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QUESTION
\begin{description}

\item[(i)]
Use the elimination method to find the inverse of the matrix
$$\textbf{A}=\left(\begin{array}{ccc}1&amp;1&amp;-1\\3&amp;2&amp;1\\-1&amp;0&amp;1\end{array}\right)$$
and verify that $\textbf{AA}^{-1}=\textbf{I}$.

\item[(ii)]
The inverse of any non-singular matrix \textbf{C} satisfies the
equations $\textbf{CC}^{-1}=\textbf{C}^{-1}\textbf{C}=\textbf{I}$.
Deduce that $$(\textbf{C}^{-1})^T=(\textbf{C}^T)^{-1}.$$

\item[(iii)]
Using parts (i) and (ii), or otherwise, solve the equations
$\textbf{A}^T\textbf{X=b}$, where $\textbf{A}$ is the matrix
defined in (i) and
$\textbf{b}=\left(\begin{array}{ccc}4\\1\\4\end{array}\right)$.

\end{description}




ANSWER
\begin{description}

\item[(i)]
$
\hspace{1cm}\left(\begin{array}{ccc|ccc}1&amp;1&amp;-1&amp;1&amp;0&amp;0\\
3&amp;2&amp;1&amp;0&amp;1&amp;0\\ -1&amp;0&amp;1&amp;0&amp;0&amp;1\end{array}\right),\\
\rightarrow\left(\begin{array}{ccc|ccc}1&amp;1&amp;-1&amp;1&amp;0&amp;0\\
0&amp;-1&amp;4&amp;-3&amp;1&amp;0\\ 0&amp;1&amp;0&amp;1&amp;0&amp;1\end{array}\right),$\ \ row$\
2-3\times$row$ \ 1\\
\rightarrow\left(\begin{array}{ccc|ccc}1&amp;1&amp;-1&amp;1&amp;0&amp;0\\
0&amp;-1&amp;4&amp;3&amp;-1&amp;0\\ 0&amp;1&amp;0&amp;1&amp;0&amp;1\end{array}\right),$ $(-1)\times$row$ \
2\\ \rightarrow\left(\begin{array}{ccc|ccc}1&amp;0&amp;3&amp;-2&amp;1&amp;0\\
0&amp;1&amp;-4&amp;3&amp;-1&amp;0\\ 0&amp;0&amp;4&amp;-2&amp;1&amp;1\end{array}\right),$ row$\ 1-\ $row$\
2,$\ row$\ 3-\ $row\ 2$\\
\rightarrow\left(\begin{array}{ccc|ccc}1&amp;0&amp;3&amp;-2&amp;1&amp;0\\
0&amp;1&amp;-4&amp;3&amp;-1&amp;0\\
0&amp;0&amp;1&amp;-\frac{1}{2}&amp;\frac{1}{4}&amp;\frac{1}{4}\end{array}\right),\
(\frac{1}{4})\times $row\ 3$\\ \rightarrow
\left(\begin{array}{ccc|ccc}1&amp;0&amp;0&amp;-\frac{1}{2}&amp;\frac{1}{4}&amp;-\frac{3}{4}\\
0&amp;1&amp;0&amp;1&amp;0&amp;1\\
0&amp;0&amp;1&amp;-\frac{1}{2}&amp;\frac{1}{4}&amp;\frac{1}{4}\end{array}\right),$
row$\ 1-3\times$row$\ 3,\ $row$\ 2+4\times$row$\ 3$

Therefore $A^{-1}=
\left(\begin{array}{ccc}-\frac{1}{2}&amp;\frac{1}{4}&amp;-\frac{3}{4}\\1&amp;0&amp;1\\
-\frac{1}{2}&amp;\frac{1}{4}&amp;\frac{1}{4}\end{array}\right)\\ AA^{-1}=
\left(\begin{array}{ccc}1&amp;1&amp;-1\\ 3&amp;2&amp;1\\ -1&amp;0&amp;1\end{array}\right)
\left(\begin{array}{ccc}-\frac{1}{2}&amp;\frac{1}{4}&amp;-\frac{3}{4}\\
1&amp;0&amp;1\\ -\frac{1}{2}&amp;\frac{1}{4}&amp;\frac{1}{4}\end{array}\right)=
\left(\begin{array}{ccc}1&amp;0&amp;0\\ 0&amp;1&amp;0\\ 0&amp;0&amp;1\end{array}\right)$

\item[(ii)]
$CC^{-1}=I$ therefore $(CC^{-1})^T=I^T;\ (C^{-1})^TC^T=I\\
C^{-1}C=I$ therefore $(C^{-1}C)^T=I^T;\ (C^T)(C^{-1})^T=I$\\ Hence
$(C^{-1})^T=(C^T)^{-1}$

\item[(iii)]
$A^TX=\textbf{b}$, solution is
$X=(A^T)^{-1}\textbf{b}=(A^{-1})^T\textbf{b}$, using (ii).

Therefore
$$X=\left(\begin{array}{ccc}-\frac{1}{2}&amp;1&amp;-\frac{1}{2}\\
\frac{1}{4}&amp;0&amp;\frac{1}{4}\\
-\frac{3}{4}&amp;1&amp;\frac{1}{4}\end{array}\right)
\left(\begin{array}{c}4\\ 1\\ 4\end{array}\right)=
\left(\begin{array}{c}-2+1-2 \\ 1+0+1\\ -3+1+1\end{array}\right)=
\left(\begin{array}{c}-3\\2 \\-1\end{array}\right)$$

\end{description}





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