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Problems #1 and #7, pg.47 are graded.
Problem #1,pg.47:Determine if the system has a nontrivial solution or not:
$$\left\{ \begin{array}{ll}
2x_1-5x_2+8x_3&=0 \\
-2x_1-7x_2+x_3&=0 \\
4x_1+2x_2+7x_3&=0.
\end{array} \right.$$
Solution: We form the augmented matrix associated with this sytem of equations and compute
$$\begin{bmatrix}
2&-5&8&0 \\
-2&-7&1&0 \\
4&2&7&0
\end{bmatrix} \sim \begin{bmatrix}1 & 0 & \frac{17}{8} & 0 \\ 0&1&-\frac{3}{4}&0 \\ 0&0&0&0 \end{bmatrix}.$$
Now interpret the reduced echelon form as a system to get
$$\left\{ \begin{array}{rr}
x_1 + \dfrac{17}{8}x_3&=0 \\
x_2 - \dfrac{3}{4}x_3&=0 \\
0&=0
\end{array} \right.$$
or equivalently
$$\left\{ \begin{array}{ll}
x_1 &= -\dfrac{17}{8}x_3 \\
x_2 &= \dfrac{3}{4}x_3 \\
0&= 0.
\end{array} \right.$$
Thus we see the solution vector $\vec{x}$ has the form
$$\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -\frac{17}{8} x_3 \\ \frac{3}{4}x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} -\frac{17}{8} \\ \frac{3}{4} \\ 1 \end{bmatrix},$$
where $x_3$ is a free variable. Hence the system has nontrivial solutions (one for each nonzero value that $x_3$ can take).
Problem #7,pg.47: Describe all solutions of $A\vec{x}=\vec{0}$ in parametric vector form, where
$$A = \begin{bmatrix} 1&3&-3&7 \\ 0&1&-4&5 \end{bmatrix}.$$
Solution: We write the matrix equation $A\vec{x}=\vec{0}$ as a vector equation in the following way:
$$x_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 3 \\ 1 \end{bmatrix} + x_3 \begin{bmatrix} -3 \\ -4 \end{bmatrix} + x_4 \begin{bmatrix} -3 \\ -4 \end{bmatrix} = \vec{0}.$$
We solve this vector equation by writing the augmented matrix and compute
$$\begin{bmatrix} 1&3&-3&7&0 \\ 0&1&-4&5&0 \end{bmatrix} \sim \begin{bmatrix} 1&0&9&-8&0 \\ 0&1&-4&5&0 \end{bmatrix} .$$
Hence we see that this is equivalent to the system of equations
$$\left\{ \begin{array}{ll}
x_1 + 9x_3-8x_4 &= 0 \\
x_2-4x_3+5x_4 &= 0,
\end{array} \right.$$
or equivalently
$$\left\{ \begin{array}{ll}
x_1 = -9x_3+8x_4 \\
x_2 = 4x_3-5x_4.
\end{array} \right.$$
Therefore we may write the solution vector $\vec{x}$ as
$$\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} -9x_3+8x_4 \\ 4x_3-5x_4 \\ x_3 \\ x_4 \end{bmatrix}=x_3 \begin{bmatrix} -9 \\ 4 \\ 1 \\ 0\end{bmatrix} + x_4 \begin{bmatrix} 8\\-5\\0\\1 \end{bmatrix},$$
which is the solution expressed in parametric vector form, with parameters (i.e. free variables) $x_3$ and $x_4$.
