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\begin{center}
\textbf{18.06 Session 2}
\end{center}

\begin{problem}
In the two-by-two system of linear equations below, what multiple of the first equation should be subtracted from the second equation when using the method of elimination? Convert this system of equations to matrix form, apply elimination (what are the pivots?), and use back substitution to find a solution.  Try to check your work before looking up the answer.

$$\begin{array}{l}
2x + 3y = 5 \\
6x + 15y = 12 \end{array}$$ 
\end{problem}

\begin{solution}
One subtracts \textbf{3} times the first equation from the second equation in order to eliminate the $x$ terms. 

To convert the given system of equations to matrix form, use the general format $A \vb x = \vb b$:

$$\begin{array}{l}
2x + 3y = 5 \\
6x + 15y = 12 \end{array} \longrightarrow
\twobytwo 2 3 6 {15} \twobyone x y = \twobyone 5 {12}.$$

We then apply elimination on matrix $A$. Using the first pivot (the number 2 in the upper left corner of $A$), we subtract three times the first row from the second row to get:

$$A = \twobytwo 2 3 6 {15} \longrightarrow U = \twobytwo {\mathbf 2} 3 0 {\mathbf 6}  $$

\noindent where $U$ is an upper triangular matrix with pivots 2 and 6. We then have a new equation of the form $U \vb x = \vb c:$

$$\twobytwo 2 3 0 6  \twobyone x y = \twobyone 5 {-3}. $$ 

\medskip

\noindent To solve our new equation, we use back substitution: 

$$6y = -3 \longrightarrow \framebox{${y = -\frac{1}{2}}$} $$

\noindent and

$$2x+3y= 5 \longrightarrow 2x + 3\left(-\frac{1}{2}\right) = 5 \longrightarrow 2x = 5 + \frac{3}{2} = \frac{13}{2} \longrightarrow \framebox{${x = \frac{13}{4}}$}$$

We know that our solution fulfills the first equation; let's make sure that our values fulfill the second equation as a check on our work:

$$6x + 15y = 6 \left(\frac{13}{4} \right) + 15 \left ( \frac{-1}{2}\right)  = \frac{78 - 30}{4} = 12 \hspace{4pt} \checkmark $$

\end{solution}


\begin{problem} (2.3 \#29.  {\em Introduction to Linear Algebra:} Strang) Find the triangular matrix $E$ that reduces ``\textit{Pascal's matrix}" to a smaller Pascal:

\vspace{10pt}

$$E \left [ \begin{array}{rrrr} 
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 2 & 1 & 0 \\
1 & 3 & 3 & 1 \\  \end{array} \right ]   
=
\left [ \begin{array}{rrrr} 
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 1 & 2 & 1 \\  \end{array} \right ].
$$

\vspace{10pt}

Which matrix $M$ (multiplying several $E$'s) reduces Pascal all the way to $I$?

\end{problem}

\begin{solution} 
$$E = \left [ \begin{array}{rrrr} 
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 1 \\  \end{array} \right ] 
$$

\noindent One can eliminate the second column with the matrix

$$\left [ \begin{array}{rrrr} 
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 1 \\  \end{array} \right ] 
$$

\noindent and the third column with the matrix

$$\left [ \begin{array}{rrrr} 
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 &  0 \\
0 & 0 & -1 & 1 \\  \end{array} \right ] 
$$

\noindent Multiplying these together, we get

$$M =\left [ \begin{array}{rrrr} 
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & -1 & 1  \\  \end{array} \right ] 
\left [ \begin{array}{rrrr} 
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 1 \\  \end{array} \right ]
\left [ \begin{array}{rrrr} 
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 1 \\  \end{array} \right ] 
= 
\left [ \begin{array}{rrrr} 
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
1 & -2 & 1 & 0 \\
-1 & 3 & -3 & 1 \\  \end{array} \right ].$$
\end{solution}

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