Enter The Matrix

Generalization

We have to calculate- $\Delta_{n\times n}=\begin{bmatrix} a_{ 11 } & a_{ 12 } & \dots & a_{ 1n } \\ { a }_{ 21 } & \ddots & & \\ \vdots & & \ddots & \\ { a }_{ n1 } & & & a_{ nn } \end{bmatrix}_{ n\times n }$ Which is equal to- $\Delta_{n\times n}=(a_{11}\times M_{11})+(a_{12}\times M_{12}+\cdots+(a_{1n}\times M_{1n}))$ $M_{1k}$ is the determinent of cofactor matrix of $a_{1k}$ which is a $(n-1)\times (n-1)$ matrix.
Hence if we name our calculating function as CalDet(n) ,
We would get,
CalDet(n)= n*(CalDet(n-1)+1)+(n-1)=n*(CalDet(n-1)+2)-1
This is a recursive function with 'CalDet(1)=0` (we don't need to perform any arithmetic operation to calculate determinant of a $1\times1$ matrix!)
Which in python gives -

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>>> def CalDet(n):
    if (n==1):
        return 0
    else:
        return n*(CalDet(n-1)+2)-1


>>> CalDet(10)
9864099

Explaination of CalDet(n)= n*(CalDet(n-1)+1)+(n-1)
$n \text{ (...there are } n \text{ } a_{1k}\text{ s)}\times$ $\color{#D61F06}{(}$ CalDet(n-1) $\color{grey}{\left(\text{...number of operations needed to calculate determinant of }\left[M_{1k}\right]_{(n-1)\times(n-1)}\right)} +$

$1\quad\color{grey}{\left(\text{...multiplication of }a_{1k}\times M_{1k} \right)}$ $\color{#D61F06}{)}$ $\color{grey}{\quad\quad\dots\text{this first term is for calculation of }A_{1k}}$ $+(n-1) \color{grey}{\left(...\text{ for } \sum{}A_{1k}\right)}$

$O(det(A_{n \times n} )) = O(det(A_{(n-1) \times (n-1)})) + 2n - 1$

$O(det(A_{10 \times 10} )) = 9864099$