Math 343 Midterm I Fall 006 sections 00 003 Instructor: Scott Glasgow
1 Assuming A B are invertible matrices of the same size, prove that ( ) 1 1 AB B A 1 = (11) B A 1 1 is the inverse of AB if only if ( )( ) ( )( ) AB B A = B A AB = I, (1) to whit we first note that, by the associative property of matrix multiplication, ( )( ) = ( ) AB B A A BB A ( )( ) = ( ) B A AB B A A B (13) Then, by the definition of the inverses, in particular that an inverse is both a right a left inverse, we have, respectively, that 1 1 ( AB)( B A ) = A( BB ) A = AIA = A( IA ) 1 1 ( B A )( AB) = B ( A A) B = B IB = B ( IB) (14) In (14) we have also used the associative property of matrix multiplication again Using now the fact that the identity matrix is in fact the multiplicative identity we get ( )( ) ( ) AB B A = A IA = AA ( )( ) ( ) B A AB B IB B B = = (15) Finally we use again the definition of the inverses In particular, using that an inverse is both a right a left inverse, we have, respectively, that
( )( ) 1 1 1 AB B A AA I ( )( ) = = 1 1 1 B A AB = B B = I, (16) which is the required (1) Use Gaussian elimination, noting along the way the various relevant row operations their relationship to the (evolving calculation of) the determinant, to show that 1 x x = 1 z z ( )( )( ) det 1 y y x y y z z x (17) Be sure to note along the way the various relevant row operations to your Gaussian elimination their accurate relationship to the (evolving calculation of) the determinant We indicate the various Gaussian elimination row operations in the upper right-h corner of the matrix (whose determinant is being calculated), making sure to use the correct relationship of such to the evolving determinant calculation: ( y x) ( z x) R R1 R/ R3 R1 R3/ det 1 det 0 det 0 1 R3 R 1 x x 1 x x 1 x x y y = y x y x = ( y x)( z x) y+ x = 1 z z 0 z x z x 0 1 z x + R3 R 1 x x = ( y x)( z x) det 0 1 y+ x = ( y x)( z x) ( 1 1 ( z y) ) = ( x y)( y z)( z x) 0 0 z y (18) In the final steps above we used that a) the determinant of a triangular matrix is the product of its diagonal entries, b) some obvious algebraic properties of multiplication (of scalars)
3 By using the permutation definition of the determinant, prove that if square matrix B is the same as square matrix A, except that one of B s rows is a scalar multiple k of the corresponding row of A, then det B = kdet A (19) Let it be the j th row of the matrix B that is the same as the scalar k multiplied by the j th row of the matrix A Then, by definition of the determinant, the relationship between the rows (of matrix A matrix B ), we have d( p) d( p) ( ) ( ) ( ) det B: = 1 b b b = 1 a ka a 1 p(1) jp( j) np( n) 1 p(1) jp( j) np( n) p:[ n] [ n] p:[ n] [ n] d( p) ( 1 ) 1 p(1) jp( j) np( n) : det p:[ n] [ n] = k a a a = k A 4 Prove that, if the matrix A is invertible, the system A x = bhas one only one solution x, namely x = A 1 b (Warning: there are two things to prove here, namely a) that if the system has a solution, then it can only be x = A 1 b, that b) x = A 1 bactually does solve the system Here then you will have addressed the one only one issues in reverse order: you first show that a) there is at most one solution, b) that there is in fact one solution (rather than none) In parts a) 1 b) you will use that A is A s left right inverse, respectively) If the system has a solution x, then, for any such x,we may write (without implicitly lying), then, by left application of associative property of matrix multiplication, obtain that (110) A x = b (111) ( ) ( ) 1 1 1 A A A A A I 1 A to (111), as well as by the b= x = x = x = x (11) 1 In (11) we also used that A is a left inverse of A, as well as the fact that the so-called identity matrix I is in fact a multiplicative identity Here then we have just showed that if (111) has a solution, it s got to be x = A 1 b Thus we have showed that (111) has at most one solution But our demonstration does not yet preclude their being no solution To do that, we confirm that, for the only promising cidate x = A 1 b, we get
1 1 ( ) ( ) Ax = A A b = AA b= Ib= b, (113) so that our cidate was successful (Here we have used the associative property of 1 matrix multiplication, the fact that A is a right inverse of A, as well as the fact that the so-called identity matrix I is in fact a multiplicative identity ) Thus we have showed that the system has one only one solution, namely x = A 1 b 5 The graph of the function contains the points ( 1,1 ), ( 1, 1) (,4) y = ax + bx+ c (114) Find the values of the constants a, b, c by forming a relevant augmented matrix performing Gauss-Jordan elimination on it The stated facts dictate the three equations () () a 1 + b 1 + c= a+ b+ c = 1 ( ) ( ) a 1 + b 1 + c = a b+ c= 1 ( ) ( ) a + b + c= 4a b+ c= 4 (115) We row reduce the associated augmented matrix to reduced row echelon form as follows: R R1 R/( ) 1 1 1 1 R3 4R1 1 1 1 1 R3/( 3) 1 1 1 1 1 1 1 1 0 0 0 1 0 1 4 1 4 0 6 3 0 0 1 0 R1 R3 R1 R 1 1 1 1 1 1 0 3 1 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1 0 0 1 R3 R (116) The latter indicates that a =, b= 1, c =
6 Assume that both the matrix B the matrix C are inverses of the matrix A Show that B C are just two aliases for the same matrix, ie show that in fact B = C The descriptions of B C dem that AB = BA = I = AC = CA (117) Using the associative property of matrix multiplication in two different ways on the product BAC we get so that indeed ( ) ( ) BAC = B AC = BI = B BAC = BA C = IC = C, B = BAC = C B = C (118) (119) as claimed Note that in (118) we also used that a) C is a right inverse of A, b) B is a left inverse of A, that c) the identity matrix acts as both a right left multiplicative identity 7 The eigenvalues of a (square) matrix B turn out to be the zeroes λ of the characteristic polynomial PB ( λ) : = det( λi B), ie the eigenvalues of B are the roots λ of the characteristic equation det( λi B) = 0 (where I denotes the identity matrix of the same size as B ) Find the eigenvalues of the matrix B = 6 3 (10) What is the determinant of this matrix? What is the product of its eigenvalues?
The characteristic equation is λ 0= det( λi B) = det = λ λ 3 6 6 λ 3 ( )( ) ( )( ) ( )( ) = λ 5λ + 6 1= λ 5λ 6= λ + 1 λ 6, (11) so that the eigenvalues are clearly 1 6 The determinant of the matrix is 3 6 6 1 6 1 6 = 6also ( )( ) ( )( ) = =, while the product of the eigenvalues is ( )( ) 8 Find the inverse of the matrix a b A = c d (1) by row reducing A I to 1 I A Assume the parameters abc,,, ddo not take on any special values, nor have a special relationship among them that is row reduce naively, without worrying about any divisions by hidden zeros The naïve row reduction mentioned proceeds as follows: ( bc) ( ) ar cr1 ad bc R1 br a b 1 0 a b 1 0 AI = c d 0 1 0 ad bc c a R1/ a R1/ a R1/ a a ad 0 ad ab ad bc 0 d b 0 ad bc c a 0 ad bc c a 1 0 1 d b 1 = IA 0 1 ad bc c a (13) 9 When is the product of two symmetric matrices symmetric? Prove this (Assume that the product of the two matrices makes sense, ie assume the two matrices are the same size)
The product AB of two matrices A B is, by definition, symmetric if only if AB = ( AB) T On the other h we have proved that ( ) T T T AB B A =, which, under the present circumstances, is the product BA Thus, if the matrices A B are symmetric, their product is symmetric if only if AB = BA, ie if only if A B commute 10 As in problem 7, let B be the matrix let PX ( ): ( X I)( X 6I) the identity Calculate PB ( ) ( B I)( B 6I) 6 3, (14) = +, where X is any by matrix, where I denotes = + (it s easier to calculate as written do not exp the polynomial factors unless you want to work harder than necessary) Does the result surprise you (given the outcomes of problem 7)? We have 1 0 3 B+ I = 6 3 + 0 1 = 6 4 6 0 4 B 6 I =, 6 3 + = 0 6 6 3 (15) so that ( ) ( ) ( ) ( ) ( ) + ( ) ( ) + ( ) 3 4 3 4 + 6 3 + 3 PB ( ) = ( B+ I)( B 6I) = 6 4 6 3 = 6 4 46 6 4 3 (16) 1 + 1 6 6 0 0 = = 4 + 4 1 1 0 0