# Derivative of a determinant

• Aug 27th 2009, 10:00 PM
altave86
Derivative of a determinant
Which is a formula for calculating the derivative of a determinant with respect to one of its elements? How do I prove this?

I know the result has something to do with the inverse, but I can't just quite grasp the proof and can't find it anywhere
• Aug 27th 2009, 11:47 PM
red_dog
Let $\displaystyle D(x)=\begin{vmatrix}f_{11}(x) & f_{22}(x) & \dots & f_{1n}(x)\\f_{21}(x) & f_{22}(x) & \ldots & f_{2n}(x)\\\ldots & \ldots & \ldots & \ldots\\f_{i1}(x) & f_{i2}(x) & \ldots & f_{in}(x)\\\ldots & \ldots & \ldots & \ldots\\f_{n1}(x) & f_{n2}(x) & \ldots & f_{nn}(x)\end{vmatrix}$

Then

$\displaystyle D'(x)=\sum_{i=1}^n\begin{vmatrix}f_{11}(x) & f_{22}(x) & \dots & f_{1n}(x)\\f_{21}(x) & f_{22}(x) & \ldots & f_{2n}(x)\\\ldots & \ldots & \ldots & \ldots\\f_{i1}'(x) & f_{i2}'(x) & \ldots & f_{in}'(x)\\\ldots & \ldots & \ldots & \ldots\\f_{n1}(x) & f_{n2}(x) & \ldots & f_{nn}(x)\end{vmatrix}$
• Aug 28th 2009, 03:56 AM
halbard
If $\displaystyle \mathbf A$ is an invertible matrix which depends on a real parameter $\displaystyle t$, and if $\displaystyle \frac{\mathrm d\mathbf A}{\mathrm dt}$ exists, then

$\displaystyle \frac{\mathrm d}{\mathrm dt}(\det\mathbf A)=(\det\mathbf A)\mathop{\textrm{tr}}\biggl(\frac{\mathrm d\mathbf A}{\mathrm dt}\mathbf A^{-1}\biggr)$, where $\displaystyle \mathop{\textrm{tr}}(\mathbf B)$ is the trace of $\displaystyle \mathbf B$.

I have discovered a truly marvellous proof of the above statement, which unfortunately my brain is too small to contain.