308F/G Winter 2013, Quiz 3

  1. This was problem 18 from section 3.5 in the text. Note that the general solution to this problem is \[x_{1}=-x_{3}+2x_{4}\] \[x_{2}=-2x_{3}+3x_{4}\] \[x_{3}=free\] \[x_{4}=free.\] Hence, any solution $x$ to the system must have the form \[x=\left(\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right) =\left(\begin{array}{c} -x_{3}+2x_{4} \\ -2x_{3} +3x_{4} \\ x_{3} \\ x_{4} \end{array}\right)\] Now we separate the vector into two, corresponding to the free variables $x_{3}$ and $x_{4}$, so the above is \[=x_{3}\left(\begin{array}{c} -1 \\ -2 \\ 1 \\ 0 \end{array}\right)+x_{4}\left(\begin{array}{c} 2 \\ 3 \\ 0 \\ 1 \end{array}\right)\] so now we've shown that the set of solutions are spanned by $\left(\begin{array}{c} -1 \\ -2 \\ 1 \\ 0 \end{array}\right)$ and $\left(\begin{array}{c} 2 \\ 3 \\ 0 \\ 1 \end{array}\right)$. I didn't require you to show that these were a basis, but if you wanted to show this, you'd just have to show that these two vectors were independent, which they are because, in the case of two vectors, being dependent is equivalent to one vector being the multiple of the other, which is not the case here. In any case, this method always produces a basis. Hence, the dimension of the solution set is just the size of this basis, which is 2.
  2. This second problem was problem 23a from section 3.4, but in addition, I asked you to find the dimension of the span of the vectors. First, arrange the vectors as columns in a matrix, so we get \[A=(v_{1}|v_{2}|v_{3}|v_{4})=\left(\begin{array}{cccc} 1 & 2 & 3 & 1 \\ 2 & 5 & 7 & 1 \\ 1 & 0 & 1 & 3 \end{array}\right).\] Now, after putting in reduced echelon form, you should get \[\left(\begin{array}{cccc} 1 & 0 & 1 & 3 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right).\] Hence, this tells us that $v_{3}$ and $v_{4}$ are redundant vectors and that $v_{1},v_{2}$ are our basis (again, I didn't require you to show that these two vectors are a basis, but this method always produces basis vectors, so there's no need to worry about this part unless I explicitly ask you to show this). Hence the dimension of the span of $v_{1},v_{2},v_{3},v_{4}$ is just 2.
  3. The final question was problem 7 and 9 from section 3.5. The set $\{v_{1},v_{2}\}$ does not span $\mathbb{R}^{3}$, because it only has two vectors and $\mathbb{R}^{3}$ has dimension 3 (if they did span, then $\mathbb{R}^{3}$ would have dimension at most 2, which is not possible). The set $S=\{v_{1},v_{2},v_{3},v_{4}\}$ is not linearly independent since there are four vectors, which is more than the dimension of the space $\mathbb{R}^{3}$). I didn't require you to give these explanations, but on an exam if I asked for you to explain, these would be adequate answers).