3e-spring2011-chapter_1_tf

3e-spring2011-chapter_1_tf - True or False Ch 1.TF.11 T;...

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Unformatted text preview: True or False Ch 1.TF.11 T; Find rref. Ch 1.TF.12 T; Find rref Ch 1.TF.13 F; Consider the 4 × 3 matrix A that contains all zeroes, except for a 1 in the lower left corner. Ch 1.TF.14 F; Note that A Ch 1.TF.15 F; The rank is 1. Ch 1.TF.16 F; The product on the left-hand side has two components. −3 Ch 1.TF.17 T; Let A = −5 −7 0 0 , for example. 0 2 1 = 2A 2 1 for all 2 × 2 matrices A. 7 4 1 Ch 1.TF.18 T; We have 2 = 2 5 − 8 . 9 6 3 Ch 1.TF.19 T; The last component of the left-hand side is zero for all vectors x. Ch 1.TF.20 T; A = 3 4 0 , for example. 0 1 10 0 1 and B = 0 0 00 always end up with a 00 1 0 , for example. We can apply elementary row operations 00 matrix that has all zeros in the first column. 0 Ch 1.TF.21 F; Let A = 0 0 to A all we want, we will Ch 1.TF.22 T; If u = av + bw and v = cp + dq + er, then u = acp + adq + aer + bw. Ch 1.TF.23 F; The system x = 2, y = 3, x + y = 5 has a unique solution. Ch 1.TF.24 F; Let A = 01 , for example. 00 2 0 1 and b = 3 , for example. 5 1 1 Ch 1.TF.25 F; Let A = 0 1 Ch 1.TF.26 T, by Exercise 1.3.44. Ch 1.TF.27 F; Find rref to see that the rank is always 2. Ch 1.TF.28 T; Note that v = 1v + 0w. 45 Chapter 1 0 2 1 , for example. ,w = ,v = 1 0 0 Ch 1.TF.29 F; Let u = Ch 1.TF.30 T; Note that 0 = 0v + 0w 1 1 . . . . F; If A 2 = 0, then x = 2 is a solution to A.0 . However, since rank(A) = 3, rref A.0 = 3 3 00 0.0 . , meaning that only 0 is a solution to Ax = 0. . 10 00 Ch 1.TF.31 1 0 0 0 0 1 0 0 Ch 1.TF.32 F; If b = 0, then having a row of zeroes in rref(A) does not force the system to be inconsistent. Ch 1.TF.33 T; By Example 3c of Section 1.3, the equation Ax = 0 has the unique solution x = 0. Now note that A(v − w) = 0, so that v − w = 0 and v = w. Ch 1.TF.34 T; Note that rank(A) = 4, by Theorem 1.3.4 Ch 1.TF.35 F; Let u = 2 1 0 , v= , w= , for example.. 0 0 1 −2t Ch 1.TF.36 T; We use rref to solve the system Ax = 0 and find x = −3t , where t is an arbitrary constant. t −2 Letting t = 1, we find [u v w] −3 = −2u − 3v + w = 0, so that w = 2u + 3v . 1 Ch 1.TF.37 F; Let A = B = 10 , for example. 01 0 0 . Running the elementary 0 1 into I and then I into B . 1 0 ... 0 1 ... Ch 1.TF.38 T; Matrices A and B can both be transformed into I = ... ... ... 0 0 0 operations backwards, we can transform I into B . Thus we can first transform A Ch 1.TF.39 T; If v = au + bw, then Av = A(au + bw) = A(au) + A(bw) = aAv + bAw. Ch 1.TF.40 T; check that the three defining properties of a matrix in rref still hold. F; If b = 0, then having a row of zeroes in rref(A) does not force the system to be inconsistent. . Ch 1.TF.41 T; Ax = b is inconsistent if and only if rank A.b = rank(A)+1, since there will be an extra leading . one in the last column of the augmented matrix: (See Figure 1.16.) 46 True or False Figure 1.16: for Problem T/F 41. Ch 1.TF.42 T; The system Ax = b is consistent, by Example 3b, and there are, in fact, infinitely many solutions, by Theorem 1.3.3. Note that Ax = b is a system of three equations with four unknowns. . Ch 1.TF.43 T; Recall that we use rref A.0 . . to solve the system Ax = 0. Now, rref A.0 . = . rref(A).0 . = . . . . rref(B ).0 = rref B .0 . Then, since rref(A).0 = rref(B ).0 , they must have the same solutions. . . . . 12 . If we remove the first column, then the remaining matrix fails to be in rref. 00 Ch 1.TF.44 F; Consider Ch 1.TF.45 T; First we list all possible matrices rref(M ), where M is a 2 × 2 matrix, and show the corresponding solutions for M x = 0: rref(M ) 10 01 1a 00 01 00 00 00 solutions of M x = 0 {0} −at , for an arbitrary t t t , for an arbitrary t 0 R2 Now, we see that if rref(A) = rref(B ), then the systems Ax = 0 and Bx = 0 must have different solutions. Thus, it must be that if the two systems have the same solutions, then rref(A) = rref(B ). 47 ...
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This note was uploaded on 04/19/2011 for the course MTH 203 taught by Professor Staff during the Spring '08 term at Grand Valley State University.

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