LU Decomposition Vs Gaussian Elimination

In my last two lectures, I talked about Gaussian elimination and LU decomposition methods to solve a linear system of equations. I should have emphasized that while performing the row reduction operations, one can switch the rows as well.
The LU decomposition looks more complicated than Gaussian elimination. Do we use LU decomposition because it is computationally more efficient than Gaussian elimination to solve a set of n equations given by [A][X]=[B]?. For a square matrix [A] of $n \times n$ size, the computational time $CT_{DE}$ to decompose the [A]matrix to [L][U]form is given by
$CT_{DE}=T \bigg( \frac{8n^3}{3}+4n^2-\frac{20n}{3}\bigg)$, where $T=clock cycle time^2$.

The computational time $CT_{FS}$ to solve by forward substitution [L][Z]=[B] is given by$CT_{FS}=T(4n^2-4n)$The computational time $CT_{BS}$ to solve by backward
substitution [U][X]=[Z] is given by $CT_{BS}=T(4n^2+12n)$

The total computational time required to solve by LU decomposition is given by
$CT_{LU}=CT_{DE}+CT_{FS}+CT_{BS}$
$ = T \bigg( \frac{8n^3}{3}+4n^2-\frac{20n}{3}\bigg)+T(4n^2-4n)+T(4n^2+12n)$
$ = T\bigg( \frac{8n^3}{3}+12 n^2+\frac{4n}{3} \bigg)$

If we now consider Gaussian elimination the computational time taken by Gaussian elimination. The computational time $CT_{FE}$ for the forward elimination part is given by

$CT_{FE}= T \bigg( \frac{8n^3}{3}+8n^2-\frac{32n}{3}\bigg)$

The computational time $CT_{BS}$required is $T(4n^2+12n)$ and the total computational time is given by $ CT_{GE}= CT_{FE}+CT_{BS}= T\bigg( \frac{8n^3}{3}+12 n^2+\frac{4n}{3} \bigg)$ which is the same as LU decomposition.


This part of analysis leads to a confusion whether to use Gaussian elimination or LU decomposition. Let us now look at finding the inverse of a matrix. If we consider a $n \times n $ matrix. For calculations of each column of the inverse of the [A] matrix, the coefficient matrix [A] matrix in the set of equation [A][X]=[B] does not change. So if we use the LU decomposition method, the [A]= [L][U] decomposition needs to be done only once, the forward substitution (Equation 1) n times, and the back substitution (Equation 2) n times.
$CT_{INv LU} =1 \times CT_{LU} +n \times CT_{FS}+n \times CT_{BS}$
$ = T\bigg ( \frac{32n^3}{3}+12n^2-\frac{20n}{3}\bigg)$
In comparison, if Gaussian elimination method was used to find the inverse of a matrix, the forward elimination as well as the back substitution will have to be done n times. The total computational time $CT_{INv GE}$ required to find an inverse of a matrix is given by
$CT_{INv GE}= n \times CT_{FE}+n \times CT_{BS}$
$ = T\bigg( \frac{8n^4}{3}+12 n^3+\frac{4n^2}{3}\bigg)$

Clearly for large $n$, $CT_{INv GE} >> CT{INv LU}$ as $CT_{INv GE}$ has a dominating $n^4$ term where as $CT_{INv LU}$ has a $n^3$ term.