JEE-Mains 2015 (4/30)

One can find the answer by direct substitution, but there is another way to do this:

$\begin{aligned}\because \Big(\frac A3\Big)\Big(\frac {A^T}3\Big)&=I\end{aligned}$

So $\frac13A$ is orthogonal matrix.

$\begin{aligned}\frac13A&=\left[\begin{matrix} \frac13&\frac23&\frac23\\\frac23&\frac13&-\frac23\\\frac a3&\frac23&\frac b3 \end{matrix}\right]\end{aligned}$

We can see the column vector $c_1=(\frac13,\frac23,\frac a3),c_2=(\frac23,\frac13,\frac23),c_3=(\frac23,-\frac23,\frac b3)$ , these three vector are pairwise orthogonal, so $c_1\cdot c_2=0,c_2\cdot c_3=0$ , leads to $a=-2,b=-1$ .

Simply Find The Element Corresponding to First Row And Third Column After Multiplication and put it equal to zero it will yield a+2b = -4 only one option satisfies