Question

If $A=\left[\begin{array}{cc}co{s}^2\alpha & cos\alpha sin\alpha \\ cos\alpha sin\alpha & si{n}^2\alpha \end{array}\right]$ and $B=\left[\begin{array}{cc}co{s}^2\beta & cos\beta sin\beta \\ cos\beta sin\beta & si{n}^2\beta \end{array}\right]$ are two matrices such that the product $AB$ is null matrix, then $\alpha -\beta$ is

• A. $0$
• B. multiple of $\pi$
• C. an odd multiple of $\pi /2$
• D. None of the above

Answer 1

Answer: (C)

Given, $AB=O$ $\therefore \left[\begin{array}{cc}co{s}^2\alpha & cos\alpha sin\alpha \\ cos\alpha sin\alpha & si{n}^2\alpha \end{array}\right]$
$\times \left[\begin{array}{cc}co{s}^2\beta & cos\beta sin\beta \\ cos\beta sin\beta & si{n}^2\beta \end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\left[\begin{array}{cc}cos\alpha cos\beta cos\left(\alpha -\beta \right)& cos\alpha sin\beta cos\left(\alpha -\beta \right)\\ cos\beta sin\alpha cos\left(\alpha -\beta \right)& sin\alpha sin\beta cos\left(\alpha -\beta \right)\end{array}\right]$
$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\Rightarrow cos\left(\alpha -\beta \right)=0$
$\Rightarrow \alpha -\beta$ is an odd multiple of $\pi /2$