MCQ
If $y = f(x) = \frac{{ax + b}}{{cx - a}}$, then $x$ is equal to
- A$1/f(x)$
- B$1/f(y)$
- C$yf(x)$
- ✓$f(y)$
$⇒ x(cy - a) = b + ay$
$⇒ x = \frac{{ay + b}}{{cy - a}} = f(y)$.
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Statement $-2 :$ For any matrix $A,$ $\det \left( {{A^T}} \right) = {\rm{det}}\left( A \right)$ and $\det \left( { - A} \right) = - {\rm{det}}\left( A \right)$ Where $\det \left( A \right) = A$. Then :