text: .
  
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  #11, pg.32: Determine if $\vec{b}$ is a linear combination of $\vec{a}_1,\vec{a}_2,$ and $\vec{a}_3$ where $$\vec{a}_1 = \left[ \begin{array}{ll} 1 \\ -2 \\ 0 \end{array} \right], \vec{a}_2=\left[ \begin{array}{ll} 0 \\ 1 \\ 2 \end{array} \right], \vec{a}_3 = \left[ \begin{array}{ll} 5 \\ -6 \\ 8 \end{array} \right], \vec{b}=\left[ \begin{array}{ll} 2 \\ -1 \\ 6 \end{array} \right].$$ Solution: To write $\vec{b}$ as a linear combination of $\vec{a}_1, \vec{a}_2,$ and $\vec{a}_3$ means we must find the weights $x_1,x_2,x_3$ (if they exist) that satisfy $$x_1 \vec{a}_1 + x_2 \vec{a}_2 + x_3\vec{a}_3 = \vec{b}.$$ When the vector algebra is simplified (carry out the vector sums) we get the equation $$\left[ \begin{array}{ll} x_1 + 5x_3 \\ -2x_1 +x_2 - 6x_3 \\ 2x_2 + 8x_3 \end{array} \right] = \left[ \begin{array}{ll} 2 \\ -1 \\ 6 \end{array} \right],$$ which is a system of linear equations. We will solve this system using the augmented matrix $$\begin{array}{ll} \left[ \begin{array}{llll} 1 & 0 & 5 & 2\\ -2 & 1 & -6 & -1 \\ 0 & 2 & 8 & 6 \end{array} \right] &\stackrel{r_2^* = r_2+2r_1}{\sim} \left[ \begin{array}{llll} 1 & 0 & 5 & 2 \\ 0 & 1 & 4 & 3 \\ 0 & 2 & 8 & 6 \end{array} \right] \\ &\stackrel{r_3^* = r_3-2r_2}{\sim} \left[ \begin{array}{ll} 1 & 0 & 5 & 2 \\ 0 & 1 & 4 & 3 \\ 0 & 0 & 0 & 0 \end{array} \right] \\ \end{array}$$ If we interpret this augmented marix as a system, we get $$\left\{ \begin{array}{ll} x_1 + 5x_3 = 2 \\ x_2 + 4x_3 = 3 \\ 0=0 \end{array} \right.$$ or equivalently $$\left\{ \begin{array}{ll} x_1 = 2-5x_3 \\ x_2 = 3-4x_3 \\ 0 = 0 \end{array} \right.$$ We are free to choose any value of $x_3$ so we will choose $x_3=0$ and doing so yields the solution $$\left\{ \begin{array}{ll} x_1 = 2 \\ x_2 = 3 \\ x_3 = 0. \end{array} \right.$$ This implies that we should be able to write the vector $\vec{b}$ as $$2 \vec{a}_1 + 3\vec{a}_2 + 0\vec{a}_3 = \left[ \begin{array}{ll} 2 \\ -4 \\ 0 \end{array} \right] + \left[ \begin{array}{ll} 0 \\ 3 \\ 6 \end{array} \right] = \left[ \begin{array}{ll} 2 \\ -1 \\ 6 \end{array} \right]= \vec{b},$$ as was to be shown.
  NOTE: Every different choice of $x_3$ yields a different linear combination that yields $\vec{b}$. Here we call $x_3$ a free variable.
  Additional Problem (A): Solve the matrix equation $A \vec{x}=\vec{b}$ where $A = \left[ \begin{array}{ll} 1 & 0 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right]$ and $\vec{b}=\left[ \begin{array}{ll} b_1 \\ b_2 \\ b_3 \end{array} \right]$.
  NOTE: Many people wrote the matrix in this problem as $\left[ \begin{array}{ll} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right]$. This may because I made a typo in class...I will not penalize correct answers that start with this matrix instead of the actual problem.
  Solution: We know the solution of this equation to be equivalent to the solution of the system whose augmented matrix is $$\begin{array}{ll} \left[ \begin{array}{ll} 1 & 0 & 1 & b_1 \\ 0 & -1 & 1 & b_2 \\ 0 & 0 & -1 & b_3 \end{array} \right] &\stackrel{r_1^* = r_1+r_3}{\stackrel{r_2^*=r_2+r_3}{\sim}} \left[ \begin{array}{ll} 1 & 0 & 0 & b_1+b_3 \\ 0 & -1 & 0 & b_2+b_3 \\ 0 & 0 & -1 & b_3 \end{array} \right] \\ & \stackrel{r_2^*=-r_2}{\stackrel{r_3^*=-r_3}{\sim}} \left[ \begin{array}{llll} 1 & 0 & 0 & b_1+b_3 \\ 0 & 1 & 0 & -b_2-b_3 \\ 0 & 0 & 1 & -b_3 \end{array} \right] \end{array}$$ hence we see that the solution is given by $$\vec{x} = \left[ \begin{array}{ll} b_1+b_3 \\ -b_2-b_3 \\ -b_3. \end{array} \right]$$
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