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Lec 17: Inverse of a matrix and Cramer's rule

We are aware of algorithms that allow to solve linear systems and invert a matrix. It turns out that determinants make possible to ¯nd those by explicit formulas. For instance, ifAis ann£ninvertible matrix, then A

¡1=1

det(A)2 6 664A

11A21¢¢¢An1

A A

1nA2n¢¢¢Ann3

7

775:(1)

Note that the (i;j) entry of matrix (1) is the cofactorAji(notAij!). In fact the entry is Aji det(A)as we multiply the matrix by1 det(A). [We can divide by det(A) since it is not 0 for an invertible matrix.] Curiously, in spite of the simple form, formula (1) is hardly applicable for ¯ndingA¡1whennis large. This is because computing det(A) and the cofactors requires too much time for suchn. Notice that det(A) can be found as soon as we know the cofactors, because of the cofactor expansion formula.

Example.

Find the inverse, if it exists, for

A=2

40 1 2

¡2 3¡1

4 0 13

5

We have:

A

11=¯¯¯¯3¡1

0 1¯

¯¯¯= 3; A12=¡¯¯¯¯¡2¡1

4 1¯

¯¯¯=¡2; A13=¯¯¯¯¡2 3

4 0¯

¯¯¯=¡12:

Find the determinant by the expansion along the ¯rst row: det(A) =a11A11+a12A12+a13A13= 0¢3 + 1¢(¡2) + 2¢(¡12) =¡26: Since det(A)6= 0, we conclude thatAis invertible, and we can continue computing cofactors 1: A

21=¡¯¯¯¯1 2

0 1¯

¯¯¯=¡1; A22=¯¯¯¯0 2

4 1¯

¯¯¯=¡8; A23=¡¯¯¯¯0 1

4 0¯

¯¯¯= 4;

A

31=¯¯¯¯1 2

3¡1¯

¯¯¯=¡7; A32=¡¯¯¯¯0 2

¡2¡1¯

¯¯¯=¡4; A33=¯¯¯¯0 1

¡2 3¯

¯¯¯= 2:

By formula (1)

A

¡1=¡1

26
2

43¡1¡7

¡2¡8¡4

¡12 4 23

5 =2

4¡3

26
1 26
7 26
1 13 4 13 2 13 6 13 ¡2 13 ¡1 13 3 5 The method of ¯ndingA¡1using the augmented matrix [AjI3] seems to be faster for the previous example. It worth mentioning that in case of 2£2 matrixAformula (1) is especially simple:

IfA=·a b

c d¸ and det(A) =ad¡bc6= 0;thenA¡1=1 ad¡bc· d¡b

¡c a¸

1 If the determinant were 0, we would stop here and say thatAis singular (there is no need to

¯nd rest cofactors).

1 Make sure thatAA¡1=I2(thus you will prove formula (1) for the casen= 2). For example,·2 1

4 3¸

¡1 =1 2·

3¡1

¡4 2¸

3

2¡1

2¡2 1¸

Now describe the Cramer's rule for solving linear systemsA¹x=¹b. It is assumed thatAis a square matrix and det(A)6= 0 (or, what is the same,Ais invertible). Then, as we know, the linear system has a unique solution. The rule says that this solution is given by the formula x

1=det(A1)

det(A); x2=det(A2) det(A); :::; xn=det(An) det(A);(2) whereAiis the matrix obtained fromAby replacing theithcolumn ofAby¹b. [Don't confuse with cofactorsAij!]

Example.

Solve the linear system

3x1+x2¡2x3= 4

¡x1+ 2x2+ 3x3= 1

2x1+x2+ 4x3=¡2:

We have (check all calculations!)

det(A) =¯

¯¯¯¯¯3 1¡2

¡1 2 3

2 1 4¯

¯¯¯¯¯= 35

Since det(A)6= 0, we can use the Cramer's rule. Let's ¯nd determinants ofA1;A2;A3: det(A1) =¯

¯¯¯¯¯4 1¡2

1 2 3

¡2 1 4¯

¯¯¯¯¯= 0;det(A2) =¯

¯¯¯¯¯3 4¡2

¡1 1 3

2¡2 4¯

¯¯¯¯¯= 70;det(A3) =¯

¯¯¯¯¯3 1 4

¡1 2 1

2 1¡2¯

¯¯¯¯¯=¡35:

Now by formula (2):

x 1=0 35
= 0; x2=70 35
= 2; x3=¡35 35
=¡1:

Thus 0;2;¡1 is the solution to our system.

As before, in case of the linear system with two equations and two variables the solution is particularly simple. Consider the system ax+by=e cx+dy=f with unknownsxandy. Ifad¡bc6= 0, then by Cramer's rule x=de¡bf ad¡bc; y=af¡ce ad¡bc: Make sure that these satisfy to the above system (thus you will prove Cramer's rule for 2£2 case). For example, the system x+ 3y= 0

2x+ 7y= 1

has the solutionx=¡3 1 =¡3,y=1 1 = 1. 2quotesdbs_dbs11.pdfusesText_17