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3 and 4 .Determinants and Matrices
hard
Let integers $a , b \in[-3,3]$ be such that $a + b \neq 0$. Then the number of all possible ordered pairs $(a, b)$, for which $\left|\frac{z-a}{z+b}\right|=1$ and $\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|$ $=1, z \in C$, where $\omega$ and $\omega^2$ are the roots of $x^2+x+$ $1=0$, is equal to________
A$10$
B$11$
C$12$
D$13$
(JEE MAIN-2025)
Solution
$a, b \in I,-3 \leq a, b \leq 3, a+b \neq 0$
$|z-a|=|z+b|$
$\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1$
$\Rightarrow\left|\begin{array}{ccc}z & z & z \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1$
$\Rightarrow z\left|\begin{array}{ccc}1 & 1 & 1 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1$
$\Rightarrow z\left|\begin{array}{ccc}1 & 0 & 0 \\ \omega & z+\omega^2-\omega & 1-\omega \\ \omega^2 & 1-\omega^2 & z+\omega-\omega^2\end{array}\right|=1$
$\Rightarrow z^3=1$
$\Rightarrow z=\omega, \omega^2, 1$
Now
$|1-a|=|1+b|$
$\Rightarrow 10 \text { pairs }$
$|z-a|=|z+b|$
$\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1$
$\Rightarrow\left|\begin{array}{ccc}z & z & z \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1$
$\Rightarrow z\left|\begin{array}{ccc}1 & 1 & 1 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1$
$\Rightarrow z\left|\begin{array}{ccc}1 & 0 & 0 \\ \omega & z+\omega^2-\omega & 1-\omega \\ \omega^2 & 1-\omega^2 & z+\omega-\omega^2\end{array}\right|=1$
$\Rightarrow z^3=1$
$\Rightarrow z=\omega, \omega^2, 1$
Now
$|1-a|=|1+b|$
$\Rightarrow 10 \text { pairs }$
Standard 12
Mathematics
