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Problems #12, pg.40 and the additional problem (A) are graded.
Problem #12, pg.40: Write the augmented matrix for the linear system that corresponds to the matrix equation $A\vec{x}=\vec{b}$ for $A=\begin{bmatrix}1&2&-1 \\ -3&-4&2 \\ 5&2&3 \end{bmatrix}$ and $\vec{b}=\begin{bmatrix} -2 \\ 4 \\ 12 \end{bmatrix}$. Then solve the system and write the solution as a vector.
Solution: The augmented matrix we seek is $\begin{bmatrix} 1 & 2& -1 & 1 \\ -3 & -4 & 2 & 2 \\ 5 & 2 & 3 & -3 \end{bmatrix}$. We will find a solution by row reduction. So compute
$$\begin{array}{ll}
\begin{bmatrix} 1&2&-1&1\\ -3&-4&2&2 \\ 5&2&3&-3 \end{bmatrix} &\stackrel{\stackrel{r_2^*=r_2+3r_1}{r_3^*=r_3-5r_1}}{\sim} \begin{bmatrix} 1&2&-1&1 \\ 0&2&-1&5 \\ 0&-8&8&-8 \end{bmatrix} \\
&\stackrel{r_3^*=r_3+4r_2}{\sim} \begin{bmatrix} 1&2&-1&1 \\ 0&2&-1&5 \\ 0&0&4&12 \end{bmatrix} \\
&\stackrel{r_3^*=\frac{1}{4}r_3}{\sim} \begin{bmatrix} 1&2&-1&1 \\ 0&2&-1&5 \\ 0&0&1&3 \end{bmatrix} \\
&\stackrel{\stackrel{r_1^*=r_1+r_3}{r_2^*=r_2+r_3}}{\sim} \begin{bmatrix} 1&2&0&4 \\ 0&2&0&8 \\ 0&0&1&3 \end{bmatrix} \\
&\stackrel{r_1^*=r_1-r_2}{\sim} \begin{bmatrix} 1&0&0&-4 \\ 0&2&0&8 \\ 0&0&1&3 \end{bmatrix} \\
&\stackrel{r_2^*=\frac{1}{2}r_2}{\sim} \begin{bmatrix} 1&0&0&-4 \\ 0&1&0&4 \\ 0&0&1&3 \end{bmatrix}.
\end{array}$$
Therefore we see that the solution is $x_1=-4, x_2=4,x_3=3$. So we now write the solution vector $\vec{x}$:
$$\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -4 \\ 4 \\ 3 \end{bmatrix}.$$
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 $\vec{b}=\left[ \begin{array}{ll}
b_1 \\
b_2 \\
b_3
\end{array} \right]$.
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
$$\vec{x} = \left[ \begin{array}{ll}
b_1+b_3 \\
-b_2-b_3 \\
-b_3.
\end{array} \right]$$
