If $F\left( x \right) = \left[ {\begin{array}{*{20}{c}} {\cos x}&{ - \sin x}&0\\ {\sin x}&{\cos x}&0\\ 0&0&1 \end{array}} \right]\;{\rm{and\;}}G\left( y \right) = \left[ {\begin{array}{*{20}{c}} {\cos x}&0&{\sin x}\\ 0&1&0\\ { - \sin x}&0&{\cos x} \end{array}} \right],$ then ${\left[ {F\left( x \right)G\left( y \right)} \right]^{ - 1}}$ is equal to

$F\left( {{x^{ - 1}}} \right)G\left( {{y^{ - 1}}} \right)$

$G\left( {{y^{ - 1}}} \right)F\left( {{x^{ - 1}}} \right)$

JEE MATHEMATICS MCQ: MATRICES

1 Medium

1 of 1

Application +4 | -1

F\left( x \right) = \left[ {\begin{array}{cccccccccccccccccccc} {\cos x}&{ - \sin x}&0\\ {\sin x}&{\cos x}&0\\ 0&0&1 \end{array}} \right]\;\mathrm{and\;}G\left( y \right) = \left[ {\begin{array}{cccccccccccccccccccc} {\cos x}&0&{\sin x}\\ 0&1&0\\ { - \sin x}&0&{\cos x} \end{array}} \right],

then

{\left[ {F\left( x \right)G\left( y \right)} \right]^{ - 1}}

is equal to

View answer and explanation

Correct option: C

G\left( { - y} \right)F\left( { - x} \right)

Option C (

G\left( { - y} \right)F\left( { - x} \right)

) is correct because the inverse of the product of two matrices is the product of their inverses in reverse order.