Question

(a) If A b, what are the derivptives Oxi/abj with A fixed? (b) What are the derivatives of Oxi/Ajk with b fixed? 3

Answer 1

(a) We have

AIkk k=1

Thus, for any fixed , we get

Obi ab に!

Since

Obi the Kronecker's δ 8, ,

we find

k=1 に!

Recall that we fixed ; writing

$y_k:={\frac{\partial x_k}{\partial b_j}}$

we get

y1 =е

Therefore,

$\begin{pmatrix}y_1\\ \cdot\\\cdot\\\cdot\\ y_n\end{pmatrix}=A^{-1}{\bf e_j}$

which means

${\frac{\partial x_i}{\partial b_j}}=A^{-1}_{ij}$

(b) We have

$b_i=\sum_{l=1}^nA_{i,l}x_l$

Thus, for any fixed , we get

$\begin{align*}{\frac{\partial b_i}{\partial A_{j,k}}}&=\sum_{l=1}^n{\frac{\partial A_{i,l}}{\partial A_{j,k}}} x_l+\sum_{l=1}^nA_{il}{\frac{\partial x_l}{\partial A_{j,k}}} \end{align*}$

Since ${\bf b}$ is fixed, we have

${\frac{\partial b_i}{\partial A_{j,k}}}=0$

Also, because

${\frac{\partial A_{i,l}}{\partial A_{j,k}}}=0~~~~~~\forall~(i,l)\neq (j,k),~~~~~~~~~~{\frac{\partial A_{j,k}}{\partial A_{j,k}}}=1$

we find that if $i\neq j$ then

$\begin{align*}0&={\frac{\partial b_i}{\partial A_{j,k}}}\\ &=\sum_{l=1}^n{\frac{\partial A_{i,l}}{\partial A_{j,k}}} x_l+\sum_{l=1}^nA_{il}{\frac{\partial x_l}{\partial A_{j,k}}}\\ &=\sum_{l=1}^nA_{il}{\frac{\partial x_l}{\partial A_{j,k}}}\end{align*}$

and if $\begin{align*}i=j\end{align*}$ then

$\begin{align*}0&={\frac{\partial b_i}{\partial A_{j,k}}}\\ &=\sum_{l=1}^n{\frac{\partial A_{i,l}}{\partial A_{j,k}}} x_l+\sum_{l=1}^nA_{il}{\frac{\partial x_l}{\partial A_{j,k}}}\\ &=x_k+\sum_{l=1}^nA_{jl}{\frac{\partial x_l}{\partial A_{j,k}}}\end{align*}$

Recall that we fixed ; writing

$y_l:={\frac{\partial x_l}{\partial A_{j,k}}}$

we get

$\begin{align*}-x_k&=\sum_{l=1}^nA_{jl}y_l\\ 0&=\sum_{l=1}^nA_{il}y_l~~~~~~~~~~~~~i\neq j\end{align*}$

Equivalently,

$A\begin{pmatrix}y_1\\ \cdot\\\cdot\\\cdot\\ y_n\end{pmatrix}=\begin{pmatrix}0\\\cdot\\\cdot\\0\\-x_k\\0\\\cdot\\\cdot\\0\end{pmatrix}=-x_k{\bf e_j}$

Therefore,

$\begin{pmatrix}y_1\\ \cdot\\\cdot\\\cdot\\ y_n\end{pmatrix}=-x_kA^{-1}{\bf e_j}$

which means

${\frac{\partial x_i}{\partial A_{j,k}}}=-x_kA^{-1}_{ij}$