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Problems #11 from pg. 32 and additional problem (A) are graded.
#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
$$\vec{x} = \left[ \begin{array}{ll}
b_1+b_3 \\
-b_2-b_3 \\
-b_3.
\end{array} \right]$$
