Hi Kristen,

First I need some notation and terminology. Suppose you have a square $n$ by $n$ matrix $M.$ An $n - 1$ by $n - 1$ determinant obtained by deleting one row and one column from $M$ is a minor of $M.$ If you delete row $i$ and column $j$ to form the minor it is called the $ij$ minor of $M$ and written $M_{i,j}.$ Thus, for example if \[M = \left( \begin{array}{rrr}
1 & 3 & 2\\ -1& 4 & 1\\ 0 & -2 & 4 \end{array} \right)\]

then

\[M_{1,1} = \left| \begin{array}{rr} 4 & 1\\ -2 & 4 \end{array} \right| \]

and

\[M_{2,3} = \left| \begin{array}{rr} 1 & 3\\ 0 & -2 \end{array} \right|. \]

One more term to define. For an $n$ by $n$ square matrix $M$ the $i, j$ cofactor $C_{i j}$ is defined by

\[C_{i, j} = (-1)^{i + j} M_{i ,j}.\]

Thus the $i, j$ cofactor is the {i, j} minor if $i + j$ is positive and it is the negative of the $i, j$ minor if {i + j} is odd.

Now consider the matrix

\[M = \left( \begin{array}{rrr}
a_{1 ,1} & a_{1, 2} & a_{1, 3}\\ a_{2, 1}& a_{2, 2} & a_{2, 3}\\ a_{3, 1} & a_{3, 2} & a_{3, 3}\end{array} \right).\]

The determinant of $M$ can be evaluated by

\[det(M) = a_{1, 1} C_{1, 1} + a_{1, 2} C_{1, 2} + a_{1, 3} C_{1, 3} \]

so for the matrix above

\[M = \left( \begin{array}{rrr}
1 & 3 & 2\\ -1& 4 & 1\\ 0 & -2 & 4 \end{array} \right)\] I get

\[det(M) = 1 \times C_{1, 1} + 3 \times C_{1, 2} + 2 \times C_{1,3} = 1 \times (16 + 2) + 3 \times (4 - 0) + 2 \times (2 - 0) = 18 + 12 + 4 = 34.\]

Here I have expanded across the first row. It is worthwhile noting that

\[ -1 \times C_{2, 1} + 4 \times C_{2, 2} + 1 \times C_{2,3} \]

is 34 also as well as

\[ 0 \times C_{3, 1} + (-2) \times C_{1, 2} + 4 \times C_{1,3} \]

In fact for any $3$ by $3$ matrix $M$ and any $i = 1, 2 \mbox{ or } 3$

\[det(M) =a_{i, 1} C_{i, 1} + a_{i, 2} C_{i, 2} + a_{i, 3} C_{i, 3}.\]

You can also expand along any column if you wish so for $j = 1, 2 \mbox{ or } 3$

\[det(M) = a_{1, j} C_{1, j} + a_{2, j} C_{2, j} + a_{3, j} C_{3, j}.\]

I hope this helps,
Penny