# Prove the derivative of the determinant equals the sum of the derivatives of ea/row

## Homework Statement

PROVE:
If A(t) is nxn with elements which are differentiable functions of t
Then:
$\frac{d}{dt}$(det(A))=$\sum$det(Ai(t))
where Ai(t) is found by differentiating the ith row only.

## Homework Equations

I know I should prove this by induction on n

## The Attempt at a Solution

Consider the matrix A1 being a 1x1 matrix
So n=1
the derivative of the determinant is the same as the derivative of that one row, therefore the theorem holds for n=1

assume the proof will hold true for n=k call this matrix Ak

now prove the theorem holds true for n=k+1
$\frac{d}{dt}$(det(Ak+1)) =$\frac{d}{dt}$(det(A1)+$\frac{d}{dt}$(det(Ak))
AND
$\sum$(det(Ak+1) = $\sum$det(A1)+$\sum$det(Ak)

Is this it?
have I proved it?

I had posted the exact same question a while ago and was too lazy to solve it.

I can't understand how you justify your first equation.
I can assure you the proof is much longer.

The determinant of a 1x1 matrix is equal to that one entry in the matrix isnt it?
so the derivative of the determinant is equal to the derivative of the one function in A.

HallsofIvy
Homework Helper

## Homework Statement

PROVE:
If A(t) is nxn with elements which are differentiable functions of t
Then:
$\frac{d}{dt}$(det(A))=$\sum$det(Ai(t))
where Ai(t) is found by differentiating the ith row only.

## Homework Equations

I know I should prove this by induction on n
Sounds like an excellent plan!

## The Attempt at a Solution

Consider the matrix A1 being a 1x1 matrix
So n=1
the derivative of the determinant is the same as the derivative of that one row, therefore the theorem holds for n=1
Good.

assume the proof will hold true for n=k call this matrix Ak

now prove the theorem holds true for n=k+1
$\frac{d}{dt}$(det(Ak+1)) =$\frac{d}{dt}$(det(A1)+$\frac{d}{dt}$(det(Ak))
Now, you have lost me. A "k+1 by k+ 1" determinant is NOT the sum of a 1 by 1 matrix and a k by k determinant which what you appear to be saying since you just use the "sum rule" for the derivative.

And $\sum$(det(Ak+1) = $\sum$det(A1)+$\sum$det(Ak)

Is this it?
have I proved it?
What is true is that a determinant can be "expanded" on one row. That is, we can, for example, expand by the first row. The k+1 by k+1 determinant is the sum of each entry in the first row time (plus or minus) the k by k matrix got by removing the first row and appropriate column. Use that to do a proof by induction.

I knew there was something fishy about my proof.
It just seemed too simple