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Problems #9 and #13 from pg.32 are graded.
Problem 9: Write a vector equation equivalent to the given system of equations:
$$\left\{ \begin{array}{ll}
x_2+5x_3&=0 \\
4x_1+6x_2-x_3&=0 \\
-x_1+3x_2-8x_3&=0.
\end{array} \right.$$
Solution: Let $\vec{a}_1=\begin{bmatrix} 0 \\ 4 \\ -1 \end{bmatrix}, a_2 = \begin{bmatrix} 1 \\ 6 \\ 3 \end{bmatrix},$ and $\vec{a}_3 = \begin{bmatrix} 5 \\ -1 \\ -8 \end{bmatrix}$. Then it is clear that the vector equation
$$x_1 \vec{a}_1 + x_2\vec{a_2} + x_3\vec{a}_3 = \vec{0}$$
is a vector equation that is equivalent to the given system of equations.
Problem 13: Determine if $\vec{b}=\begin{bmatrix} 3 \\ -7 \\ -3 \end{bmatrix}$ is a linear combination of the vectors formed by the columns of $A = \begin{bmatrix} 1 & -4 & 2 \\ 0 & 3 & 5 \\ -2 & 8 & -4 \end{bmatrix}$.
Solution: We will express this question in the form of a vector equation and then solve that vector equation. Write $\vec{a}_1=\begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}, \vec{a}_2 = \begin{bmatrix} -4 \\ 3 \\ 8 \end{bmatrix},$ and $\vec{a}_3 = \begin{bmatrix} 2 \\ 5 \\ -4 \end{bmatrix}$. Now we see that the question of whether $\vec{b}$ is a linear combination of the columns of $A$ is simply asking whether or not the vector equation
$$x_1\vec{a}_1 + x_2\vec{a}_2 + x_3\vec{a}_3 = \vec{b}$$
has a solution or not. We know that all vector equations can reduce to a system of equations, and in this case the system we must solve is
$$\left\{ \begin{array}{ll}
x_1-4x_2+2x_3&=3 \\
3x_2+5x_3&=-7 \\
-2x_1+8x_2-4x_3 &= -3,
\end{array} \right.$$
for which we will write an augmented matrix and reduce to reduced echelon form to get
$$\begin{bmatrix} 1 & -4 & 2 & 3 \\ 0 & 3 & 5 & -7 \\ -2 & 8 & -4 & -3 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & \frac{26}{3} & 0 \\ 0 & 1 & \frac{5}{3} & 0 \\ 0 &0 & 0 & 1 \end{bmatrix}$$
and we see the system of equations has no solution. Therefore there is no way to write the vector $\vec{b}$ as a linear combination of $\vec{a}_1, \vec{a}_2,$ and $\vec{a}_3$.
