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Problems #8 and #12 from pg.10 are graded.
Problem 8: Use row operations to solve the system of equations whose augmented matrix is $\begin{bmatrix} 1 & -5 & 4 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \end{bmatrix}$.
Solution: We will reduce this matrix to reduced echelon form. Compute
$$\begin{array}{ll}
\begin{bmatrix} 1 & -5 & 4 & 0 & 0 \\ 0 & 1 & 0 & 1 &
0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \end{bmatrix} &\stackrel{r_2^*=r_2-\frac{1}{2}r_4}{=} \begin{bmatrix} 1 & -5 & 4 & 0 & 0 \\ 0 & 1 & 0 & 0 &
0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \end{bmatrix} \\
&\stackrel{r_1^*=r_1-\frac{4}{3}r_3}{=} \begin{bmatrix} 1 & -5 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 &
0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \end{bmatrix} \\
&\stackrel{r_1^*=r_1+5r_2}{=} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \end{bmatrix} \\
&\stackrel{\stackrel{r_3^*=\frac{1}{3}r_3}{r_4^*=\frac{1}{2}r_4}}{=} \begin{bmatrix} 1&0&0&0&0 \\ 0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \end{bmatrix}.
\end{array}$$
Therefore we see that the system associated with the original augmented matrix has solution $x_1=x_2=x_3=x_4=0$.
Problem #12: Solve the system
$$\left\{ \begin{array}{ll}
x_1-5x_2+4x_3 = -3 \\
2x_1-7x_2+3x_3=-2 \\
-2x_1+x_2+7x_3=-1.
\end{array} \right.$$
Solution: We will solve this by writing its associated augmented matrix and then reducing that matrix to reduced ecehlon form. The associated augmented matrix is $\begin{bmatrix} 1&-5&4&-3 \\ 2&-7&3&-2 \\ -2&1&7&-1 \end{bmatrix}$. Now compute
$$\begin{array}{ll}
\begin{bmatrix} 1&-5&4&-3 \\ 2&-7&3&-2 \\ -2&1&7&-1 \end{bmatrix} &\stackrel{\stackrel{r_2^*=r_2-2r_1}{r_3^*=r_3+2r_1}}{=} \begin{bmatrix} 1 & -5 & 4 & -3 \\ 0 & 3 & -5 & 4 \\ 0 & -9 & 15 & -7 \end{bmatrix} \\
&\stackrel{r_3^*=r_3+3r_2}{=} \begin{bmatrix} 1 & -5 & 4 & -3 \\ 0&3&-5&4 \\ 0&0&0&5 \end{bmatrix} \\
&\stackrel{r_3^*=\frac{1}{5}r_3}{=} \begin{bmatrix} 1 & -5 & 4 & -3 \\ 0&3&-5&4 \\ 0&0&0&1 \end{bmatrix} \\
&\stackrel{\stackrel{r_1^*=r_1+3r_3}{r_2^*=r_2-4r_3}}{=} \begin{bmatrix} 1 & -5 & 4 & 0 \\ 0 & 3 & -5 & 0 \\ 0 &0&0&1 \end{bmatrix} \\
&\stackrel{r_1^*=r_1+\frac{5}{3}r_2}{=} \begin{bmatrix} 1&0&-\frac{13}{3}&0 \\ 0&3&-5&0 \\ 0&0&0&1 \end{bmatrix} \\
&\stackrel{r_2^*=\frac{1}{3}r_2}{=} \begin{bmatrix} 1&0&-\frac{13}{3}&0 \\ 0&1&-\frac{5}{3}&0 \\ 0&0&0&1 \end{bmatrix}.
\end{array}$$
Now it is clear that the system has no solution because the system of equations associated with the reduced echelon form is
$$\left\{ \begin{array}{ll}
x_1 - \dfrac{13}{3}x_3 = 0 \\
x_2 - \dfrac{5}{3}x_3 = 0 \\
0 = 1,
\end{array} \right.$$
which has no solution because no choice of $x_1,x_2,$ and $x_3$ can make the third equation true.
