



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


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