[PDF] Section 5.1 2 If a 3 by 3 matrix has detA = find det(2A) and det(?A

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Section 5.1

2If a 3 by 3 matrix has detA=12

, nd det(2A) and det(A) and detA2and det(A1).

Solution: det(2A) = 4, det(A) =12

, detA2=14 , and det(A1) = 2.

3True or false, with a reason if true or a counterexample if false:

(a) The determinant ofI+Ais 1 + detA. False, example withA=Ibeing the two by two identity matrix. Then det(I+A) = det(2I) =

4 and 1 + detA= 2.

(b) The determinant ofABCisjAjjBjjCj. True, the determinant of a product is the product of the determinants. (c) The determinant of 4Ais 4jAj. False, the determinant of 4Ais 4njAjifAis annbynmatrix. (d) The determinant ofABBAis zero.

False, exampleA=0 0

0 1 ,B=a b c d . Then

ABBA=0 0

c d 0b 0d =0b c0 with determinantbc. This is not zero unlessb= 0 orc= 0.

4Which row exchanges show that these \reverse identity matrices"J3andJ4havejJ3j=1

butjJ4j= +1? det 2

40 0 1

0 1 0

1 0 03

5 =1 but det2 6

640 0 0 1

0 0 1 0

0 1 0 0

1 0 0 03

7

75= +1:

In the rst case exchanging the rst and third rows makesJ3the identity matrix. In the second case exchanging the rst and fourth, and then the second and third row makesJ4the identity. An even number of row changes (forJ4) does not change the determinant, an odd number (forJ3) changes the sign. Since the determinant of the identity matrix is 1 in any dimension, this shows the claim.

12The inverse of a 2 by 2 matrix seems to have determinant = 1:

detA1= det1adbc db c a =adbcadbc= 1: What is wrong with this calculation? What is the correct detA1? Multiplying annbynmatrix by a factor multiplies the determinant by then-th power of this factor, in this case by

1(adbc)2, not1adbc. The correct calculation is

detA1= det1adbc db c a =adbc(adbc)2=1adbc:

28True or false (give a reason if true or a 2 by 2 example if false):

(a) IfAis not invertible thenABis not invertible. 1 2

True,jABj=jAjjBj= 0jBj= 0.

(b) The determinant ofAis always the product of its pivots. False, there might be an odd number of row exchanges. ExampleA=0 1 1 0 has pivots 1 and 1, but determinant1. (c) The determinant ofABequals detA- detB.

False, exampleA=1 0

0 0 andB=0 0 01 both have determinant 0, butAB=Ihas determinant 1.

Section 5.2

3Show that detA= 0, regardless of the ve nonzeros marked byx's:

A=2

4x x x

0 0x 0 0x3 5 What are the cofactors of row 1? What is the rank ofA? What are the 6 terms in detA. The rst two columns ofAare both multiples of (1;0;0), so they are dependent,Ais not invertible, and so detA= 0. The cofactors of row 1 are C 11=0x 0x = 0; C12=0x 0x = 0; C13=0 0 0 0 = 0: As seen above, the rst and second column are dependent. The rst and third column are independent, though, so the rank is 2. The 6 terms in detAare all zero.

5Place the smallest number of zeros in a 4 by 4 matrix that will guarantee detA= 0. Place

as many zeros as possible while still allowing detA6= 0. Placing 4 zeros in the rst row (or any other row or column) will guarantee detA= 0. Placing zeros in all o-diagonal entries will still allow detA6= 0 (if all the diagonal entries are non- zero).

23With 2 by 2 blocks in 4 by 4 matrices, you cannot always use block determinants:

A B 0D =jAjjDjbutA B C D

6=jAjjDj jCjjBj:

(a) Why is the rst statement true? SomehowBdoesn't enter.

Cofactor expansion along the rst column gives

A B 0D =a11 a

22b21b22

0d11d12

0d21d22

a21 a

12b11b12

0d11d12

0d21d22

=a11a22d 11d12 d 21d22
a21a12d 11d12 d 21d22
= (a11a22a21a12)d 11d12 d 21d22
=jAjjDj: (b) Show by example that equality fails (as shown) whenCenters. 3 ForA=D= 0 andB=C=Iwe havejAjjDj jBjjCj= 01 =1, but A B C D

0 0 1 0

0 0 0 1

1 0 0 0

0 1 0 0

= 1: (No need for calculations here, this matrix can be transformed into the identity matrix by two row exchanges, so the determinant is the same as the identity matrix.)

Section 6.1

2Find the eigenvalues and eigenvectors of these two matrices:

A=1 4 2 3 andA+I=2 4 2 4 The eigenvalues ofAare -1 and 5, those ofA+Iare 0 and 6. The corresponding eigenvectors are (2;1) and (1;1). A+Ihas the sameeigenvectors asA. Its eigenvalues are greaterby 1.

5Find the eigenvalues ofAandB(easy for triangular matrices) andA+B:

A=3 0 1 1 andB=1 1 0 3 andA+B=4 1 1 4 Eigenvalues of bothAandBare 3 and 1, eigenvalues ofA+Bare 3 and 5. Eigenvalues ofA+Bare not equal toeigenvalues ofAplus eigenvalues ofB.

21 The eigenvalues ofAequal the eigenvalues ofAT.This is because det(AI) equals

det(ATI). That is true because (AI)T=ATIand the determinant does not changeunder transposition of matrices. Show by an example that the eigenvectors ofAandATare

notthe same.

Example:A=0 1

0 0 has only 0 as an eigenvalue, with eigenvectorx=1 0 . However, A

Tx=0 0

1 0 1 0 =0 1 is not a multiple of1 0 , so1 0 is not an eigenvector ofAT.

29(Review) Find the eigenvalues ofA,B, andC:

A=2

41 2 3

0 4 5

0 0 63

5 andB=2

40 0 1

0 2 0

3 0 03

5 andC=2

42 2 2

2 2 2

2 2 23

5 Eigenvalues ofAare 1, 4, and 6, eigenvalues ofBare 2 andp3, eigenvalues ofCare 0, 0, and 6.

Section 6.2

2IfAhas1= 2 with eigenvectorx1=1

0 and2= 5 withx2=1 1 , useSS1to ndA.

No other matrix has the same's andx's.

4

A=SS1=1 1

0 1 2 0 0 5 11 0 1 =2 5 0 5 11 0 1 =2 3 0 5

11True or false: If the eigenvalues ofAare 2, 2, 5 then the matrix is certainly

(a) invertible

True, because the eigenvalues are non-zero.

(b) diagonalizable False, it might not be because of the repeated eigenvalue 2. (c) not diagonalizable False, it might be diagonalizable, e.g., it could just be the diagonal matrix with diagonal entries 2, 2, and 5.

14The matrixA=3 1

0 3 is not diagonalizable because the rank ofA3Iis one. Change one entry to makeAdiagonalizable. Which entries could you change? Changing the 1 to a 0 obviously makes the matrix diagonal, and thus diagonalizable. Changing any of the 3's to a dierent number makes it diagonalizable because it will have two dierent eigenvalues. Changing the 0 to something else will also lead to two dierent eigenvalues and make the matrix diagonalizable.

16(Recommended) Find andSto diagonalizeA1=:6:9

:4:1 . What is the limit of kas k! 1? What is the limit ofSkS1? In the columns of the limiting matrix you see the eigenvectors to= 1.

Eigenvalues are 1 and:3, so =1 0

0:3 . Corresponding eigenvectors are:9 :4 and1 1 soS=:91 :4 1 . The limit of kis1 0 0 0 , the limit ofAk=SkS1is11:3:9:9 :4:4quotesdbs_dbs50.pdfusesText_50

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