Solutions to Assignment 2

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1 MATH 235/W08 Determinants Solutions to Assignment 2 Hand in questions 1,2,3,4,5,6,7,8 Several solutions are found using MATLAB. The MATLAB file hwolkowi/henr contains a diary command and a echo on command. 1. Evaluate the determinants of the following matrices: (a) (b) (c) (d) 1 i 1 + i 2 i i 3 1 i π a b c 2a + 3d 2b + 3e 2c + 3f, where a, b, c, d, e, f C. d e f BEGIN SOLUTION: disp( solutions for assign 2 in W08 ) solutions for assign 2 in W08 disp( ) disp( Prob 1. Evaluate the det of the following four matrices ) 1

2 Prob 1. Evaluate the det of the following four matrices A1=[ 1 i 1+i 2-i i 3 1-i A1 = i i i i i ] det(a1) i disp( or by expanding along row 2 we get ) or by expanding along row 2 we get -(2-i)*det([i 1+i;3 1-i]) +3*det([1 1+i;-1+i 1-i]) i A2=[ pi A2 = ] det(a2)

3 72 disp( exchange rows 1,2 and rows 3,4 to get a triangular matrix ) exchange rows 1,2 and rows 3,4 to get a triangular matrix A2copy=A2([ ],:) A2copy = prod(diag(a2copy)) 72 pause a=sym( a ) a = a b=sym( b ) b = b c=sym( c ) c = c d=sym( d ) d = 3

4 d e=sym( e ) e = e f=sym( f ) f = f A3=[ a b c 2*a+3*d 2*b+3*e 2*c+3*f d e f A3 = [ a, b, c] [ 2*a+3*d, 2*b+3*e, 2*c+3*f] [ d, e, f] ] disp( row 2 is a linear combination of rows 1 and 3 - so det is 0 - verify ) row 2 is a linear combination of rows 1 and 3 - so det is 0 - verify det(a3) 0 A4=[ A4 =

5 ] det(a4) -144 disp( exchange two rows ) exchange two rows A4copy=A4([ ],:) A4copy = disp( exchange two cols ) exchange two cols disp( the sign after the two exchanges is + ) the sign after the two exchanges is + A4copy=A4copy(:,[ ]) A4copy = det(a4copy) 5

6 144 disp( by expanding along row 1 repeatedly: det is product along diagonal ) by expanding along row 1 repeatedly: det is product along diagonal prod(diag(a4copy)) 144 END SOLUTION. 2. Find the value of the determinants of the following matrices. Explain your answer in terms of cofactor expansion: (a) (b) The matrix A below: A = The ijth entry of A is 1 if i + j = 101; all other entries are 0. BEGIN SOLUTION: disp( Prob 2. Evaluate the det of the following two matrices - use cofactors ) Prob 2. Evaluate the det of the following two matrices - use cofactors A1=zeros(4); A1(1,4)=1; A1(2,3)=-1; A1(3,2)=1; A1(4,1)=-1 A1 = 6

7 det(a1) disp( or by cofactors: det(a1)= a_14 * C14 ) or by cofactors: det(a1)= a_14 * C14 disp( and expansion along first row of A14: C14 = - (-1)(0+1) = 1 ) and expansion along first row of A14: C14 = - (-1)(0+1) = 1 disp( so det(a1) = 1 ) so det(a1) = 1 the matlab commands: for n=2:8, A2=zeros(n); A2(n:n-1:end-(n-1))=ones(n,1); disp([ dimension of A and det(a) are:,num2str([n det(a2)])]) end give the output: dimension of A and det(a) are: 2-1 dimension of A and det(a) are: 3-1 dimension of A and det(a) are: 4 1 dimension of A and det(a) are: 5 1 dimension of A and det(a) are: 6-1 dimension of A and det(a) are: 7-1 dimension of A and det(a) are: 8 1 In the above problem where A2 = A is n = , we can conjecture that the sign alternates in steps of 2 from dimension n = 2. Therefore, 7

8 since 50 = 100/2 is even, we should get that det(a) = 1. The above alternating sign statement can be proved by looking at the four possible cases for k k matrices: A 1k i.e. 1 + k even and det(a 1k ) = k even and det(a 1k ) = k odd and det(a 1k ) = k odd and det(a 1k ) = 1 END SOLUTION. 3. Evaluate the following determinants by row-reducing to an upper triangular matrix: (a) (b) (c) BEGIN SOLUTION: disp( solutions for assign 2 in W08 ) solutions for assign 2 in W08 disp( ) disp( Prob 3. Evaluate the det - use row reduction ) Prob 3. Evaluate the det - use row reduction 8

9 A1orig=[ A1orig = ] A1=A1orig; det(a1) -72 A1=[A1 sum(a1,2)] A1 = A1(2,:)=A1(2,:)-A1(1,:) A1 = A1(3,:)=A1(3,:)-A1(1,:) A1 =

10 A1(4,:)=A1(4,:)-A1(1,:) A1 = A1(4,:)=A1(4,:)-A1(3,:) A1 = disp( sign change ) sign change A1=A1([ ],:) A1 = A1(4,:)=A1(4,:)-2*A1(3,:) A1 = disp( det is now minus prod. along diagonal ) det is now minus prod. along diagonal -prod(diag(a1)) 10

11 -72 A2orig= [ A2orig = ] A2=A2orig; det(a2) 24 A2=[A2 sum(a2,2)] A2 = A2(2,:)=A2(2,:)-2*A2(1,:) A2 = A2(3,:)=A2(3,:)+A2(1,:) A2 = 11

12 A2(3,:)=A2(3,:)-A2(2,:) A2 = A2(4,:)=A2(4,:)-A2(2,:) A2 = A3orig=[ A3orig = ] A3=A3orig; A3=[A3 sum(a3,2)] A3 = 12

13 disp( sign change ) sign change A3=A3([ ],:) A3 = A3(2,:)=A3(2,:)+2*A3(1,:) A3 = disp( second sign change ) second sign change A3=A3([ ],:) A3 = A3(3,:)=A3(3,:)+3*A3(2,:) A3 =

14 disp( third sign change ) third sign change A3=A3([ ],:) A3 = A3(4,:)=A3(4,:)+4*A3(3,:) A3 = disp( fourth sign change ) fourth sign change A3=A3([ ],:) A3 = A3(5,:)=A3(5,:)+5*A3(4,:) A3 =

15 disp( det is prod. of diag els ) det is prod. of diag els prod(diag(a3)) 1 END SOLUTION. 4. Suppose that A M 4 4 (C) i.e. a 4 4 matrix with complex entries. Suppose that the columns of A are given by the 4 vectors v 1, v 2, v 3, v 4 and det(a) = 9. Find: det ( v 1 v 4 v 3 v 2 ) ; det ( 6v1 + 2v 4 v 2 v 3 3v 1 + v 4 ) BEGIN SOLUTION: (a) Two columns are interchanged; therefore, det is -9, i.e. a sign change. (b) Column 1 is 6v 1 + 2v 2 which is twice column 4. Therefore, det is 0. END SOLUTION. 5. Find the determinant of the matrix representation (with respect to the standard basis) of the linear operator T(p(t)) = p(4t 1), where T : P 2 P 2 i.e. T is a linear operator on the vector space of polynomials of degree 2. BEGIN SOLUTION: We need to find the matrix representation. (a) T(1) = 1, i.e. the constant polynomial p(t) = 1 remains constant. We get column 1 is: (b) T(t) = 4t 1, i.e. the variable t is replaced by 4t 1. We get column 2 is: (c) T(t 2 ) = (4t 1) 2 = 16t 2 8t + 1. We get column 3 is: Therefore the matrix representation is triangular The determinant is 64. END SOLUTION. 15

16 6. Use the determinant ( of an( appropriate matrix to find the area of the triangle defined by and ) 2) BEGIN SOLUTION: We could use the matrix formed from the vectors as rows or columns; we get, 6 56 /2 = 50/2 = 25. END SOLUTION. 7. Let A be an n n matrix with integer entries and with det(a) = 1. Are the entries of A 1 integers? Why? BEGIN SOLUTION: Yes, the inverse has integer entries. This arises 1 by applying Cramers rule or the formula for the inverse det(a) adj(a). Therefore the integer entries are divided by 1. END SOLUTION. 8. MATLAB: You will need the two Teaching Code file cofactor.m available from the course webpage or Use the diary command to save your input/output in MATLAB, e.g. diary inoutassign2.txt will save your output in the file inoutassign2.txt. (a) Create a random 5 5 matrix A using the MATLAB randn command. (Use help randn in MATLAB if you need to.) (b) Evaluate the determinant using the MATLAB det command. Then exchange rows 3 and 5: B = A(:,[ ]). Use MATLAB to verify that det B = - det A. (c) Now let C be the matrix obtained from A by adding 3 times row 4 to row 1. Again, use MATLAB to verify that now the determinant is unchanged. (d) The file cofactor.m calculates the matrix of cofactors of a square matrix. Construct an integer matrix from A using D=round(2*A). Then find the cofactor matrix using COF=cofactor(D). The adjunct matrix is ADJ=COF. Find the inverse of D using ADJ. (e) Generate a random integer vector b to solve the system Dx=b using Cramer s rule. Verify that you have the solution using the \ command in MATLAB. BEGIN SOLUTION: A=randn(5,5); det(a) 16

17 det(a([ ],:)) det(a([ ],:)) +det(a) 0 C=A; C(1,:)=C(1,:)+ 3*C(4,:); det(c)-det(a) 0 D=round(2*(A+1)) D = COF=cofactor(D) COF = ADJ=COF ADJ = 17

18 Dinv=ADJ/det(D) Dinv = Dinv*D b=round(2*randn(5,1)); for i=1:5 Di=D; Di(:,i)=b; x(i)=det(di)/det(d); Di=D; Di(:,i)=b; x(i)=det(di)/det(d); Di=D; Di(:,i)=b; x(i)=det(di)/det(d); Di=D; Di(:,i)=b; x(i)=det(di)/det(d); Di=D; Di(:,i)=b; x(i)=det(di)/det(d); end x 18

19 x = D\b-x e-015 * diary off END SOLUTION. 9. For each of the following, find all values of x for which the matrix is invertible. (a) cos x 1 sin x sin x 3 cos x (b) 1 1 x 1 x x x x x 10. Theorem 9, p. 205, of Lay interprets deta in terms of area and volume. This generalizes to Theorem: The k volume of a k parallelepiped defined by vectors v 1,..., v k in R n is det(a T A) where A is the n k matrix whose column vectors are v 1,..., v k. When k = n, this reduced to deta. Note: 2-volume = area; 2-parallelepiped= parallelogram, etc. Apply this Theorem to answer the following problems: (a) Find the area of the parallelogram defined by the vectors and

20 (b) Find the volume of the parallelpiped defined by the vectors , 1 1, [ ] A B 11. (a) Prove that det = (det A)(det D), if A and D are square 0 D blocks not necessarily of the same size. (Hint: Expand along the last row of D and use induction.) [ ] A B (b) Let a 2n 2n matrix M be in the form M =, where C D each block is an n n matrix. Suppose that A is invertible and that AC = CA. Prove that det M = det(ad CB). (Hint: Post-multiply M by the triangular matrix [ ] In A 1 B, whose determinant has value 1.) 0 I n 20

Determinants Chapter 3 of Lay

Determinants Chapter 3 of Lay Determinants Chapter of Lay Dr. Doreen De Leon Math 152, Fall 201 1 Introduction to Determinants Section.1 of Lay Given a square matrix A = [a ij, the determinant of A is denoted by det A or a 11 a 1j