{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

3e-spring2011-chapter_1_tf

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online