Question-183863

Question Number 183863 by Michaelfaraday last updated on 31/Dec/22

Answered by qaz last updated on 31/Dec/22

(d/dθ){ determinant (( ((A,B),(C,D) ), ((E,F),(G,H) )))} =(d/dθ){ determinant ((A,B),(C,D))∙ determinant ((E,F),(G,H))} = determinant ((A,B),(C,D))^′ ∙ determinant ((E,F),(G,H))+ determinant ((A,B),(C,D))∙ determinant ((E,F),(G,H))^′ =( determinant (((A′),(B′)),(C,D))+ determinant ((A,B),((C′),(D′))))∙ determinant ((E,F),(G,H))+ determinant ((A,B),(C,D))∙( determinant (((E′),(F′)),(G,H))+ determinant ((E,F),((G′),(H′))))

$$\frac{{d}}{{d}\theta}\left\{\begin{vmatrix}{\begin{pmatrix}{{A}}&{{B}}\\{{C}}&{{D}}\end{pmatrix}}&{\begin{pmatrix}{{E}}&{{F}}\\{{G}}&{{H}}\end{pmatrix}}\end{vmatrix}\right\} \\ $$$$=\frac{{d}}{{d}\theta}\left\{\begin{vmatrix}{{A}}&{{B}}\\{{C}}&{{D}}\end{vmatrix}\centerdot\begin{vmatrix}{{E}}&{{F}}\\{{G}}&{{H}}\end{vmatrix}\right\} \\ $$$$=\begin{vmatrix}{{A}}&{{B}}\\{{C}}&{{D}}\end{vmatrix}^{'} \centerdot\begin{vmatrix}{{E}}&{{F}}\\{{G}}&{{H}}\end{vmatrix}+\begin{vmatrix}{{A}}&{{B}}\\{{C}}&{{D}}\end{vmatrix}\centerdot\begin{vmatrix}{{E}}&{{F}}\\{{G}}&{{H}}\end{vmatrix}^{'} \\ $$$$=\left(\begin{vmatrix}{{A}'}&{{B}'}\\{{C}}&{{D}}\end{vmatrix}+\begin{vmatrix}{{A}}&{{B}}\\{{C}'}&{{D}'}\end{vmatrix}\right)\centerdot\begin{vmatrix}{{E}}&{{F}}\\{{G}}&{{H}}\end{vmatrix}+\begin{vmatrix}{{A}}&{{B}}\\{{C}}&{{D}}\end{vmatrix}\centerdot\left(\begin{vmatrix}{{E}'}&{{F}'}\\{{G}}&{{H}}\end{vmatrix}+\begin{vmatrix}{{E}}&{{F}}\\{{G}'}&{{H}'}\end{vmatrix}\right) \\ $$

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