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Unformatted text preview: Solutions to problems from section 2.1 Throughout this assignment you were to assume the each matrix expression is deﬁned. 4. Compute A − 5I3 and (5I3 )A when 9 −1 3 A = −8 7 −6 . −4 1 8 Solution: 4 −1 3 9 −1 3 500 A − 5I3 = −8 2 −6 . 7 −6 − 0 5 0 = −8 005 −4 1 3 −4 1 8 45 −5 15 (5I3 )A = 5(I3 A) = 5A = −40 35 −30 . −20 5 40 6. Compute AB in two diﬀerent ways: (a) by the deﬁnition, where Ab1 and Ab2 are computed separately, and (b) by the rowcolumn rule for computing AB . 4 −2 A = −3 0 , 3 5 Solution: 4 −2 (a) Since Ab1 = −3 0 3 5 we see 0 4 −2 1 = −3 and Ab2 = −3 0 2 13 3 5 0 14 AB = Ab1 Ab2 = −3 −9 13 4 3 −1 14 = −9 4 1 3 2 −1 B= . (b) 4 −2 AB = −3 0 3 5 10. Let A = B = C. Solution: AB = 2 −3 −4 6 84 55 = 1 −7 −2 14 = 2 −3 −4 6 5 −2 3 1 = AC 2 −3 ,B= −4 6 1 3 2 −1 4(1) − 2(2) 4(3) − 2(−1) 0 14 = −3(1) + 0(2) −3(3) + 0(−1) = −3 −9 3(1) + 5(2) 3(3) + 5(−1) 13 4 5 −2 . Verify that AB = AC and yet 3 1 84 , and C = 55 12. Let A = 3 −6 . Construct a 2 × 2 matrix B such that AB is the zero matrix. Use two −1 2 diﬀerent nonzero columns for B . ab Solution: You can either guess and check for a while, or you can set B = so that cd AB = 3 −6 −1 2 ab cd = 3a − 6c 3b − 6d −a + 2c −b + 2d Thus we’re solving the homogeneous linear system 3a − 6c = 0 −a + 2c = 0 3b − 6d = 0 −b + 2d = 0 which has augmented matrix 3 0 −6 0 −1 0 2 0 0 3 0 −6 0 −1 0 2 which says that a = 2c and Thus any solution of the form 2c 2d cd with distinct, nonzero c and d will do. 16. Suppose A, B , and C are any matrices for which the indicated sums and products are deﬁned. True or False? Justify your answer. a. If A and B are 3 × 3 and B = [ b1 b2 b3 ], then AB = [ Ab1 + Ab2 + Ab3 ]. Solution: FALSE. [ Ab1 + Ab2 + Ab3 ] is a 3 × 1 matrix and AB is a 3 × 3 matrix. Thus they can never be equal. b. The second row of AB is the second row of A multiplied on the right by B . Solution: TRUE. In class we noticed that rowi (AB ) = rowi (A)B for all i. In this notation, the statement above says row2 (AB ) = row2 (A)B . c. (AB )C = (AC )B . Solution: FALSE. For example, if A = I2 , B = (AB )C = whereas (AC )B = 10 01 1 0 0 −1 11 01 = 1 0 0 −1 11 01 = 1 1 0 −1 10 01 11 01 1 0 0 −1 = 11 , and C = 01 11 01 1 0 0 −1 1 0 0 −1 = then b = 2d. 0 0 0 0 NR1 =R1 +3R2 NR3 =R3 +3R4 −→ 0 0 −1 0 0 0 0 −1 0 2 0 0 0 0 0 2 0 0 0 0 1 −1 0 −1 [You should realize this statement is false since matrix multiplication is noncommutative.] d. (AB )T = AT B T . Solution: FALSE. For example 01 00 whereas 01 00
T 11 11 11 11
T T = 11 00 T = 10 10 , = 00 10 11 11 = 00 11 e. The transpose of a sum of matrices equals the sum of their transposes. Solution: TRUE. This is part (b) of theorem 3 on page 115. 22. Show that if the columns of B are linearly dependent, then so are the columns of AB . Solution 1: The columns of a matrix M are linearly dependent if and only if the equation M x = 0 has a nontrivial solution (see the blue box on page 66). So if the columns of B are linearly dependent, then there exists a nonzero vector u with B u = 0. But then (AB )u = A(B u) = A0 = 0. Thus x = u is a nontrivial solution to the equation (AB )x = 0, which implies the columns of AB are also linearly dependent. Solution 2: The columns of a matrix M are linearly dependent if and only if the matrix transformation x → M x is not onetoone (theorem 12 part (b) on page 89). So if the columns of B are linearly dependent, then there exists two vectors v = u with B u = B v. But then (AB )u = A(B u) = A(B v) = (AB )v. Thus the matrix transformation x → (AB )x is not onetoone, which implies the columns of AB are linearly dependent. 32. Show that AIn = A when A is an m × n matrix. Solution: The column deﬁnition of matrix multiplication says that the ith column of AIn is the product of A with the ith column of In . But the ith column of In is what we’re calling ei . Thuw we have AIn = [ Ae1 Ae2 · · · Aen ] But Aei is the ith column of A (you can see this by inspection, or it follows from theorem 10 on page 83). Thus AIn = A. 34. Give a formula for (AB x)T , where x is a vector and A and B are matrices of appropriate sizes. Solution: Using theorem 3 part (d) on page 115 we have (AB x)T = xT (AB )T = xT B T AT . ...
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 Fall '08
 AXENOVICH

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