Math Matrix Theory - Spring 2012

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1 Math Matrix Theory - Spring 202 HW #2 Solutions Which of the following are true? Why? If not true, give an example to show that If true, give your reasoning (a) Inverse of an elementary matrix is elementary True If we set up the augmented matrix to find the inverse of some elementary matrix A, it will be of the form: 0 0 A 0 0 Since A is elementary, it takes exactly one elementary row operation (call it e 0 ) to move from A to I Since the inverse of e 0 is also an elementary row operation, taking (e 0 ) of the above augmented matrix gives 0 0 B 0 0, where B is an elementary matrix Furthermore, B = A, since we have the identity on the left So the statement is true (b) Inverse of a triangular matrix, if it exists, is triangular True Following the same procedure as in (a), we have an n n n n augmented matrix of the form: a a 2 a 3 a n 0 0 a 22 a 23 a 2n 0 a 33 a 3n 0 0 a nn 0 First, for each row, send R i a ii R i, getting leading s in each row:

2 a 2 a n a a a a 2n 0 a 22 a 22 a nn After this, as long as we only do row operations by subtracting multiples of lower rows, we can preserve the right triangular matrix on the right side and also reduce the left side to the identity matrix Thus, the inverse of a right triangular matrix (if it exists) is indeed a right triangular matrix The proof for left triangular follows similarly (c) Inverse of a symmetric matrix, if it exists, is symmetric True For any symmetric matrix A, we have A T Then AB = BA = I It will suffice to show B = B T : = A Call B the inverse of A (if an inverse exists) B = IB = I T B = (AB) T B = B T A T B = B T AB = B T (AB) = B T I = B T So indeed, the inverse of an invertible symmetric matrix is also symmetric (d) Inverse of a diagonal matrix, if it exists, is diagonal a 0 a True For any n n diagonal matrix A = 2 0, we then have A = 0 a n a a 2 0 and A is clearly also a diagonal matrix as long as a i 0 for all i Since A is only invertible if a i 0 for all i, the statement is proven to be true (e) Sum of elementary atrices is elementary False and 0 are both elementary matrices, but a n, is not an elementary matrix (f) Sum of diagonal matrices is diagonal True (Of course, must have the provision that they are the same size matrix) Let A = a 0 a a n, 2

3 and B = b 0 b Then A + B = a + b 0 a 2 + b 2 0 0, which is clearly a b n diagonal matrix a n + b n 2 Is it true that if A and B are two 3 3 diagonal matrices then AB = BA? If true prove it If not, give an example to show that it is not true Is this true for 3 3 triangular matrices? Do you think it is true for n n diagonal matrices? Why? Is it true for symmetric matrices? a b Let A = 0 a 2 0, and B = 0 b 2 0 a 3 b 3 a b b a Then AB = 0 a 2 b 2 0, while BA = 0 b 2 a 2 0 Clearly, AB = BA, so the statement is true a 3 b 3 b 3 a 3 It is not necessarily true for triangular matrices: let A = 0 and B = Then AB = 0 3 2, but BA = This counterexample disproves the second statement The third is true; a i b i = b i a i always, where a i and b i are the i-th diagonal positions of the n n diagonal matrices A and B For symmetric matrices in general, the statement is not true- recall from last homework that it was possible to construct two symmetric matrices A and B for which AB BA! 3

4 Determine if the matrices A = 2 0 and B = are invertible If invertible, find the inverse Since det A = 7, A is invertible We set up the augmented matrix calculation: ; R R ; R 2 R 2 + 2R ; R 3 R 3 + 3R : ; R 7R + 3R 2 ; R 2 7R 3 8R 2 : ; R 7R + 2R 3 ; R 2 7R 2 4R 3 ; R 3 7 R 3 : ; R 7 R ; R 2 7 R 2 : ; so our inverse matrix is A 2 7 = As for B, we see that det B = 0, so B has no inverse 4 Compute the determinant of A = 2 4 A = a a 2 b b 2 c c 2 Use this to compute the determinant of 4

5 det A = (bc 2 b c ) a(c 2 b 2 ) + a 2 (c b) = bc(c b) a(c b)(c + b) + a 2 (c b) = (c b)(bc + a( c b) + a 2 ) = (c b)(a c)(a b) So letting a = 2, b = 3, and c = 5, we get the given matrix A So det A = 2( 7)( 5) = 70 5 A and B are square matrices such that AB is invertible Prove that B must be invertible without using determinants Since AB is invertible, we know that for any vector x, (AB)y = 0 if and only if y = 0 Similarly, since A is invertible, Ay = 0 if and only if y = 0 Now, suppose for sake of contradiction that B is not invertible Then there must exist a vector x such that x 0 and Bx = 0 Then (AB)x = A(Bx) = 0, since Bx = 0 This contradicts the invertibility of AB, since x 0 Therefore, we conclude that B must be invertible 5

Math 54 HW 4 solutions

Math 54 HW 4 solutions 2.2. Section 2.2 (a) False: Recall that performing a series of elementary row operations A is equivalent to multiplying A by a series of elementary matrices. Suppose that E,...,