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\textbf{Exercises on solving $\mb A \vb x = \vb b$ and row reduced form $\mb R$}
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\begin{problem} (3.4 \#13.(a,b,d)  {\em Introduction to Linear Algebra:} Strang) Explain why these are all false:

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\noindent (a) The complete solution is any linear combination of $\vb x_p$ and $\vb x_n.$

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\noindent(b) The system $A \vb x = \vb b$ has at most one particular solution.

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\noindent(d) If $A$ is invertible there is no solution $\vb x_n$ in the nullspace.
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\begin{solution} \hspace{2pt}

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\noindent(a) The coefficient of $\vb x_p$ must be one. 

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\noindent(b) If $\vb x_n \in \mathbf{N}(A)$ is in the nullspace of $A$ and $\vb x_p$ is one particular solution, then $\vb x_p + \vb x_n$ is also a particular solution.

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\noindent(d) There's always $\vb x_n = 0.$
\end{solution}

\begin{problem} (3.4 \#28.) Let $$\mb U = \left [ \begin{array}{ccc} 
1 & 2 & 3 \\
0 & 0 & 4  \end{array} \right ] \text{ and }\vb c = \twobyone 58.$$ 

\noindent Use Gauss-Jordan elimination to reduce the matrices $[U\hspace{6pt} 0]$ and $[U\hspace{6pt} \vb c]$ to $[R\hspace{6pt} 0]$ and $[R\hspace{6pt} \vb d]$. Solve $\mb R \vb x = \vb 0$ and $\mb R \vb x = \vb d$. 

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\noindent Check your work by plugging your values into the equations $\mb U \vb x = \vb 0$ and $\mb U \vb x = \vb c$.
\end{problem}

\begin{solution} 
First we transform $[U\hspace{6pt} 0]$ into $[R\hspace{6pt} 0]$:

$$[U\hspace{6pt} 0] =  \left [ \begin{array}{rrrr} 
1 & 2 & 3 & 0 \\
0 & 0 & 4 & 0  \end{array} \right ] \longrightarrow \left [ \begin{array}{rrrr} 
1 & 2 & 3 & 0 \\
0 & 0 & 1 & 0  \end{array} \right ] \longrightarrow \left[ \begin{array}{rrrr} 
1 & 2 & 0 & 0 \\
0 & 0 & 1 & 0  \end{array} \right ] = [R\hspace{6pt} 0]. $$

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\noindent We now solve $\mb R \vb x = \vb 0$ via back substitution:

$$\left [ \begin{array}{rrr} 
1 & 2 & 0\\
0 & 0 & 1 \end{array} \right ] \threebyone {x_1} {x_2} {x_3} = \twobyone 0 0
\longrightarrow \left [ \begin{array}{r} 
x_1 + 2 x_2  = 0 \\
x_3 = 0  \end{array} \right ] \longrightarrow \vb x = \threebyone 2 {-1} 0, $$

\noindent where we used the free variable $x_2 = -1$. ($c\vb x$ is a solution for all $c$.) We check that this is a correct solution by plugging it into $\mb U \vb x = \vb 0$:

$$ \left [ \begin{array}{rrr} 
1 & 2 & 3  \\
0 & 0 & 4 \end{array} \right ] \threebyone 2 {-1} 0  = \twobyone 0 0 \hspace{4pt} \checkmark$$ 

\noindent Next, we transform $[U\hspace{6pt} \vb c]$ into $[R\hspace{6pt} \vb d]$:

$$[U\hspace{6pt} \vb c] = \left[ \begin{array}{rrrr} 
1 & 2 & 3 & 5 \\
0 & 0 & 4 & 8  \end{array} \right ]  \longrightarrow \left[ \begin{array}{rrrr} 
1 & 2 & 3 & 5 \\
0 & 0 & 1 & 2  \end{array} \right ]  \longrightarrow \left[ \begin{array}{rrrr} 
1 & 2 & 0 & -1 \\
0 & 0 & 1 & 2  \end{array} \right ] = [R\hspace{6pt} \vb d].$$

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\noindent We now solve $\mb R \vb x = \vb d$ via back substitution:

$$\left [ \begin{array}{rrr} 
1 & 2 & 0\\
0 & 0 & 1 \end{array} \right ] \threebyone {x_1} {x_2} {x_3} = \twobyone {-1} 2
\longrightarrow \left [ \begin{array}{r} 
x_1 + 2 x_2  = -1 \\
x_3 = 2  \end{array} \right ] \longrightarrow \vb x = \threebyone {-3} {1} 2,$$ 

\noindent where we used the free variable $x_2 = 1$. Finally, we check that this is the correct solution by plugging it into the equation $\mb U \vb x = \vb c$:

$$ \left [ \begin{array}{rrr} 
1 & 2 & 3  \\
0 & 0 & 4 \end{array} \right ] \threebyone {-3} {1} 2  = \twobyone 5 8 \hspace{4pt} \checkmark$$ 
\end{solution}

\begin{problem} (3.4 \#36.) Suppose $A \vb x = \vb b$ and $C \vb x = \vb b$ have the same (complete) solutions for every $\vb b.$ Is it true that $A = C?$
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\begin{solution} \textbf{Yes.} In order to check that $A=C$ as matrices, it is enough to check that $A \vb y = C \vb y$ for all vectors $\vb y$ of the correct size (or just for the standard basis vectors, since multiplication by them ``picks out the columns"). So let $\vb y$ be any vector of the correct size, and set $\vb b = A \vb y.$ Then $\vb y $ is certainly a solution to $A \vb x = \vb b,$ and so by our hypothesis must also be a solution to $C \vb x = \vb b;$ in other words, $C \vb y = \vb b = A \vb y.$

\end{solution}


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