# Derivative of a determinant whose entries are functions

Happy New Year, everyone! I do not understand a remark in Adams' Calculus (page 628 $7^{th}$ edition). This remark is about the derivative of a determinant whose entries are functions as quoted below.

$$Since every term in the expansion of a determinant of any order is a product involving one element from each row, the general product rule implies that the derivative of an $n\times n$ determinant whose elements are functions will be the sum of $n$ such $n\times n$ determinants, each with the elements of one of the rows differentiated. For the $3\times 3$ case we have

$$\frac{d}{dt}\begin{vmatrix} a_{11}(t) & a_{12}(t) & a_{13}(t) \\ a_{21}(t) & a_{22}(t) & a_{23}(t) \\ a_{31}(t) & a_{32}(t) & a_{33}(t) \end{vmatrix}=\begin{vmatrix} a'_{11}(t) & a'_{12}(t) & a'_{13}(t) \\ a_{21}(t) & a_{22}(t) & a_{23}(t) \\ a_{31}(t) & a_{32}(t) & a_{33}(t) \end{vmatrix}+\begin{vmatrix} a_{11}(t) & a_{12}(t) & a_{13}(t) \\ a'_{21}(t) & a'_{22}(t) & a'_{23}(t) \\ a_{31}(t) & a_{32}(t) & a_{33}(t) \end{vmatrix}+\begin{vmatrix} a_{11}(t) & a_{12}(t) & a_{13}(t) \\ a_{21}(t) & a_{22}(t) & a_{23}(t) \\ a'_{31}(t) & a'_{32}(t) & a'_{33}(t) \end{vmatrix}."$$

It is not difficult to check this equality by simply expanding both sides. However, the remark sounds like using some clever trick to get this result. Can anyone explain it to me, please? Thank you!

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The author is probably referring to the fact that the determinant is given by:

$$\sum_{i,j,k=1}^n\varepsilon_{ijk}a_{1i}a_{2j}a_{3k}$$

where $\varepsilon_{ijk}$ is $1$ if $(ijk)$ is an even permutation of $(123)$, $-1$ if $(ijk)$ is an odd permutation of $(123)$ and $0$ if two or more of $i,j,k$ are equal. Differentiating this expression immediately gives:

$$\sum_{i,j,k=1}^n\varepsilon_{ijk}(a_{1i}'a_{2j}a_{3k}+a_{1i}a_{2j}'a_{3k}+a_{1i}a_{2j}a_{3k}')$$

which is easily seen to be the sum of the three determinants given, using the same formula again.

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Thank you for your response! – LaTeXFan Jan 1 '14 at 8:46

That remarks has said most of what it needs to explain.However, I think a more precise explaination for the example is necessary.Hence, I'll cite one.
i) $a_{11}(t).a_{23}(t).a_{32}(t)$ is abitrary term in expansion of left determinant
ii) $(a_{11}(t).a_{23}(t).a_{32}(t))' = a'_{11}(t)a_{23}(t).a_{32}(t)+a_{11}(t)a'_{23}(t).a_{32}(t)+a_{11}(t)a_{23}(t).a'_{32}(t)$
iii) $a'_{11}(t)a_{23}(t).a_{32}(t)$ is a term in expansion of the first determinant on the left of equation. This is a determinant in which the first row is differentiated
$a_{11}(t)a'_{23}(t).a_{32}(t),a_{11}(t)a_{23}(t).a'_{32}(t)$ are similar.

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Thank you for your response! Surely it is a right example. However, it is not clear whether the two sides are equal without expanding them. – LaTeXFan Jan 1 '14 at 8:43

The determinant is like a generalized product of vectors (in fact, it is related to the outer product). So considering the rows as factors in this generalized product, this formula reflects the product rule of differentiation.

If $D(a,b,c)$ is generally a function of vectors that is linear in each argument, and you apply it to vector functions in one variable, then

\begin{align} &D(a(t+h),b(t+h),c(t+h))-D(a(t),b(t),c(t))\\[0.5em] &=\ D(a(t+h),b(t+h),c(t+h))-D(a(t),b(t+h),c(t+h))\\ &\ +D(a(t),b(t+h),c(t+h))-D(a(t),b(t),c(t+h))\\ &\ +D(a(t),b(t),c(t+h))-D(a(t),b(t),c(t))\\[0.5em] &=\ D([a(t+h)-a(t)],b(t+h),c(t+h)) +D(a(t),[b(t+h)-b(t)],c(t+h)) +D(a(t),b(t),[c(t+h)-c(t)]) \end{align}

and from that the claimed generalized product rule can be obtained.

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Thank you for your response! – LaTeXFan Jan 1 '14 at 9:33

If a $n$x$n$ matrix $A(t)=A_{ij}(t)$ is differentiable, then $d(det(A(t))/dt$ can be performed as follows (for brevity, we shall take the particular case $n=3$). The determinant of a matrix $A(t)=A_{ij}(t)$ is given by $$det(A_{ij}(t))=e_{ijk}A_{i1}(t)A_{j2}(t)A_{k3}(t),$$ where $e_{ijk}$ is the Cartesian alternator. Let us take the derivative of this expression with respect to the parameter $t$, $$\frac{d}{dt}det(A_{ij}(t))=e_{ijk}(A_{i1}(t),tA_{j2}(t)A_{k3}(t)+A_{i1}(t)A_{j2}(t),tA_{k3}(t)+A_{i1}(t)A_{j2}(t)A_{k3}(t),t),$$ where "$,t$" stands for derivation. Let us now consider the quantity $$C_{k3}(t)=e_{ijk}A_{i1}(t)A_{j2}(t)=\frac{1}{2}(e_{ijk}A_{i1}(t)A_{j2}(t)+e_{jik}A_{j1}(t)A_{i2}(t))=\frac{1}{2}e_{ijk}(A_{i1}(t)A_{j2}(t)-A_{i2}(t)A_{j1}(t)),$$ which is the same as $$C_{k3}(t)=\frac{1}{2}e_{ijk}e_{lm3}A_{il}(t)A_{jm}(t),$$ and in general, $$C_{kn}(t)=\frac{1}{2}e_{ijk}e_{lmn}A_{il}(t)A_{jm}(t).$$ This is the so-called adjugate ($adj$) of the matrix $A(t)$ (notice that this formula is also useful to find the inverse of $A(t)$ whenever it be no singular). Thus, the sum for $\frac{d}{dt}det(A_{ij}(t))$ above, may be expressed by $$\frac{d}{dt}det(A_{ij}(t))=\frac{1}{2}e_{ijk}e_{lmn}A_{il}(t)A_{jm}(t)A_{kn}(t),t=C_{kn}(t)A_{kn}(t),t,$$ which is clearly the trace'' $tr$ of this product. So that, in matrix notation we have $$\frac{d(detA(t))}{dt}=tr(adj(A(t))\frac{d(A(t))}{dt}).$$ Indeed this formula is also valid for any finite dimension $n$, and its proof is a little more elaborated. We have exposed here the case for $n=3$ for reason of brevity. Also note that $C_{kn}^T=C_{nk}$ is a cofactor of the matrix $A(t).$

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