308F/G Winter 2013, Quiz 2

  1. This was problem 20a from section 3.3. Determine if $v=\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right)$ is in the span of $v_{1}\left(\begin{array}{c} 1 \\ 2 \\ 0 \end{array}\right)$ and $v_{2}=\left(\begin{array}{c} 0 \\ 1 \\ -1 \end{array}\right)$. To do this, we solve the equation \[a_{1}v_{1}+a_{2}v_{2}=v\] for $a_{1}$ and $a_{2}$. The augmented matrix for this problem is \[\left(\begin{array}{cc|c} 1 & 0 & 1 \\ 2 & 1 & 1 \\ 0 & -1 & 1 \end{array}\right).\] To solve, we do row reduction, subtracting twice the first row from the second to get \[\left(\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & -1 & 1 \end{array}\right)\] then we add the second row to the third and multiply the second row by $-1$ to get \[\left(\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right)\] and so the solution is $a_{1}=1$ and $a_{2}=-1$, and thus \[v=v_{1}-v_{2}.\]
  2. This is problem 3 from section 3.4. Find a basis for the subspace of solutions of $\mathbb{R}^{4}$ to the equation $x_{1}-x_{2}+x_{3}-3x_{4}=0$. Observe that this is just a system of one equation with 4 unknowns, and there are 3 free variables and one dependent variable. In particular, we can solve for $x_{1}$ in terms of the other three variables: \[x_{1}=x_{2}-x_{3}+3x_{4}\] so that if $x=\left(\begin{array}{c} x_{1} \\ x_{2} \\ x_{3}\\x_{4}\end{array}\right)$ is a solution to the above equation, then \[x=\left(\begin{array}{c} x_{1} \\ x_{2} \\ x_{3}\\x_{4}\end{array}\right) = \left(\begin{array}{c} x_{2}-x_{3}+3x_{4} \\ x_{2} \\ x_{3}\\x_{4}\end{array}\right) =x_{2}\left(\begin{array}{c} 1 \\ 1 \\ 0 \\ 0 \end{array}\right) +x_{3}\left(\begin{array}{c} -1 \\ 0 \\ 1 \\ 0 \end{array}\right) +x_{4}\left(\begin{array}{c} 3 \\ 0 \\ 0 \\ 1 \end{array}\right). \] These are linearly independent because each vector has a $1$ in a position in which all other vectors are zero, and hence no vector can be written as a linear combination of the others.