Back to the class
Problems #7 and #13 from pg.10 are graded.
Problem 7: Find the row equivalent reduced echelon matrix for the matrix
$$\left[ \begin{array}{llll}
1 & 7 & 3 & -4 \\
0 & 1 & -1 & 3 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & -2
\end{array} \right]$$
Solution: The system associated with this matrix has no solution because of the third row (do you see why?). However we will continue to perform reduction to reduced echelon form anyway (did not need to do this for credit). Compute using row operations
$$\begin{array}{ll}
\left[ \begin{array}{llll}
1 & 7 & 3 & -4 \\
0 & 1 & -1 & 3 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & -2
\end{array} \right] &\stackrel{\mathrm{swap \hspace{2pt}} r_3 \leftrightarrow r_4}{\sim} \left[ \begin{array}{llll}
1 & 7 & 3 & -4 \\
0 & 1 & -1 & 3 \\
0 & 0 & 1 & -2 \\
0 & 0 & 0 & 1
\end{array} \right] \\
&\stackrel{r_1^*=r_1-7r_2}{\sim} \left[ \begin{array}{llll}
1 & 0 & 10 & -25 \\
0 & 1 & -1 & 3 \\
0 & 0 & 1 & -2 \\
0 & 0 & 0 & 1
\end{array} \right] \\
&\stackrel{\stackrel{r_1^*=r_1-10r_3}{r_2^*=r_2+r_3}}{\sim} \left[ \begin{array}{llll}
1 & 0 & 0 & -5 \\
0 & 1 & 0 & 1\\
0&0&1&-2 \\
0&0&0&1
\end{array} \right] \\
&\stackrel{\stackrel{r_1^*=r_1+5r_4}{\stackrel{r_2^* = r_2 - r_4}{r_3^*=r_3+2r_4}}}{\sim} \left[ \begin{array}{llll}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array} \right]
\end{array}$$
Notice this computation can be performed in Wolfram Alpha.
Problem 13: Solve the following system:
$$\left\{ \begin{array}{ll}
x_1 - 3x_3 = 8 \\
2x_1 + 2x_2 + 9x_3 = 7 \\
x_2 + 5x_3 = -2.
\end{array} \right.$$
Solution: Recall if we line up the variables $x_i$ in columns consistently, we can write this system as an augmented matrix: $$\left[ \begin{array}{ll}
1 & 0 & -3 & 8 \\
2 & 2 & 9 & 7 \\
0 & 1 & 5 & -2
\end{array} \right].$$
So we will find the row equivalent reduced echelon matrix associated with this augmented matrix to solve the system. Compute
$$\begin{array}{ll}
\left[ \begin{array}{ll}
1 & 0 & -3 & 8 \\
2 & 2 & 9 & 7 \\
0 & 1 & 5 & -2
\end{array} \right] &\stackrel{\mathrm{swap \hspace{2pt} r_2 \leftrightarrow r_3}}{\sim} \left[ \begin{array}{llll}
1 & 0 & -3 & 8 \\
0 & 1 & 5 & -2 \\
2 & 2 & 9 & 7
\end{array} \right] \\
&\stackrel{r_3^* = r_3 - 2r_1}{\sim} \left[ \begin{array}{llll}
1 & 0 & -3 & 8 \\
0 & 1 & 5 & -2 \\
0 & 2 & 15 & -9
\end{array} \right] \\
&\stackrel{r_3^* = r_3 - 2r_2}{\sim} \left[ \begin{array}{llll}
1 & 0 & -3 & 8 \\
0 & 1 & 5 & -2 \\
0 & 0 & 5 & -5
\end{array} \right] \\
&\stackrel{r_2^* = r_2 - r_3}{\sim} \left[ \begin{array}{llll}
1 & 0 & -3 & 8 \\
0 & 1 & 0 & 19\\
0 & 0 & 5 & -5
\end{array} \right] \\
&\stackrel{r_1^* = r_1 + \frac{3}{5}r_3}{\sim} \left[ \begin{array}{llll}
1 & 0 & 0 & 5 \\
0 & 1 & 0 & 3 \\
0 & 0 & 5 & -5
\end{array} \right] \\
&\stackrel{r_3^* = \frac{1}{5}r_3}{\sim} \left[ \begin{array}{llll}
1 & 0 & 0 & 5 \\
0 & 1 & 0 & 3 \\
0 & 0 & 1 & -1
\end{array} \right].
\end{array}$$
Notice that this computation can be performed in Wolfram Alpha.
From the reduced echelon form we see that the solution of the system is
$$\left\{ \begin{array}{ll}
x_1 &= 5 \\
x_2 &= 3 \\
x_3 &= -1.
\end{array} \right.$$
