## Determinant

$$ \det(M) = m_{11}M_{1,1} - m_{12}M_{1,2} +(-1)^{n+1}\dots m_{1n}M_{1,n}$$

where $m_{ij}$ is the element of the matrix in row $i$ and column $j$ and $M_{i,j}$ denotes the determinant of the sub matrix of $M$ formed by deleting the $i^{th}$ row and $j^{th}$ column. This is also called the $(i,j)$ - minor.

This is an operation that is easier understood through example than through description!

## Calculation of determinant 1

The determinant of $M$ is given by

$$\begin{align} \det(M) &= 1\det\left(\begin{array}{cc}5 & -2 \\ 5 & 2\end{array}\right) - 2\det\left(\begin{array}{cc}3 & -2 \\ 2 & 2\end{array}\right) + 4\det\left(\begin{array}{cc}3 & 5 \\ 2 & 5\end{array}\right) \\ &= 1((5)(2) - (-2)(5)) - 2((3)(2) - (-2)(2)) + 4((3)(5) - (5)(2))\\ &= 20 - 20 +20 \\ &= 20\end{align}$$

## Calculation of determinant 2

The interesting aspect of this question is expanding the derivative about the second row instead of the first; this will substantially reduce the amount of work:

The determinant is given by $$ \begin{align} \det(A) &= -m_{21}M_{21} + m_{22}M_{22} - m_{23}M_{23} \\ &= 0\cdot M_{21} + 0\cdot M_{22} - 3M_{23} \\ &= -3((2)(1) - (4)(4)) \\ &= 42 \end{align} $$