Question

The product of matrices A = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\theta }&{\cos \theta \sin \theta } \\\ {\cos \theta \sin \theta }&{{{\sin }^2}\theta } \end{array}} \right]and $B = \left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\phi }&{\cos \phi \sin \phi } \\

{\cos \phi \sin \phi }&{{{\sin }^2}\phi }
\end{array}} \right]$is null matrix if$\theta - \phi = ?$$
a.{\text{ }}2n\pi ,n \in Z \\
b.{\text{ }}n\dfrac{\pi }{2},n \in Z \\
c.{\text{ }}\left( {2n + 1} \right)\dfrac{\pi }{2},n \in Z \\
d.{\text{ }}n\pi ,n \in Z \\
$

Answer 1

Hint- Null matrix is a matrix if all the elements of the matrix are zero. Multiplication of two matrices is given as a null matrix so the value of multiplication will be equal to Zero matrix.
Given matrix are
A = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\theta }&{\cos \theta \sin \theta } \\\ {\cos \theta \sin \theta }&{{{\sin }^2}\theta } \end{array}} \right], B = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\phi }&{\cos \phi \sin \phi } \\\ {\cos \phi \sin \phi }&{{{\sin }^2}\phi } \end{array}} \right]
Now find out the product of matrices i.e.$\left( {AB} \right)$
\left( {AB} \right) = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\theta }&{\cos \theta \sin \theta } \\\ {\cos \theta \sin \theta }&{{{\sin }^2}\theta } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\phi }&{\cos \phi \sin \phi } \\\ {\cos \phi \sin \phi }&{{{\sin }^2}\phi } \end{array}} \right]
Now apply the matrix multiplication rule, both matrices have 2 rows and 2 columns so the multiplication of these two matrices also have two rows and two columns.

{{{\cos }^2}\theta {{\cos }^2}\phi + \cos \theta \sin \theta \cos \phi \sin \phi }&{{{\cos }^2}\theta \cos \phi \sin \phi + \cos \theta \sin \theta {{\sin }^2}\phi } \\\ {\cos \theta \sin \theta {{\cos }^2}\phi + {{\sin }^2}\theta \cos \phi \sin \phi }&{\cos \theta \sin \theta \cos \phi \sin \phi + {{\sin }^2}\theta {{\sin }^2}\phi } \end{array}} \right]$$ Now this above matrix is also written as $$\left( {AB} \right) = \left[ {\begin{array}{*{20}{c}} {\cos \theta \cos \phi \left( {\cos \theta \cos \phi + \sin \theta \sin \phi } \right)}&{\cos \theta \sin \phi \left( {\cos \theta \cos \phi + \sin \theta \sin \phi } \right)} \\\ {\sin \theta \cos \phi \left( {\cos \theta \cos \phi + \sin \theta \sin \phi } \right)}&{\sin \theta \sin \phi \left( {\cos \theta \cos \phi + \sin \theta \sin \phi } \right)} \end{array}} \right]$$ Now as we know $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ so use this property we have $$\left( {AB} \right) = \left[ {\begin{array}{*{20}{c}} {\cos \theta \cos \phi \cos \left( {\theta - \phi } \right)}&{\cos \theta \sin \phi \cos \left( {\theta - \phi } \right)} \\\ {\sin \theta \cos \phi \cos \left( {\theta - \phi } \right)}&{\sin \theta \sin \phi \cos \left( {\theta - \phi } \right)} \end{array}} \right]$$ Now we have to convert the matrix to a null matrix, so we have to convert all the elements of the above matrix to zero. Therefore substitute $$\cos \left( {\theta - \phi } \right) = 0$$………….. (1) $$ \Rightarrow \left( {AB} \right) = \left[ {\begin{array}{*{20}{c}} {\cos \theta \cos \phi \cos \left( {\theta - \phi } \right)}&{\cos \theta \sin \phi \cos \left( {\theta - \phi } \right)} \\\ {\sin \theta \cos \phi \cos \left( {\theta - \phi } \right)}&{\sin \theta \sin \phi \cos \left( {\theta - \phi } \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\\ 0&0 \end{array}} \right]$$ Now from equation (1) We know that the value of cosine is zero for$$\left( {2n + 1} \right)\dfrac{\pi }{2}$$, where $$n \in Z$$ $$\cos \left( {\theta - \phi } \right) = 0 = \cos \left( {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right)$$, where $$n \in Z$$ So, on comparing $$\left( {\theta - \phi } \right) = \left( {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right)$$, $$n \in Z$$ Hence option (c) is correct. Note- In such types of questions the key concept we have to remember is that the null matrix is a matrix such that all the elements of the matrix are zero, so in this questions first find out the matrix multiplication, then simplify the matrix using basic trigonometric properties which is stated above, then substitute one of the element to zero which is common in all the elements of the matrix AB, then use the property of cosine which is stated above, we will get the required null matrix.