Proof of why $ d|X| = |X|\text{tr}(X^{-1}dX) $?

Question

I am trying to understand why the identity

$$ d|X| = |X|\text{tr}(X^{-1}dX) $$

How can we mathematically prove this?

I am not sure how to do this, so I tried to verify this for a small example:

$$ X = \begin{bmatrix} x & y \\ z & w \end{bmatrix} $$

Step 1: Calculate $|X|$

$$ |X| = xw - yz $$

Step 2: Calculate $d|X|$

$$ d|X| = \frac{\partial |X|}{\partial x}dx + \frac{\partial |X|}{\partial y}dy + \frac{\partial |X|}{\partial z}dz + \frac{\partial |X|}{\partial w}dw $$

Calculate each partial derivative:

$$ \frac{\partial |X|}{\partial x} = w $$ $$ \frac{\partial |X|}{\partial y} = -z $$ $$ \frac{\partial |X|}{\partial z} = -y $$ $$ \frac{\partial |X|}{\partial w} = x $$

Substituting these back into theequation:

$$ d|X| = w\,dx - z\,dy - y\,dz + x\,dw $$

Step 3: Now, to verify $d|X| = |X|\text{tr}(X^{-1}dX)$:

Calculate $X^{-1}$

$$ X^{-1} = \frac{1}{xw-yz} \begin{bmatrix} w & -y \\ -z & x \end{bmatrix} $$

Step 4: Express $dX$

$$ dX = \begin{bmatrix} dx & dy \\ dz & dw \end{bmatrix} $$

Step 5: Calculate $X^{-1}dX$

$$ X^{-1}dX = \frac{1}{xw-yz} \begin{bmatrix} w & -y \\ -z & x \end{bmatrix} \begin{bmatrix} dx & dy \\ dz & dw \end{bmatrix} $$

$$ = \frac{1}{xw-yz} \begin{bmatrix} w\,dx - y\,dz & w\,dy - y\,dw \\ -z\,dx + x\,dz & -z\,dy + x\,dw \end{bmatrix} $$

Step 6: Calculate $\text{tr}(X^{-1}dX)$

$$ \text{tr}(X^{-1}dX) = \frac{1}{xw-yz} (w\,dx - y\,dz - z\,dy + x\,dw) $$

Step 7: Calculate $|X|\text{tr}(X^{-1}dX)$

$$ |X|\text{tr}(X^{-1}dX) = (xw-yz) \cdot \frac{1}{xw-yz} (w\,dx - y\,dz - z\,dy + x\,dw) $$

$$ = w\,dx - y\,dz - z\,dy + x\,dw $$

After a lot of manipulation, I hope I have done this correctly. But can someone please help me prove this?

Answer 1

HINT: By cofactor expansion, the coefficient of $dx_{ij}$ in $d|X|$ is the $ij$-cofactor of $X$. Now recall that $|X|X^{-1}$ is the transpose of the cofactor mateix.