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and fill in the chart below
M11 = 1 M12 = −5 M13=−9 M21 =−1 M22 =−11 M23=−7 M31 =−4 M32 = 4 M33 = 4 .
Change the minors into cofactors, by multiplying by−1 those minors with i+j equal to an odd number. Finally transpose the result to form the adjoint matrix, so that
⇒ A−1 = 1
|A|adj(A) =− 1 16
1 1 −4
5 −11 −4
−9 7 4
.
As with all calculations, it is easy to make a mistake. Therefore, having found A−1, the next thing you should do is check your result by showing that AA−1 =I,
− 1 16
1 2 3
−1 2 1
4 1 1
1 1 −4
5 −11 −4
−9 7 4
=− 1 16
−16 0 0
0 −16 0
0 0 −16
=I.
Activity 8.6 Use this method to find the inverse of the matrix A=
1 2 3 0 4 0 5 6 7
Check your result.
Remember: the adjoint matrix only contains the cofactorsof A; the (i, j) entry is the cofactor Cji of A. The entries only multiply the cofactors when calculating the
determinant of A, |A|.
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8.4. Overview
Example 8.6 Use Cramer’s rule to find the solution of the linear system x+ 2y+ 3z = 7
−x+ 2y+z =−3 4x + y + z = 5 In matrix form Ax=b this system is,
1 2 3
−1 2 1
4 1 1
x y z
=
7
−3 5
We first check that |A| 6= 0. This is the same matrix A we used Example 8.5 to find the inverse of a matrix on page 125; |A|=−16. Then applying Cramer’s rule, we find x by evaluating the determinant of the matrix obtained from A by replacing column 1 with b,
x=
7 2 3
−3 2 1
5 1 1
|A| =−16
−16= 1 and in the same way we obtain y and z.
y=
1 7 3
−1 −3 1
4 5 1
|A| = 48
−16 =−3 z =
1 2 7
−1 2 −3
4 1 5
|A| = −64
−16 = 4 which can be easily checked by substitution into the original equations (or multiplying Ax).
R Read and work through the proof of this theorem in the text A-H, where it is labelled Theorem 3.43. Be sure you understand how the proof works.
Summary of Cramer’s rule. To find xi, (1) replace column i of A byb,
(2) evaluate the determinant of the resulting matrix, (3) divide by |A|.
Activity 8.7 Can you think of any other methods you can use to obtain the solution to Example 8.6?
Overview
In this chapter we have shown how to obtain the determinant of a square matrix, the real number intrinsically associated with the matrix. We looked at properties of the determinant and how its value is affected by changing the matrix using row operations.
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We then used this information to obtain the result that a square matrix A is invertible if and only if |A| 6= 0. This led in turn to the method of finding the inverse of A using cofactors (the adjoint matrix) and to Cramer’s rule.
Learning outcomes
At the end of this chapter and the relevant reading you should be able to:
find the determinant of a square matrix
use and understand the determinant as a means of determining invertibility of a square matrix
find the inverse of a matrix using cofactors
solve a consistent system of n linear equations in n unknowns using Cramer’s rule.
In addition you should know that:
There are three methods to solve Ax=b if A is n×n and |A| 6= 0:
(1) Gaussian elimination.
(2) Find A−1, then x=A−1b.
(3) Cramer’s rule.
There is one method to solve Ax=b if A ism×n and m6=n, or if |A|= 0:
(1) Gaussian elimination.
There are two methods to find A−1: (1) Using cofactors for the adjoint matrix.
(2) By row reduction of (A|I) to (I|A−1).
If A is an n×n matrix, then the following statements are equivalent (Theorem 7.3 and Theorem 8.5):
(1) A is invertible.
(2) Ax=b has a unique solution for anyb∈Rn. (3) Ax=0 has only the trivial solution,x=0.
(4) The reduced row echelon form of A isI. (5) |A| 6= 0.
Test your knowledge and understanding
Work Exercises 3.3–3.9 and 3.11 in the text A-H. The solutions can be found at the end of the textbook.
Work Problem 3.10 in the text A-H. (Carry out the required proof using determinants.) You will find the solution on the VLE.
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8.4. Comments on selected activities
Comments on selected activities
Feedback to activity 8.1 C13= 13.
Feedback to activity 8.2
|M|=−1(8−3)−2(0−3) + 1(0−2) =−1 Feedback to activity 8.3
You should either expand by column 1 or row 2. For example, using column 1:
|M|=−1(8−3) + 1(6−2) =−1.
Feedback to activity 8.5
|A|=
1 2 −1 4
0 5 −1 6
0 −3 3 −6
0 2 2 −1
=
5 −1 6
−3 3 −6
2 2 −1
At this stage you can expand the 3×3 matrix using a cofactor expansion, or continue a bit more with row operations:
|A|= 3
1 −1 2
5 −1 6
2 2 −1
= 3
1 −1 2
0 4 −4
0 4 −5
= 3
4 −4 4 −5
= 3(−4) =−12
Feedback to activity 8.6
|A|=−326= 0 A−1 = 1
|A|adj(A) =− 1 32
28 4 −12
0 −8 0
−20 4 4
= 1 8
−7 −1 3
0 2 0
5 −1 −1
.
Feedback to activity 8.7
One way is to use Gaussian elimination. Another method is to use the inverse matrix.
You foundA−1 in Example 8.5 on page 125. Now use it to calculate the solution, x=A−1b.
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Chapter 9
Rank, range and linear systems
Introduction
In this short chapter we aim to extend and consolidate what we have learned so far about systems of equations and matrices, and tie together many of the results of the previous chapters. We will intersperse an overview of the previous chapters with two new concepts, the rank of a matrix and the range of matrix.
This chapter will serve as a synthesis of what we have learned so far in anticipation of a return to these topics later in the guide.
Aims
The aims of this chapter are to:
synthesise what we have learned so far about linear systems introduce the concepts of the rank and range of a matrix
Essential reading
R Anthony, M. and M. Harvey. Linear Algebra: Concepts and Methods. Chapter 4.
Further reading
If you would like to have another source containing the material covered in this chapter, you can look up the key concepts and definitions (as listed in the synopsis which
follows) in any elementary linear algebra textbook, such as the two listed in Chapter 1, using either the table of contents or the index.
Synopsis
We start by introducing the rank of a matrix. We then show how the rank of a matrix is connected with the set of solutions of linear systems corresponding to the matrix and relate the number of free parameters in the general solution (when it exists) to the rank of the corresponding matrix. Finally, we define the range of a matrix.