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Answered on 06 Apr Learn Unit III: Calculus
Sadika
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Answered on 06 Apr Learn Unit III: Calculus
Sadika
To determine the principal value of
Answered on 06 Apr Learn Unit III: Calculus
Sadika
Let's denote \( \tan^{-1} \left( \frac{x - 1}{x - 2} \right) \) as \( \alpha \) and \( \tan^{-1} \left( \frac{x + 1}{x + 2} \right) \) as \( eta \).
Given that \( \tan^{-1} \left( \frac{x - 1}{x - 2} \right) + \tan^{-1} \left( \frac{x + 1}{x + 2} \right) = \frac{\theta}{4} \), we can use the tangent addition formula:
\[ \tan(\alpha + eta) = \frac{\tan \alpha + \tan eta}{1 - \tan \alpha \cdot \tan eta} \]
Substitute \( \tan \alpha = \frac{x - 1}{x - 2} \) and \( \tan eta = \frac{x + 1}{x + 2} \):
\[ \tan(\alpha + eta) = \frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \cdot \frac{x + 1}{x + 2}} \]
\[ \tan(\alpha + eta) = \frac{\frac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2)}}{1 - \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)}} \]
\[ \tan(\alpha + eta) = \frac{x^2 + x - 2 + x^2 - x - 2}{(x - 2)(x + 2) - (x^2 - 1)} \]
\[ \tan(\alpha + eta) = \frac{2x^2 - 4}{x^2 + 4 - x^2 + 1} \]
\[ \tan(\alpha + eta) = \frac{2x^2 - 4}{5} \]
Given that \( \tan(\alpha + eta) = \frac{\theta}{4} \), we have:
\[ \frac{2x^2 - 4}{5} = \frac{\theta}{4} \]
\[ 8x^2 - 16 = 5\theta \]
\[ 8x^2 = 5\theta + 16 \]
\[ x^2 = \frac{5\theta + 16}{8} \]
\[ x = \pm \sqrt{\frac{5\theta + 16}{8}} \]
So, the value of \( x \) depends on the value of \( \theta \).
In LaTeX code:
\[ x = \pm \sqrt{\frac{5\theta + 16}{8}} \]
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 06 Apr Learn Unit III: Calculus
Sadika
To find the principal value of tan−1(1)tan−1(1), we need to determine the angle whose tangent is equal to 1.
Since tangent is defined as the ratio of the opposite side to the adjacent side in a right triangle, we can consider a right triangle where the angle whose tangent is 1 is one of its acute angles.
In a right triangle, if the ratio of the opposite side to the adjacent side is 1, then the opposite side and the adjacent side are equal in length. Therefore, we have a triangle with legs of equal length.
The angle whose tangent is 1 corresponds to a 45-degree angle (or π44π radians) in standard position.
So, the principal value of tan−1(1)tan−1(1) is π44π radians.
In LaTeX code: tan−1(1)=π4tan−1(1)=4π
Answered on 06 Apr Learn Matrices
Sadika
The identity matrix of order n is denoted by I∩
It is a square matrix with dimensions n×nn×n where all the elements on the main diagonal (from the top left to the bottom right) are 1, and all other elements are 0.
Answered on 06 Apr Learn Matrices
Sadika
A square matrix is a matrix that has the same number of rows and columns. In other words, it is a matrix where the number of rows is equal to the number of columns.
For example, a 3×33×3 matrix and a 4×44×4 matrix are both square matrices because they have 3 rows and 3 columns, and 4 rows and 4 columns, respectively.
Square matrices are commonly encountered in various mathematical contexts, such as linear algebra, where they are used to represent linear transformations, systems of linear equations, and many other mathematical structures.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 06 Apr Learn Matrices
Sadika
To find the number of all possible matrices of order 3×33×3 with each entry being either 0 or 1, we can consider that each entry in the matrix has 2 choices (0 or 1), and there are a total of 3×3=93×3=9 entries in the matrix.
Therefore, the total number of possible matrices is 2929, because for each entry, there are 2 choices, and we multiply these choices together for all 9 entries.
So, the number of all possible matrices of order 3×33×3 with each entry being either 0 or 1 is 29=51229=512.
Answered on 06 Apr Learn Matrices
Sadika
Two matrices A=[aij]A=[aij] and B=[bij]B=[bij] are said to be equal if they have the same dimensions (i.e., the same number of rows and columns) and if each corresponding entry of matrix AA is equal to the corresponding entry of matrix BB.
Formally, two matrices AA and BB are equal if and only if:
In other words, matrices AA and BB are equal if they have the same size and if the corresponding entries in each position are equal.
Answered on 06 Apr Learn Matrices
Sadika
In a skew-symmetric matrix, also known as an antisymmetric matrix, the diagonal elements are all equal to zero.
Formally, a matrix AA is skew-symmetric if it satisfies the condition AT=−AAT=−A, where ATAT denotes the transpose of matrix AA.
In a skew-symmetric matrix, for any diagonal element aiiaii, it must satisfy aii=−aiiaii=−aii. The only number that satisfies this condition is zero.
Therefore, every diagonal element of a skew-symmetric matrix is zero.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 06 Apr Learn Matrices
Sadika
If a matrix AA is both symmetric and skew-symmetric, then AA will be the zero matrix.
Let's denote AA as the matrix:
A=[aij]A=[aij]
Symmetric Matrix: A matrix is symmetric if it is equal to its transpose. Mathematically, AT=AAT=A. In other words, for every ii and jj, aij=ajiaij=aji.
Skew-Symmetric Matrix: A matrix is skew-symmetric if its transpose is equal to the negative of itself. Mathematically, AT=−AAT=−A. In other words, for every ii and jj, aij=−ajiaij=−aji.
Combining these two conditions, we have aij=ajiaij=aji and aij=−ajiaij=−aji.
The only number that satisfies both conditions simultaneously is aij=0aij=0, because it's the only number that is equal to its negative.
Therefore, in this case, every element of matrix AA must be zero, making AA the zero matrix.
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