? Recall that as long as det A = 0, the SLE Ax = b has a unique solution. According to Mr. Cramer, the values of the xj ’s in the unique solution to such
Theorem 12.1. Cramer’s Rule Let A be any square matrix of order n with det A = 0 and let b be any n × 1 column vector. Then in the unique solution to the system Ax = b, the value of the j th unknown, xj , is given by: xj = det A(j ) det A
x1 x1 x1
+ −
x2 x2
− x3 + x3 − 2x3
= = =
6 2 0
1 1 −1 1 and we found the matrices with coefficient matrix A = 1 −1 1 0 −2 6 1 −1 1 6 −1 1 1 1 A(1) = 2 −1 A(2) = 1 2 A(3) = 1 0 0 −2 1 0 −2 1
Math 1229A/B
Unit 12: Applications of the Determinant
(text reference: Section 4.3)
c V. Olds 2010
Unit 12
157
12
Applications of the Determinant
We shall finish up the course by looking at a couple of other ways that the determinant of a square matrix can be used, that is, a couple of applications of the determinant of a square matrix. The first is a method of finding the solution to the SLE Ax = b when A is a nonsingular (i.e. invertible) square matrix, without row reducing or finding the inverse matrix. Instead, we calculate det A as well as the determinant of certain other matrices obtained from A and b. After that, we will learn how to find the inverse of square matrix A, when it exists, using det A and another matrix which is obtained using the cofactors of A. Again, this gives us another method for doing something which previously we could only do by row reducing.
First, let’s look at an example of forming these matrices, to make sure you understand what the definition is saying.
Example 12.1. Find A(1), A(2) and A(3), where A is the coefficient matrix of the linear system: x1 x1 x1 Solution: We have the SLE Ax = b where 1 A= 1 1 + − x2 x2 − x3 + x3 − 2x3 = = = 6 2 0
Cramer’s Rule Consider a SLE Ax = b in which A is a square matrix of order n with det A = 0. We can form n new n × n matrices by replacing different columns of A by the column vector b. And if we do so, then we can directly find the value of xj in the unique solution to Ax = b using the determinant of one of these new matrices and the determinant of A. A fellow named Cramer developed a rule for doing this. Before we get to the rule, though, we need to define these new matrices and the notation we use to refer to them. Definition: Let Ax = b be any SLE in which A is a square matrix. We define the matrix A(j ) to be the matrix obtained by replacing column j of A with the column vector b.
159
−4 2 −6 −1
−0+0
(−4)(−1) − (−6)(2) = 4 − (−12) = 16
Finally, for A(3) we can expand along row 3 again: 1 1 6 1 6 det A(3) = det 1 −1 2 = 1 det − 0 + 0 = (1)(2) − (−1)(6) = 2 − (−6) = 8 −1 2 1 0 0 So using Cramer’s Rule, we see that the values of x1 , x2 and x3 in the unique solution to the system are: x1 = 16 det A(1) = = 4, det A 4 x2 = det A(2) 16 = = 4, det A 4 x3 = det A(3) 8 = =2 det A 4
Notice that det A = 0, so the SLE does indeed have a unique solution. Now we find the determinants of the A(j ) matrices. For A(1) we can expand along row 3: 6 1 −1 6 1 1 = 0 − 0 + (−2) det det A(1) = det 2 −1 2 −1 0 0 −2 = (−2)[(6)(−1) − 2(1)] = (−2)[−6 − 2] = (−2)(−8) = 16
Unit 12 For A(2), it will be easiest to 1 det A(2) = det 1 1 = zero out column 1 again: 6 −1 1 6 −1 2 1 = det 0 −4 2 = 1 det 0 −2 0 −6 −1
1 −1 −1 1 0 −2
and
6 b= 2 0 instance columns 6 2 0

We form the matrix A(j ) by replacing the j th column of A by the column vector b. So for to form A(1) we write the numbers from b instead of the first column of A, and then write 2 and 3 of A as usual. And so forth. We get: 6 1 −1 1 6 −1 1 1 1 1 A(1) = 2 −1 A(2) = 1 2 A(3) = 1 −1 0 0 −2 1 0 −2 1 0
This means that if these determinants are reasonably easy to find, using Cramer’s Rule can be an easier way to find the solution to a SLE than row reducing. (However if the determinants require a lot of work to calculate, then using Cramer’s Rule involves more work than row reducing. So that’s just obnoxious.) For instance determinants of 2 × 2 matrices are always easy to find, so Cramer’s Rule is a reasonably good way to solve a system of 2 equations in 2 unknowns, as long as the coefficient matrix is nonsingular. And if there are 0’s around then sometimes the determinants of larger square matrices are reasonably easy to find. The following examples show how Cramer’s Rule can be used to find the unique solution to a SLE with a nonsingular square coefficient matrix. It’s important to remember, though, that Cramer’s Rule simply doesn’t apply to a system whose coefficient matrix isn’t square, or has determinant 0. Example 12.2. Use Cramer’s Rule to find the unique solution to the SLE in Example 12.1. Solution: We have the SLE: