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\textbf{Exercises on solving $\mb A\vb x = \vb 0$: pivot variables, special solutions}
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\begin{problem} \hspace{2pt}

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\noindent (a) Find the row reduced form of:

$$ A = \left [ \begin{array}{rrrr} 
1 & 5 & 7 & 9 \\
0 & 4 & 1 & 7 \\
2 & -2 & 11 & -3 \\  \end{array} \right ] 
$$ 

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\noindent(b) What is the rank of this matrix?

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\noindent(c) Find any special solutions to the equation $A \vb x = \vb 0.$
\end{problem}

\begin{solution} \hspace{2pt}

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\noindent(a) To transform $A$ into its reduced row form, we perform a series of row operations. Different operations are possible (same answer!). First, we multiply the first row by 2 and subtract it from the third row:

$$ \left [ \begin{array}{rrrr} 
1 & 5 & 7 & 9 \\
0 & 4 & 1 & 7 \\
2 & -2 & 11 & -3 \\  \end{array} \right ]  \longrightarrow \left [ \begin{array}{rrrr} 
1 & 5 & 7 & 9 \\
0 & 4 & 1 & 7 \\
0 & -12 & -3 & -21 \\  \end{array} \right ] $$

We then multiply the second row by $\frac{1}{4}$ to make the second pivot 1:

$$ \left [ \begin{array}{rrrr} 
1 & 5 & 7 & 9 \\
0 & 4 & 1 & 7 \\
0 & -12 & -3 & -21 \\  \end{array} \right ] \longrightarrow 
\left [ \begin{array}{rrrr} 
1 & 5 & 7 & 9 \\
0 & 1 & 1/4 & 7/4 \\
0 & -12 & -3 & -21 \\  \end{array} \right ]$$

Multiply the second row by 12 and add it to the third row:

$$\left [ \begin{array}{rrrr} 
1 & 5 & 7 & 9 \\
0 & 1 & 1/4 & 7/4 \\
0 & -12 & -3 & -21 \\  \end{array} \right ] \longrightarrow \left [ \begin{array}{rrrr} 
1 & 5 & 7 & 9 \\
0 & 1 & 1/4 & 7/4 \\
0 & 0 & 0 & 0 \\  \end{array} \right ] $$

Finally, multiply the second row by 5 and subtract it from the first row:

$$\left [ \begin{array}{rrrr} 
1 & 5 & 7 & 9 \\
0 & 1 & 1/4 & 7/4 \\
0 & 0 & 0 & 0 \\  \end{array} \right ]\longrightarrow \left [ \begin{array}{rrrr} 
1 & 0 & -23/4 & 1/4 \\
0 & 1 & 1/4 & 7/4 \\
0 & 0 & 0 & 0 \\  \end{array} \right ] $$

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\noindent(b) The matrix is of \textbf{rank 2} because it has 2 pivots.

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\noindent(c) The special solutions to $A \vb x = \vb 0 $ are:

$$\fourbyone {23/4} {-1/4} 1 0 \text{ and }  \fourbyone {-1/4} {-7/4}  0 1$$


\end{solution}



\begin{problem} (3.3 \#17.b  {\em Introduction to Linear Algebra:} Strang) Find $A_1$ and 

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\noindent $A_2$ so that rank($A_1 B$) = 1 and rank($A_2 B$) = 0 for $B = \twobytwo 1 1 1 1.$ 
\end{problem}

\begin{solution}
Take $A_1 = I_2$ and $A_2 = 0_2$ 

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\noindent(or for a less trivial example $A_2 = \twobytwo 1 {-1} 1 {-1}$)
\end{solution}



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