Linear Algebra with Applications, 10th Edition

3.4

1. 1. Only in parts (a) and (e) do they form a basis.

2. 2. Hint: Since the dimension of ${ℝ}^3$ is 3, a basis must consist of exactly 3 vectors and they must be linearly independent. If the set has either more or less than 3 vectors then it cannot be a basis. If the set consists of 3 vectors it will be a basis for ${ℝ}^3$ if and only if the vectors are linearly independent.

3. 3.  

   1. (a) Hint: By Theorem 3.4.3, it suffices to show that the two vectors are linearly independent.

   2. (b) Hint: Apply Theorem 3.4.1.

   3. (c) 2

4. 4. 1

5. 5. 

   1. (c) 2;

   2. (d) a plane through $(0,0,0)$ in 3-space

6. 6. 

   1. (b) $\{(1,1,1{)}^T\}$, dimension 1;

   2. (c) $\{{(1,0,1)}^T,(0,1,1{)}^T\}$, dimension 2

7. 7. basis $\{{(1,1,0,0)}^T,{(1,-1,1,0{)}^T,(0,2,0,1)}^T\}$

8. 8. 

   1. (a) Hint: Hint. See Theorem 3.4.4.

9. 9. 

   1. (b) Hint: If $\mathbf{b}=\mathit{\text{A}}\mathbf{x}$ then how ...