Time Complexity in finding the determinant of upper triangular matrix of order $n*n$












0












$begingroup$


I am trying to prove that the time Complexity in finding the determinant of upper triangular matrix of order $n*n$ is $O (n)$



My Approach:

Let us proceed this problem by taking an upper triangular matrix of order $3*3$

Let $A_3=
begin{bmatrix}
a_{11} & a_{12} & a_{13} \
0 & a_{22} & a_{23} \
0 & 0 & a_{33}
end{bmatrix}
$

Then, $det(A_3)=a_{11}*a_{22}*a_{33}$
$implies $ two multiplications

So, for an upper triangular matrix of order $n*n$ , we require $(n-1)$ multiplications



Now,

we know: Time Complexity $[f(n)]$ is defined as total number of multiplications required,i.e,
$f(n)=$ Total number of multiplications

And Big-O notation is the highest degree(i.e, if $f(n)=3n^2 +2n+1 $ ,then in Big-O notation it is represented by $O(n^2)$)



So,

here $(n-1)$ multiplication is required
$implies f(n)=n-1$
$implies O(n)$



Although my answer is right but i don't know whether i have proceeded correctly,whether i have used the concepts correctly or not...

So, anyone please verify it






linear-algebra determinant computational-complexity







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Looks okay to me
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 13:29
















0












$begingroup$


I am trying to prove that the time Complexity in finding the determinant of upper triangular matrix of order $n*n$ is $O (n)$



My Approach:

Let us proceed this problem by taking an upper triangular matrix of order $3*3$

Let $A_3=
begin{bmatrix}
a_{11} & a_{12} & a_{13} \
0 & a_{22} & a_{23} \
0 & 0 & a_{33}
end{bmatrix}
$

Then, $det(A_3)=a_{11}*a_{22}*a_{33}$
$implies $ two multiplications

So, for an upper triangular matrix of order $n*n$ , we require $(n-1)$ multiplications



Now,

we know: Time Complexity $[f(n)]$ is defined as total number of multiplications required,i.e,
$f(n)=$ Total number of multiplications

And Big-O notation is the highest degree(i.e, if $f(n)=3n^2 +2n+1 $ ,then in Big-O notation it is represented by $O(n^2)$)



So,

here $(n-1)$ multiplication is required
$implies f(n)=n-1$
$implies O(n)$



Although my answer is right but i don't know whether i have proceeded correctly,whether i have used the concepts correctly or not...

So, anyone please verify it






linear-algebra determinant computational-complexity







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Looks okay to me
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 13:29














0












0








0





$begingroup$


I am trying to prove that the time Complexity in finding the determinant of upper triangular matrix of order $n*n$ is $O (n)$



My Approach:

Let us proceed this problem by taking an upper triangular matrix of order $3*3$

Let $A_3=
begin{bmatrix}
a_{11} & a_{12} & a_{13} \
0 & a_{22} & a_{23} \
0 & 0 & a_{33}
end{bmatrix}
$

Then, $det(A_3)=a_{11}*a_{22}*a_{33}$
$implies $ two multiplications

So, for an upper triangular matrix of order $n*n$ , we require $(n-1)$ multiplications



Now,

we know: Time Complexity $[f(n)]$ is defined as total number of multiplications required,i.e,
$f(n)=$ Total number of multiplications

And Big-O notation is the highest degree(i.e, if $f(n)=3n^2 +2n+1 $ ,then in Big-O notation it is represented by $O(n^2)$)



So,

here $(n-1)$ multiplication is required
$implies f(n)=n-1$
$implies O(n)$



Although my answer is right but i don't know whether i have proceeded correctly,whether i have used the concepts correctly or not...

So, anyone please verify it






linear-algebra determinant computational-complexity







share|cite|improve this question









$endgroup$




I am trying to prove that the time Complexity in finding the determinant of upper triangular matrix of order $n*n$ is $O (n)$



My Approach:

Let us proceed this problem by taking an upper triangular matrix of order $3*3$

Let $A_3=
begin{bmatrix}
a_{11} & a_{12} & a_{13} \
0 & a_{22} & a_{23} \
0 & 0 & a_{33}
end{bmatrix}
$

Then, $det(A_3)=a_{11}*a_{22}*a_{33}$
$implies $ two multiplications

So, for an upper triangular matrix of order $n*n$ , we require $(n-1)$ multiplications



Now,

we know: Time Complexity $[f(n)]$ is defined as total number of multiplications required,i.e,
$f(n)=$ Total number of multiplications

And Big-O notation is the highest degree(i.e, if $f(n)=3n^2 +2n+1 $ ,then in Big-O notation it is represented by $O(n^2)$)



So,

here $(n-1)$ multiplication is required
$implies f(n)=n-1$
$implies O(n)$



Although my answer is right but i don't know whether i have proceeded correctly,whether i have used the concepts correctly or not...

So, anyone please verify it






linear-algebra determinant computational-complexity




linear-algebra determinant computational-complexity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 13:26









SureshSuresh

309110




309110








  • 1




    $begingroup$
    Looks okay to me
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 13:29














  • 1




    $begingroup$
    Looks okay to me
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 13:29








1




1




$begingroup$
Looks okay to me
$endgroup$
– Shubham Johri
Dec 17 '18 at 13:29




$begingroup$
Looks okay to me
$endgroup$
– Shubham Johri
Dec 17 '18 at 13:29










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