If [ _1 = | pmatrix x & & - & - x & 1 & 1 & x pmatrix |] and [ _2 = | pmatrix x & 2 & 2 - …

Question

If $${\Delta _1} = \left| {\begin{pmatrix} x & {\sin \theta } & {\cos \theta }  \
   { - \sin \theta } & { - x} & 1  \
   {\cos \theta } & 1 & x  \ \end{pmatrix} } \right|$$ and

$${\Delta _2} = \left| {\begin{pmatrix} x & {\sin 2\theta } & {\cos 2\theta }  \
   { - \sin 2\theta } & { - x} & 1  \
   {\cos 2\theta } & 1 & x  \ \end{pmatrix} } \right|$$, $$x \ne 0$$ ;

then for all $$\theta  \in \left( {0,{\pi  \over 2}} \right)$$ :

Accepted Answer

$${\Delta _1} = \left| {\begin{pmatrix} x & {\sin \theta } & {\cos \theta }  \
   { - \sin \theta } & { - x} & 1  \
   {\cos \theta } & 1 & x  \

\end{pmatrix} } \right|$$

= x(--x2 --1) -- sin$$\theta (--xsin\[\theta $$ -- cos$$\theta $$) + cos$$\theta (--sin\[\theta $$+ xcos$$\theta $$)

= --x3 -- x + xsin2$$\theta $$ + sin$$\theta $$cos$$\theta $$ -- cos$$\theta $$sin$$\theta $$
+ xcos2$$\theta $$

= --x3 -- x + x = --x3

Similarly $${\Delta _2} =  - {x^3}$$

$${\Delta _1} + {\Delta _2} =  - 2{x^3}$$

Solutions · 1