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{\bf Question}

Show that the matrix $$\left( \begin{array}{ccc} a & b & c \\ 0 &
d & e \\ 0 & 0 & f \end{array} \right)$$ is non singular if and
only if $a,d$ and $f$ are all non zero.

Write down the inverse in that case.  Check your answer by
multiplication.

\vspace{.25in}

{\bf Answer}

$\det A = adf \not=0$ if and only id $a,\,d,\,f$ are all $\not=0$

$\ds A^{-1} = \frac{1}{adf} \left( \begin{array}{ccc} df & -bf &
be -cd \\ 0 & af & -ae \\ 0 & 0 & ad \end{array} \right)$


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