How to find Determinant of a matrix I could not understand the concept while googling. can anybody provide help?
what will be the determinant of the following matrix?
$$
\left[\begin{array}{cccc}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
9 & 10 & 11 & 12 \\
13 & 14 & 15 & 16
\end{array} \right]$$
Thanks.... :)
 A: By the way, if you've continued your determinant (in order to draw a real one) in the same vein:
$$
D =
\begin{vmatrix}
1 & 2 & 3 & 4  \\\
5 & 6 & 7 & 8  \\\
9 & 10 & 11 & 12   \\\
13 & 14 & 15 & 16
\end{vmatrix} \ ,
$$
the answer would have been easy:
$$
D =
\begin{vmatrix}
1 & 2 & 3 & 4  \\\
5 & 6 & 7 & 8  \\\
4 & 4 & 4 & 4   \\\
4 & 4 & 4 & 4
\end{vmatrix}
= 0 \ .
$$
:-)
A: So to show how to calculate the determinant of a $n \times n$ matrix can be shown in a multiple of ways, but the method that we will use here is expansion by minors. We will first start of with a $2 \times 2$ matrix and work our way to a $3 \times 3$ matrix.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
$\underline{Example~1:}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ $2 \times 2$ Case
$$\left[\begin{array}{ccc}
a_{11} & a_{12} \\ 
a_{21} & a_{22} 
\end{array} \right]$$
Idea is that you want to multiply the diagonals and subtract them. So in other words:
$$a_{11}\cdot a_{22}-a_{12}\cdot a_{21}$$
That will be the method for any $2 \times 2$ matrix. Now we will look at an example of a $3 \times 3$ matrix.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
$\underline{Example~2:}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ $3 \times 3$ Case
$$\left[ \begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array} \right]$$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
First we must know the following about determinants, otherwise using this method will subdue wrong results. Below are the algebraic sign of the elements position within the matrix:
$$\left[ \begin{array}{ccc}
+ & - & + \\
- & + & - \\
+ & - & +
\end{array} \right]$$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
$\underline {\text{Evaluation of the determinant expanding it by the minors of column 1:}}$
= $~a_{11}
\left[\begin{array}{ccc}
a_{22} & a_{23} \\ 
a_{32} & a_{33} 
\end{array} \right]
-a_{21}
\left[\begin{array}{ccc}
a_{12} & a_{13} \\ 
a_{32} & a_{33} 
\end{array} \right]
+ a_{31}
\left[\begin{array}{ccc}
a_{12} & a_{13} \\ 
a_{22} & a_{23} 
\end{array} \right]
$ 
$\Rightarrow ~ a_{11}
\left(a_{22}\cdot a_{33}-a_{23}\cdot a_{32} \right)
-a_{21}
\left(a_{12}\cdot a_{33}-a_{13}\cdot a_{32} \right)
+ ~a_{31}
\left(a_{12}\cdot a_{23}-a_{13}\cdot a_{22} \right)
$ 
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
From here, you would just simply by multiply what inside the parenthesis and then subtracting the two quantities and then you would multiply the result inside of the parenthesis by the term outside the parenthesis for each (namely, $a_{11},~a_{21},~a_{31}$).
If you need more assistance on how to evaluate the determinant, here is a reference below: 
1)  Computing Determinants
Okay, well I hope this cleared it up some on how to calculate determinants for square matrices. There are other methods that you can perform for matrices that are larger say $4 \times 4$ or even $5 \times 5$ matrices. You would want to perform row operations for those or any other clever techniques you prefer.
Good~Luck.
A: You should put your matrix in  upper triangluar form then compute the product of the diagonal entries.  This is much faster than expansion by cofactors.  
A: Determinants are defined only for squared matrix. If you are sure about your question then simply the determinant for your matrix does not exist. 
A: Look too this link
http://fatosmatematicos.blogspot.com/2011/12/o-metodo-de-dodgson-para-calcular.html
A: This problem could be generalized for arbitrary n, where n is the dimension of the matrix.
\begin{bmatrix}1&2&3&... & n\\ ...&...&...&...&...& \\n^2-(n-1)&n^2 - (n - 2)&n^2-(n - 3)& ... & n^2\end{bmatrix}
Let denote 
$
r := \begin{bmatrix}1&2&3&... & n\end{bmatrix}, 
e := \begin{bmatrix}1&1&1&... & 1\end{bmatrix}$. 
Now our matrix looks like this:
\begin{bmatrix}r + 0 * e\\ r + 1 * e\\ ... \\r + (n - 1) * e\end{bmatrix}
We know 2  facts:


*

*Determinant is the polilinear function.

*If two lines are linearly dependent, then the determinant is 0.


If you use the 1st fact and expand the determinant  you will get the sum of determinants (I will denote them $d_i$) where each of them necessary will have two linear dependent rows.
For example one of $d_i$ in the sum looks like this:
\begin{bmatrix}r \\ 1 * e\\ ... \\(n - 1) * e\end{bmatrix}.
Thus each $d_i$ equals $0$ (2nd fact) and consequantly the initial determinant is 0 as the sum of zeros.
Answer: $n = 1 \ detA = 1; n = 2 \ detA = -2; n \geq 3 \ detA = 0$.
