Problem 12
Question
If \(E(\theta)=\left[\begin{array}{cc}\cos ^{2} \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin ^{2} \theta\end{array}\right]\) and \(\theta\) and \(\phi\) differ by an odd multiple of \(\frac{\pi}{2}\), then \(E(\theta) E(\phi)\) is a (A) null matrix (B) unit matrix (C) diagonal matrix (D) None of these
Step-by-Step Solution
Verified Answer
The answer is (A) null matrix.
1Step 1: Understand Odd Multiple of \(\frac{\pi}{2}\)
First, understand that when \(\theta\) and \(\phi\) differ by an odd multiple of \(\frac{\pi}{2}\), it means \(\phi = \theta + (2n+1)\frac{\pi}{2}\), where \(n\) is an integer. This implies one angle is a right angle plus \(\theta\) or minus \(\theta\).
2Step 2: Express \(E(\phi)\) Using Trigonometric Properties
If \(\phi = \theta + \frac{\pi}{2}\) or \(\phi = \theta - \frac{\pi}{2}\), utilize the trigonometric identities: \(\cos(\theta + \frac{\pi}{2}) = -\sin(\theta)\) and \(\sin(\theta + \frac{\pi}{2}) = \cos(\theta)\) to express \(E(\phi)\). The matrix \(E(\phi)\) becomes \(\left[\begin{array}{cc}\sin ^{2} \theta & -\cos \theta \sin \theta \ -\cos \theta \sin \theta & \cos ^{2} \theta\end{array}\right]\) or similar for \(\theta - \frac{\pi}{2}\).
3Step 3: Multiply \(E(\theta)\) and \(E(\phi)\)
Calculate the product \(E(\theta) E(\phi)\) where each entry of the resulting matrix is found by multiplying the respective row in \(E(\theta)\) with the respective column in \(E(\phi)\). The key observation here is to show that upon calculating, every entry sums to zero.
4Step 4: Determine Matrix Type
The resulting matrix has all entries as zero, hence it is a null matrix. This matches option (A), confirming that when \(\theta\) and \(\phi\) are odd multiples of \(\frac{\pi}{2}\) apart, \(E(\theta) E(\phi)\) equals a null matrix.
Key Concepts
Matrix MultiplicationTrigonometric IdentitiesNull Matrix
Matrix Multiplication
Matrix multiplication is a powerful tool that combines information from two matrices to produce a third matrix. In matrix multiplication, the element of the resulting matrix is calculated by taking the dot product of rows from the first matrix with columns from the second matrix. In other words, to compute an element in the resulting matrix, multiply each element of a row from the first matrix by the corresponding element in a column of the second matrix and sum these products.
- It's important to remember that matrix multiplication is not commutative. This means that the order of multiplication matters. For two matrices, A and B, generally, \(AB eq BA\).
- Matrix dimensions must align for multiplication: if A is of size \(m \times n\) and B of \(n \times p\), then their product AB will be \(m \times p\).
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables involved. They serve as fundamental tools in simplifying expressions and solving trigonometric equations. Some key trigonometric identities used in problems involving matrices include:
- Pythagorean identities: \(\cos^2 \theta + \sin^2 \theta = 1\)
- Angle sum identities: such as \(\cos(\theta + \frac{\pi}{2}) = -\sin(\theta)\) and \(\sin(\theta + \frac{\pi}{2}) = \cos(\theta)\)
Null Matrix
A null matrix, also known as a zero matrix, is a matrix in which all elements are zero. It's a special matrix that plays a pivotal role in linear algebra and signals important characteristics when it appears as a result:
- In multiplication, any matrix multiplied by a null matrix results in the null matrix, leading to an identity similar to zero in regular arithmetic.
- In the context of system solutions, a null matrix can indicate no transformation effect, meaning that the applied operations essentially "cancel out."
Other exercises in this chapter
Problem 7
If \(A\) is a square matrix, \(B\) is a singular matrix of same order, then for a positive integer \(n,\left(A^{-1} B A\right)^{n}\) equals (A) \(A^{-n} B^{n} A
View solution Problem 10
The matrix \(A=\left[\begin{array}{cc}1 / \sqrt{2} & 1 / \sqrt{2} \\ -1 / \sqrt{2} & -1 / \sqrt{2}\end{array}\right]\) is (A) unitary (B) orthogonal (C) nilpote
View solution Problem 13
If \(A\) and \(B\) are two square matrices such that \(B=\) \(-A^{-1} B A\), then \((A+B)^{2}=\) (A) 0 (B) \(A^{2}+B^{2}\) (C) \(A^{2}+2 A B+B^{2}\) (D) \(A+B\)
View solution Problem 14
If \(A=\left[\begin{array}{cc}1 & \tan \theta / 2 \\ -\tan \theta / 2 & 1\end{array}\right]\) and \(A B=I\), then \(B=\) (A) \(\cos ^{2} \frac{\theta}{2} A\) (B
View solution