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Unformatted text preview: Math 2J Practice Midterm Solution
1. Solve the linear system x2 + x3 = 3 3 x1 + 5 x2 + 9 x3 = 2 x1 + 2 x2 + 3x3 = 3 3  2  3R1 + R2 3 1 2 3 0  1 0 0 1 1 3  11 3 1 2 3 0  1 0 0 0 1 3  11  8 Solution: 0 1 1 3 5 9 1 2 3 3 R R3  2 1 3 1 2 3 3 5 9 0 1 1 R2 + R3 x3 = 8 x2 = 11 x1 + 2 x2 + 3x3 = 3 x1 5 x2 = 11 x3 3 1 4 3  2. Find the inverse of matrix A = 2 7 6 by the following methods 1 7 2 (a) Augmented with I 3 and use reduced row echelon form to obtain A1 Solution: 1 4 3 6  2  7 1 7 2 1 4  3 1 0 0 2 R + R 0 1 0 1 0 2 0 1 0 0 1 R1 + R3 0 3 1 1 0 0 2 1 0  1 0 1  3R2 + R3 1 4  3 0 0 1 0 0 1 1 0 0 2 1 0  7  3 1 3R3 + R1 1 4 0 0 1 0 0 0 1  20  9 3  4 R2 + R1 1 0 0 2 1 0 0 1 0 0 0 1  7  3 1  28  13 3  28  13 3 1 2 1 0 A = 2 1 0  7  3 1  7  3 1 (b) Use the formula involving det( A) and adjA . Solution: A11 = (  1)
1+1 7 6 = 28 7 2 A21 = (  1) 2 +1 4 3 = 13 7 2 A31 = (  1) 3+1 4 3 =3 7 6 2 6 =2 1 2 7 1+ 3  2 A13 = (  1) = 7 1 7 A12 = (  1)
1+ 2 1 1 2+3 1 A23 = (  1) 1 A22 = (  1)
2+ 2 3 =1 2 4 = 3 7 1 3 =0 2 6 1 4 3+ 3 A33 = (  1) =1 2 7 A32 = (  1)
3+ 2  28  13 3 adjA = 2 1 0 , and  7  3 1 det( A) = 1 7 6 2 6 2 7 4 + ( 3) = 28 + 8 + 21 = 1 7 2 1 2 1 7  28  13 3 1 A = adjA = 2 1 0 det( A)  7  3 1 1 I 0 3. Show that the block matrix is invertible and find its inverse. A I Solution: A B Note: for a block matrix M = , det( M ) = det( AD  BC ) . C D I 0 det( M ) = = det I 2 = 1 0 M is invertible. A I ( ) M1 M 2 1 If M has an inverse the it must have the form M = M 3 M 3 M2 I 0 I 0 M 1 M 2 M 1 1 = We know that MM = = A I M 3 M 4 AM 1 + M 3 AM 2 + M 4 0 I M1 = I M =0 I 0 2 M 1 = AM 1 + M 3 = 0 A + M 3 = 0 M 3 =  A  A I AM 2 + M 4 = I M 4 = I 4. Combine the methods of row reduction and cofactor expansion to compute the determinant of matrix  1 4 3 2 1 3 0 3 A= 3 4 2 8  3 4 0 4 Solution: 3 2 1 4 1 3 0 3 1 3 0 3 1 3 0 3 3 2 1 4 0 7 1 13 A= =  =  3 4 2 8 3 4 2 8 0 13 2 1 3R1 + R2 3 4 0 4 R 0 13 0 13 R2 3 4 0 4 1 3R1 + R3 3R1 + R4 7 1 13 1 13 7 1 = 1 13 2 1 =  13  + 13 = 13(27)  13(27) = 0 13 2 2 1 13 0 13 5. Use Cramer's rule to compute the solutions of the system 2 x1 + x2 + x3 = 4  x1 + 2x3 = 2 3 x1 + x2 + 3 x3 = 3 Solution: 2 1 1 2 1 1 1 0 2 A = 1 0 2 A=  3 1 3 3 1 3 4 1 1 2 0 2 3 1 3 4 1 0 2 =  2 1 1 R R 3 1 3
1 2 1 0 2 1 5 =  0 1 5= =4 2R + R 1 9 0 1 9 3R + R
1 2 1 3 x1 = r A1 (b ) A = = 4 1 1 1 1 1 9  2 = 2 { 4  14} =  3 1 4 1 3 4 2 x2 = r A2 (b ) A = 2 4 1 1 2 2 3 3 3 4 1 2 2 1 2 2 1 1 1 8 5 57 = 2 4 1= 0 8 5= = 4 4 43 9 4 3 3 3 0 3 9 x3 = r A3 (b ) A = 2 1 4 1 0 2 3 1 3 4 = 1 4 2 1 1 1 5  2 = (1) { 7 + 2} =  1 3 3 1 4 4 4 3 1 6 7  an 3 6. Is 2 eigenvector of A = 3 7 If so, find the eigenvalue. ? 1 6 5 5 Solution: 1 r r r  is If x = 2 an eigenvector of A then there must exist a such that Ax = x . 1 3  6 7 1 2 1 r r 2 4 2 2 x = Ax = 3 7 3 =  = 6 5 1 2 1 5 Hence, this given vector is an eigenvector of A and the corresponding eigenvalue is = 2 . 5 2 7. Find the eigenvalues and eigenvectors of A = . 1 3 Solution: Characteristic equation: 2 2  8 + 17 = 0 (by the formula  tr ( A) + det( A) = 0 but this only work for a 2x2 matrix) by the determinant: 5 2 A  I = = ( 5  ) ( 3  ) + 2 = 15  3  5 + 2 + 2 = 2  8 + 17 = 0 1 3 8 64  4 17 =4 i 2 Eigenvalues: 1,2 = Eigenvectors: a r 1 For 1 = 4 + i , let u1 = the corresponding eigenvector we have be a 2 r r 1 2 a1 0  i = 0 1 1  i a 2 ( A  1I ) u1 = 0 ( 1  i ) a1  2a2 = 0 a2 = 1 i a1 2 r 2 u1 = 1  i r For 2 = 4  i , the corresponding eigenvector is the complex conjugate of u1 i.e. r 2 u2 = . 1 + i 8. Chapter 1 Test A (p.87) #3, 6, 8, 10
True or False: if TRUE prove it and if FALSE give an example (#3) An n n matrix A is nonsingular if and only if the reduced row echelon form of A is the identity matrix. Solution: TRUE I 1 If A : I i.e. we can row reduce A to I to get [ A  I ] :  A A has an inverse it's nonsingular. If A is nonsingular i.e. it has an inverse and the inverse is obtained by row reduced echelon form to I A : I . 2 (#6) If A and B are n n matrices, then ( A  B ) = A2  2 AB + B 2 Solution: FALSE 2 Because ( A  B ) = A2  AB  BA + B 2 and in general AB 1 1 , Ex: A = 1 1
2 2 BA 0 1 B= 0 0 1 1 0 0 ( A  B ) = = 1 2 1 1 2 0 A2  2 AB + B 2 = 2 0 Clearly, these two matrices are not equal. (#8) The produce of two elementary matrices is an elementary matrix. Solution: FALSE Because the product of two elementary matrices implies that 2 elementary operations have performed violate the definition of an elementary matrix. 1 1 1 0 0 0 0 0 0 1 0 and E = 1 0 E E = 1 0 2 0 2 Ex: Let E1 = 2 1 2 0 1 0 1 0 1 0 3 3 r r r r r r r r (#10) Let A be a 4x3 matrix with a2 = a3 . If b = a1 + a2 + a3 , then the system Ax = b will have infinite many solutions. Solution: TRUE r r r Writing A using column representation we have A = [ a1 a2 a3 ] r r Hence Ax = b becomes x 1 r r r r r r r r r x = [ a1 a2 a3 ] 2 x1a1 + x2 a2 + x3a3 = a1 + a2 + a3 x1 = 1, x2 = 1, x3 = 1 3 x solution. r r r r r b = a1 + 2a3 Since a2 = a3 r r Ax = b becomes x 1 r r r r r r r r r x = [ a1 a2 a3 ] 2 x1a1 + x2 a2 + x3a3 = a1 + 0a2 + 2a3 3 x a solution. This implies that the row echelon form of A must have a row of zero i.e. there is a free variable this system has infinitely many solution. 1 r x = is a 1 1 x1 = 1, x2 = 0, x3 = 2 1 r x = is 0 2 Test B (p.89) # 4, 5, 12 (#4) Let A be a matrix of the form A = where and are fixed scalars not 2 2 both equal to 0. 3 r (a) Explain why the system Ax = must be inconsistent. 1 Solution: r r r In general Ax = b is consistent if b can be written as a linear combination of the columns of A.
For the given system the linear combination of columns of A is 1 c1 c2 ( c1 + c2b ) + = 2 2 2 r however, the given vector b can't be expressed as a linear combination of r r Ax = b is inconsistent. 1 2 r r r (b) How can one choose a nonzero vector b so that the system Ax = b will be consistent? Explain. Solution: r 1 r A Just choose b to be a constant multiple of way when we row reduce  b this will 2 r r result in a row of zero the system will have infinite many solutions Ax = b is consistent. (#5) Given 2 2 0 1 3 1 3 1 3 2 7 B = 3 5 C = 2 7 A= 4 1 0 2 7 5 3 5 1 4  3 5 (a) Find an elementary matrix E such that EA = B Solution: 1 0 0 0 1 E= 0 1 0 0 (b) Find an elementary matrix F such that AF = B Solution: Notice C is done by a column operation not row operation i.e. C is obtained by ( 2 ) Col2 + Col1 of A 1 0 0 2 1 0 F=  0 1 0 (#12) Let A and B be 10x10 matrices that are partitioned into submatrices as follows: A A21 B 11 B21 A = 11 B= A B 21 A22 21 B22 (a) If A11 is a 6x5 matrix, and B11 is a k r , what conditions if any must k and r satisfy in order to make the block multiplication of A and B possible? Solution: k = 5 and 1 r 9 (b) Assuming the block multiplication is possible, how would the (2,2) block of the product be determined? C 11 C21 Solution: let C = AB = then C22 = A21 B12 + A22 B22 . C 21 C22 9. Chapter 2 Test A (p.113) #1, 4, 5, 7, 9
True or False: if TRUE prove it and if FALSE give an example (#1) det( AB ) = det( BA) Solution: TRUE det( AB) = det( A) det( B) = det( B) det( A) = det( BA) . (#4) det ( ( AB ) ) = det( A) det( B) .
T Solution: TRUE In general det(C T ) = det(C )
T Let C = AB det C = det ( ) ( ( AB ) ) = det(C ) = det( AB) = det( A) det( B) .
T (#5) det( A) = det( B) implies A = B . Solution: FALSE 2  2 3 0 Ex: A = B = 4 0 0 6 det( A) = 12 = det( B) but A B . (#7) A triangular matrix is nonsingular if and only if its diagonal entries are all nonzero. Solution: TRUE Suppose A is a nonsingular triangular matrix det( A) 0 And we know that for a triangular matrix the det( A) = product of diagonal entries in order for det( A) 0 all of the diagonal entries must be nonzero. The converse is straight forward just go backward from the proof above. (#9) If A and B are row equivalent matrices, then their determinants are equal. Solution: FALSE 0 2 2 0 : = Ex: A = 2 B here I switch the two rows of A to get B. 2 0 0 det( A) = 4 and det( B ) = 4 which are not equal! Test B (p.113) # 2. x 1 1 1 x 1 (#2) Given A = 1 1 x  (a) Compute the value of det(A) Solution: x 1 1 x 1 1 1 1 x A = 1 x 1 = x  + = x( x 2  1)  ( x  1) + (1  x) = x 3  3 x 1 x 1  x 1 1 1 1 x (b) For what values of x will the matrix be singular? Explain. Solution: If A = 0 then A is singular for x = 0 or 3 A will be singular. ...
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This note was uploaded on 03/11/2012 for the course MATH 2J 44360 taught by Professor Donaldson,neil during the Spring '10 term at UC Irvine.
 Spring '10
 DONALDSON,NEIL

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