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{\bf Question}

Solve the equations:
\begin{eqnarray*} x + y + z & = & 2 \\ y + 2z & = & 0 \\ 2x - y - z &
= & 1 \end{eqnarray*}
\begin{description}
\item[(a)]
by rewriting the equation in the form ${\bf Ax} = {\bf b}$ and
finding $\bf{A}^{-1}$ by the cofactor method
\item[(b)]
using the standard elimination method.
\end{description}
Which method took longer?

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
Equations can be rewritten as:

$\left[ \begin{array}{ccc}  1&1&1 \\ 0&1&2 \\ 2&-1&-1 \end{array}
\right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] =
\left[ \begin{array}{c} 2 \\ 0 \\ 1 \end{array} \right]$

$A{\bf x} = {\bf b} \Rightarrow {\bf x} = A^{-1}{\bf b}$

$|A| = \left| \begin{array}{ccc}  1&1&1 \\ 0&1&2 \\ 2&-1&-1
\end{array} \right| = \left|  \begin{array}{ccc}  3&0&0 \\ 0&1&2
\\ 2&-1&-1 \end{array} \right| = 3 \times (-1 + 2) = 3$

Call matrix of cofactors $a$
\begin{eqnarray*} a & = & \left[ \begin{array}{ccc}
\left|\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right| &
-\left|\begin{array}{cc} 0 & 2 \\ 2 & -1 \end{array}\right| &
\left|\begin{array}{cc} 0 & 1 \\ 2 & -1 \end{array}\right| \\ \\
-\left|\begin{array}{cc} 1 & 1 \\ -1 & -1 \end{array}\right| &
\left|\begin{array}{cc} 1 & 1 \\ 2 & -1 \end{array}\right| &
-\left|\begin{array}{cc} 1 & 1 \\ 2 & -1 \end{array}\right| \\ \\
\left|\begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array}\right| &
-\left|\begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array}\right| &
\left|\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right| \\
\end{array}\right]\\ & = & \left( \begin{array}{ccc} 1 & 4 & -2 \\
 0 & -3 & 3 \\ 1 & -2 & 1\end{array} \right) \end{eqnarray*}

 $\displaystyle A^{-1} = \frac{a^T}{|A|} = \left( \begin{array}{ccc}  \frac{1}{3}&0&\frac{1}{3}
  \\ \frac{4}{3}&-1&\frac{-2}{3} \\ \frac{-2}{3}&1&\frac{1}{3}
\end{array} \right)$

Finally ${\bf x} = \left[ \begin{array}{c} x \\ y \\ z \end{array}
\right] = \left( \begin{array}{ccc} \frac{1}{3}&0&\frac{1}{3}
  \\ \frac{4}{3}&-1&\frac{-2}{3} \\ \frac{-2}{3}&1&\frac{1}{3}
\end{array} \right)\left[ \begin{array}{c} 2 \\ 0 \\ 1 \end{array}
\right] = \left[ \begin{array}{c} 1 \\ 2 \\ -1 \end{array}
\right]$
\item[(b)]
elimination

$\left[ \begin{array}{ccc}  1&1&1 \\ 0&1&2 \\ 2&-1&-1 \end{array}
\right]\left[ \begin{array}{c} 2 \\ 0 \\ 1 \end{array} \right]
\rightarrow ({\rm row}\ 3 \rightarrow {\rm row}\ 3 + 2 {\rm row}\
1)$

$ \left[
\begin{array}{ccc}  1&1&1 \\ 0&1&2 \\ 0&-3&-3 \end{array}
\right]\left[ \begin{array}{c} 2 \\ 0 \\ -3 \end{array} \right]
\rightarrow ({\rm row}\ 3 \rightarrow {\rm row}\ 3 + 3 {\rm row}\
2)$

$ \left[
\begin{array}{ccc}  1&1&1 \\ 0&1&2 \\ 0&0&3 \end{array}
\right]\left[ \begin{array}{c} 2 \\ 0 \\ -3 \end{array} \right] $

Hence $\begin{array}{rcr} x+ y+z & = & 2 \\ y + 2z & = & 0 \\ 3z &
= & -3 \end{array} \Rightarrow \begin{array}{l}z = -1 \\  y = -2z
= 2, \\x = 2 - y - z = 1\end{array}$
\end{description}

Cofactor way is MUCH longer.


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