solutions - 1 The solution set is 0-5 4 1 0 0 0 0 1 span 0...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1. The solution set is span 4 1 0 0 0 0 , 0 0 0 1 0 0 , - 5 0 1 0 4 1 . 2. (a) There are 4 equations and 5 unknowns in the original system. (b) Let x i correspond to the i th column for i = 1 , 2 ,..., 5. The solutions are given by x 1 x 2 x 3 x 4 x 5 = a 1 3 1 0 0 + b 2 4 0 1 0 + - 5 6 0 0 0 for all a,b R . 3. (a) The vectors v 1 and v 2 are not scalar multiples of each other and so they are linearly independent. Any two independent vectors in R 2 form a basis and so, in particular, v 1 and v 2 form a basis. Any vector in R 2 can be written uniquely as the linear combination of vectors in a basis. Hence, the answer is yes to both questions. (b) Yes and yes for the same reason as above. (c) Yes, w can be expressed as a linear combination of v 1 ,v 2 , and v 3 . These vectors do not form a basis so such an expression for w is not unique. 4. (a) Since det[ v 1 v 2 v 3 ] 6 = 0, the vectors are independent. (b) No, a set of linearly independent vectors in R 3 can have at most 3 vectors. (c) No, a set of linearly independent vectors cannot contain the zero vector. (d) Yes, neither vector is a scalar multiple of the other. 5. The columns of the standard matrix of T are not independent; therefore, T is not one-to-one. Clearly, the first, second, and third columns form an independent set of vectors. Since the number of independent columns is the same as the dimension of the codomain, the function T is onto. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
± - 1 0 0 1 ² and that of the rotation is ± 0 - 1 1 0 ² . Thus, the standard matrix of T is the product ± 0 - 1 1 0 ²± - 1 0 0 1 ² = ± 0 - 1 - 1 0 ² . 7. The standard matrix of T is ± 3 - 1 0 2 ² . 8. (a) Let A = B = [0]. (b) Let A = [ 0 0 ] and B = [ 0 0 ]. (c) Let A = [ 0 0 ] and B = [ 0 0 0 0 ]. 9. The product DA is found by multiplying the first row of A by 2, the second row of A by 3, and the third row of A by 5. The product AD is found by multiplying the first column of A by 2, the second column of A by 3, and the third column of A by 5. 10. (a) [ 3 4 5 7 ] - 1 = ³ 7 - 4 - 5 3 ´ (b) The matrix [ A I 3 × 3 ] row reduces to 1 0 0 - 9 / 2 7 - 3 / 2 0 1 0 - 2 4 - 1 0 0 1 3 / 2 - 2 1 / 2 and so A - 1 = - 9 / 2 7 - 3 / 2 - 2 4 - 1 3 / 2 - 2 1 / 2 . 11. (a) Row 3 is replaced by Row 3 minus 4 times Row 1. (b)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/22/2008 for the course CH 339K taught by Professor Appling during the Fall '08 term at University of Texas.

Page1 / 9

solutions - 1 The solution set is 0-5 4 1 0 0 0 0 1 span 0...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online