# Abels formula and homogeneous and non homogeneous solutions

As this is my first time dealing with ODE's, I'm having alot of trouble proving some of the theorems that are given to me involving homogeneous and non homogeneous systems. For example

Theorem

Let $$\phi_{1}(t),...,\phi_{n-1}(t)$$ be solutions of the homogeneous system $$\dot x = A(t)x$$ and let $$x(t)$$ be a solution of the non homogeneous system $$\dot x = A(t)x + b(t)$$ where $$A(t),b(t)$$ are continuous. Then $$Z=\det(\phi_{1}(t),...,\phi_{n-1}(t), x(t))$$ satisfies the non homogeneous Abel's formula $$\dot Z = \text{tr}(A(t))Z+\det(\phi_{1}(t),\dots,\phi_{n-1}(t),b(t))$$

Probably my problem is with linear algebra more than ODE's but still, I haven't been able to prove it nor find a proof online. Any help would be greatly appreciated. Thanks so much in advance.

In general, suppose $$B$$ and $$Y=[y_1,\ldots,y_n]$$ are two square matrices of the same sizes. Let $$\{e_1,\ldots,e_n\}$$ be the standard basis. When $$Y$$ is nonsingular, we have \begin{aligned} &\det\left(Y^{-1}BYe_1,e_2,\ldots,e_n\right) +\det\left(e_1,Y^{-1}BYe_2,\ldots,e_n\right) +\cdots +\det\left(e_1,\ldots,e_{n-1},Y^{-1}BYe_n\right)\\ &=\operatorname{tr}(Y^{-1}BY)=\operatorname{tr}(B). \end{aligned} Hence $$\det\left(By_1,y_2,\ldots,y_n\right) +\det\left(y_1,By_2,\ldots,y_n\right) +\cdots +\det\left(y_1,\ldots,y_{n-1},By_n\right)=\operatorname{tr}(B)\det(Y)\tag{1}$$ and by a continuity argument, $$(1)$$ is true for singular $$Y$$ as well.
Now, return to your question. By the product rule, \begin{aligned} \dot{Z} &= \det\left(\dot{\phi}_1,\phi_2,\ldots,\phi_{n-1},x\right) +\det\left(\phi_1,\dot{\phi}_2,\ldots,\phi_{n-1},x\right) +\cdots +\det\left(\phi_1,\phi_2,\ldots,\phi_{n-1},\dot{x}\right)\\ &= \det\left(A\phi_1,\phi_2,\ldots,\phi_{n-1},x\right) +\det\left(\phi_1,A\phi_2,\ldots,\phi_{n-1},x\right) +\cdots +\det\left(\phi_1,\phi_2,\ldots,\phi_{n-1},Ax+b\right)\\ &= \det\left(A\phi_1,\phi_2,\ldots,\phi_{n-1},x\right) +\det\left(\phi_1,A\phi_2,\ldots,\phi_{n-1},x\right) +\cdots +\det\left(\phi_1,\phi_2,\ldots,\phi_{n-1},Ax\right)\\ &\phantom{=}+\det\left(\phi_1,\phi_2,\ldots,\phi_{n-1},b\right), \end{aligned} Put $$B=A$$ and $$[y_1,\ldots,y_n]=[\phi_1,\ldots,\phi_{n-1},x]$$ into $$(1)$$, the result follows.