Question

If $$AB = 0$$  where \[A = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\theta }&{\cos \theta \sin \theta } \\ {\cos \theta \sin \theta }&{{{\sin }^2}\theta } \end{array}} \right]\]      and \[B = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\phi }&{\cos \phi \sin \phi } \\ {\cos \phi \sin \phi }&{{{\sin }^2}\phi } \end{array}} \right]\]      then $$\left| {\theta - \phi } \right|$$  is equal to

A. $$0$$
B. $$\frac{\pi }{2}$$  
C. $$\frac{\pi }{4}$$
D. $$\pi $$
Answer :   $$\frac{\pi }{2}$$
Solution :
\[AB = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\theta }&{\cos \theta \cdot \sin \theta } \\ {\cos \theta \sin \theta }&{{{\sin }^2}\theta } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\phi }&{\cos \phi \sin \phi } \\ {\cos \phi \sin \phi }&{{{\sin }^2}\phi } \end{array}} \right]\]
\[AB = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\theta {{\cos }^2}\phi + \cos \theta \sin \theta \cos \phi \sin \phi }&{{{\cos }^2}\theta \cos \phi \sin \phi + \cos \theta \sin \theta {{\sin }^2}\phi } \\ {\cos \theta \sin \theta {{\cos }^2}\phi + \cos \phi \sin \phi \cdot {{\sin }^2}\theta }&{\cos \theta \sin \theta \cos \phi \sin \phi + {{\sin }^2}\theta {{\sin }^2}\phi } \end{array}} \right]\]
\[AB = \left[ {\begin{array}{*{20}{c}} {\cos \theta \cos \phi \cos \left( {\theta - \phi } \right)}&{\cos \theta \sin \phi \cos \left( {\theta - \phi } \right)} \\ {\sin \theta \cos \phi \cos \left( {\theta - \phi } \right)}&{\sin \theta \sin \phi \cos \left( {\theta - \phi } \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]\]
$$ \Rightarrow \,\,\cos \left( {\theta - \phi } \right) = 0$$
$$ \Rightarrow \,\,\left| {\theta - \phi } \right| = $$   odd multiple of $$\frac{\pi }{2}.$$

Releted MCQ Question on
Algebra >> Matrices and Determinants

Releted Question 1

Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$  be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$  be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then

A. $$C$$ is empty
B. $$B$$  has as many elements as $$C$$
C. $$A = B \cup C$$
D. $$B$$  has twice as many elements as elements as $$C$$
View Answer
Releted Question 2

If $$\omega \left( { \ne 1} \right)$$  is a cube root of unity, then
\[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|=\]

A. 0
B. 1
C. $$i$$
D. $$\omega $$
View Answer
Releted Question 3

Let $$a, b, c$$  be the real numbers. Then following system of equations in $$x, y$$  and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$    $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$    $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$     has

A. no solution
B. unique solution
C. infinitely many solutions
D. finitely many solutions
View Answer
Releted Question 4

If $$A$$ and $$B$$ are square matrices of equal degree, then which one is correct among the followings?

A. $$A + B = B + A$$
B. $$A + B = A - B$$
C. $$A - B = B - A$$
D. $$AB=BA$$
View Answer

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Matrices and Determinants

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