Question

Show that more than$\mathit{n}\mathbf{!}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathit{n}$ multiplications are required to compute the determinant of a$\mathbf{n}\mathbf{\times }\mathbf{n}$ matrix by Laplace expansion (for$\mathit{n}\mathbf{>}\mathbf{2}$).

Short answer

Therefore, the determinant is given by, $a_n>n!,Ɐn>2$.

Step-by-step solution

Definition 

A determinant is a unique number associated with asquare matrix.

A determinant is a scalar value that is a function of the entries of a square matrix.

It is the signed factor by which areas are scaled by this matrix. If the sign is negative the matrix reverses orientation.

To calculate the determinant

Let $a_n$be the number of multiplications required to calculate the determinant of a $n\times n$matrix.

We want to prove that role="math" localid="1660716604515" $a_n>n!,Ɐn>2$.

Induction basis. For $n=3$, we have $a_n=9$.

Assumption,

Assume that $a_n>n!$, for an existing $n>2$.

Induction step,

To calculate the determinant of a role="math" localid="1660716561869" $a\left(n+1\right)\times \left(n+1\right)$matrix, we use the Laplace expansion along any given row.

For each one of the $n+1$ entries in the row, calculating the determinant of the matrix without the row and column of the entry will take more than n ! Multiplications.

So calculating the whole determinant will indeed take more than $n!\left(n+1\right)=\left(n+1\right)!$multiplications.

Thus, $a_n>n!,Ɐn>2$.