Matrix Theory and Differential Equations Homework 6 Solutions, 0/5/6 Question Find the general solution of the matrix system: x 3y + 5z 8t 5 x + 4y z + t Express your answer in the form of a particulaolution plus the general solution of the associated homogeneous system. What is the rank of the associated homogeneous system? How many degrees of freedom does the solution have? We row reduce the following augmented matrix, writing the second equation first for convenience: 4 8 5 4 0 9 44 0 9 0 4 93 0 9 0 4 93 0 9 0 4 0 9 Here we did the following sequence of row operations: 93 R R +R, R R, R 4R +R, R R, R R. The resulting matrix is in reduced row echelon form. The free variables are the variables z and t..
We may pick parameters u and v with z u and t v and now reduce the following augmented matrix, with the equations z u and t v added: 4 0 93 0 9 0 0 0 u 0 0 0 v 0 0 (93 4u) 0 9 0 0 0 u 0 0 0 v 0 0 (93 4u) 0 0 ( + 9u) 0 0 0 u 0 0 0 v 0 0 0 (93 4u + v) 0 0 0 ( + 9u v) 0 0 0 u 0 0 0 v Here we did the following sequence of row operations: 0 0 0 (93 4u + v) 0 0 ( + 9u) 0 0 0 u 0 0 0 v R 4 R 3 + R, R 9 R 3 + R, R R 4 + R, R R 4 + R. The solution is thus: [x, y, z, t] [93 4u + v, + 9u v, u, v] X p + X h, X p [93,, 0, 0], X h [ 4u + v, 9u v, u, v] u [ 4, 9,, 0] + v[,, 0, ]. Then X p is a particulaolution of the given system, as is easily checked and X h is the general solution of the associated homogeneous system, whose rank is two. The solution has two degrees of freedom..
Question Let A and B be the following matrices: A 3 4 B Determine the sum A + B and the products A, B, AB and BA. Also show that: (A + B) A + B + AB + BA. Then we have: A AB 3 4 B 3 4 BA A + B 3 4 3 + 3 4 + 4 8 3 8 [3, ].[3, ] [3, ].[, 4] [, 4].[3, ] [, 4].[, 4] 5 + 6 5 0 8 6 + 6 [5, 3].[5, ] [5, 3].[ 3, 4] [, 4].[5, ] [, 4].[ 3, 4] 3 8, [3, ].[5, ] [3, ].[ 3, 4] [, 4].[5, ] [, 4].[ 3, 4] A +B +AB+BA (A + B) 3 4 5 + 3 0 6 4 4 + 6 [5, 3].[3, ] [5, 3].[, 4] [, 4].[3, ] [, 4].[, 4] 8 0, 9 6 + 8 3 4 + 6 5 4 9 + 8 5 8 3 + 6 4 4 + 3 8 + 3 9 + 8 0 + 3 + + 8 4 8 3 0 4 + + 9 + 8 3 8 64 + 3 8 8 4 4 3 + 64 8 3 8 6 6 48 6 6 6 48 6 [8, ].[8, 3] [8, ].[, 8] [ 3, 8].[8, 3] [ 3, 8].[, 8] A + B + AB + BA. Note that (A + B) A + B + AB, since AB BA, here. 3 4 4, 3 9,
Question 3 Let A, B and C be the following matrices: A 3 B C 5 4 8 Compute the products AB and BA. Show that the equation AX C has solution X BC and compute that solution. Solve the equation Y A C for the unknown matrix Y. AB 3 [3, ].[, 5] [3, ].[, 3] [5, ].[, 5] [5, ].[, 3] 6 + 3 0 0 5 + 6 0 0, BA 3 [, ].[3, 5] [, ].[, ] [ 5, 3].[3, 5] [ 5, 3].[, ] 6 5 + 5 5 + 6 So we have AB BA I, where I is the identity two by two matrix, the diagonal matrix with all diagonal entries and all the off-diagonal entries zero. The matrices A and B are said to be inverses of each other. We check that IX XI X, for any two by two matrix X p q : 0 0. IX 0 0 p r q s [, 0].[p, r] [, 0].[q, s] [0, ].[p, r] [0, ].[q, s] p + 0 q + 0 0 + r 0 + s p q X, XI p q 0 0 [p, q].[, 0] [p, q].[0, ] [r, s].[, 0] [r, s].[0, ] p + 0 0 + q r + 0 0 + s p q Suppose that AX C. Multiply both sides of this equation on the left by the matrix B and use associativity of matrix multiplication to give: B(AX) BC (BA)X IX X, X BC. So any solution for X must be X BC. We show that this really is a solution, again using the associativity of matrix multiplication: AX A(BC) (AB)C IC C. So the unique solution of the equation AX C is X BC. X. 4
BC 5 4 8 [, ].[5, 8] [, ].[ 4, ] [ 5, 3].[5, 8] [ 5, 3].[ 4, ] 0 8 8 5 + 4 0 + 33 9 53 We check that X obeys the equation AX C, as required: X. AX 3 9 53 [3, ].[, ] [3, ].[ 9, 53] [5, ].[, ] [5, ].[ 9, 53] 6 5 + 53 0 95 + 06 5 4 8 C. Suppose that Y A C. Multiply both sides of this equation on the right by the matrix B and use associativity of matrix multiplication to give: CB (Y A)B Y (AB) Y I Y, Y CB. So any solution for Y must be Y CB. We show that this really is a solution, again using the associativity of matrix multiplication: Y A (CB)A C(BA) CI C. So the unique solution of the equation Y A C is Y CB. CB 5 4 8 0 + 0 5 6 55 8 + 33 [5, 4].[, 5] [5, 4].[, 3] [8, ].[, 5] [8, ].[, 3] 30 39 5 We check that Y obeys the equation Y A C, as required: Y. Y A 30 39 5 3 [30, ].[3, 5] [30, ].[, ] [ 39, 5].[3, 5] [ 39, 5].[, ] 90 85 30 34 + 9 + 50 5 4 8 C. 5
Question 4 Let A and I be the following matrices: A I 0 0 The matrix I is called the two by two identity matrix. Show that unique real numbers s and t exist such that A + sa + ti 0 and find s and t. Also find real numbers u and vm such that A 3 + ua + vi 0. A [3, 5].[3, ] [3, 5].[5, 4] [, 4].[3, ] [, 4].[5, 4] 9 0 5 + 0 6 8 0 + 6 35 4 6, A +sa+ti 35 4 6 + 3s 5s s 4s + t 0 0 t + 3s + t 35 + 5s 4 s 6 + 4s + t. We want this to be the zero matrix, so we need every entry to be zero. The top right entry gives the equation 35 + 5s 0, so s. Then the bottom left entry is 4 + 4 0. Next the top left entry gives the equation: 0 + 3s + t + t, so t. Then the bottom right entry is 6 + 4s + t 6 8 + 0. So we need s and t and then all entries are zero as required. For the last part, we have A A + I 0, so A A I. Multiply both sides of this equation by A on the left to give: A 3 A A (A I) A 49A 54I A A 54I. We check by computing both sides of this equation directly: A 3 A(A ) 35 4 6 [3, 5].[, 4] [3, 5].[35, 6] [, 4].[, 4] [, 4].[35, 6] 3 0 05 + 30 56 0 + 4 3 35 54 46, A 54I 8 35 54 08 + 54 0 0 54 8 54 35 + 0 54 + 0 08 54 3 35 54 46 A 3. 6
Question 5 Find a two by two matrix A that is neither the identity matrix nor the zero matrix, yet it obeys the equation A A. By analogy with the last problem, if A p q, then we find numbers u and v, such that A + ua + vi 0. A p q p r q s A +ua+vi p + qr q(p + s) r(p + s) s + qr [p, q].[p, r] [p, q].[q, s] [r, s].[p, r] [r, s].[q, s] Put u (p + s), so p + s + u 0. Then we have: p + qr q(p + s) r(p + s) s + qr, + up uq ur us + v 0 0 v p + qr + up + v q(p + s + u) r(p + s + u) s + qr + us + v. A (p + s)a + vi p + qr + p( p s) + v 0 0 s + qr + s( p s) + v qr ps + v 0 0 qr ps + v (v (ps qr))i. So we take u (p+s) and v ps qr and then: A (p+s)a+(ps qr)i 0, or A (p + s)a (ps qr)i. Then we have the required relation, namely A A, if p + s and ps qr 0. Put p ( + t) and s ( t). Then p + s, as required. In particular A is neither the identity matrix, for which p + s, nor the zero matrix for which p + s 0, as required. Then we have: ps qr 4 ( + t)( t) qr 4 ( t 4qr). So to solve the given problem we need only the relation t + 4qr to be satisfied. Put q (x + y) and r (x y), so x q + r and y q r; then we need t + x y, or t + x + y, so the general real solution is (t, x) + y (cos(θ), sin(θ)), with y and θ arbitrary real numbers. One particularly simple solution is θ y 0, for which we have t and x 0, so p and q r s 0, so A 0 0 0. So A is neither the identity matrix, nor the zero matrix and yet it is easy to check that A A, as required: A 0 0 0 0 0 0 [, 0].[, 0] [, 0].[0, 0] [0, 0].[, 0] [0, 0].[0, 0] + 0 0 + 0 0 + 0 0 + 0 A.