Problem 168
Question
If \(I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right], J=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\), then show that \(B=I \cos \theta+J \sin \theta\).
Step-by-Step Solution
Verified Answer
The given matrices I and J when multiplied by cos θ and sin θ respectively, then added together, form the matrix B, thus it’s proven that B = I cos θ + J sin θ.
1Step 1: Write out B, I and J
We start by writing out our given matrices I and J as well as matrix B:\nI=\[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\],\nJ=\[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\],\nB=\[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\]
2Step 2: Generating expression B using I and J
Now, the goal is to show that B = I cos θ + J sin θ. We start by multiplying matrix I by cos θ and matrix J by sin θ, then adding the two results: \n\nI cos θ = \[\begin{array}{ll}\cos \theta & 0 \\ 0 & \cos \theta\end{array}\] \n\nJ sin θ = \[\begin{array}{cc}0 & \sin \theta \\ -\sin \theta & 0\end{array}\] \n\nThen we add I cos θ and J sin θ: \n\nI cos θ + J sin θ = \[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\]
3Step 3: Compare the Result with B
Comparing the matrix we derived, I cos θ + J sin θ, with the original B matrix, it is clear that these two matrices are identical. Thus, it’s proven that B = I cos θ + J sin θ. That completes the solution to the problem.
Key Concepts
Matrix AdditionMatrix MultiplicationTrigonometric Functions
Matrix Addition
Matrix addition is a straightforward operation. It involves adding corresponding elements of two matrices of the same dimension. For instance, if you have two 2x2 matrices, you simply add each element in the same position to obtain a new matrix of the same dimension.
- Ensure that both matrices have the same dimensions. You cannot add matrices of different sizes.
- For matrices \( A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} \) and \( B = \begin{bmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{bmatrix} \), the result is \( A + B = \begin{bmatrix} a_{11} + b_{11} & a_{12} + b_{12} \ a_{21} + b_{21} & a_{22} + b_{22} \end{bmatrix} \).
Matrix Multiplication
Matrix multiplication is a bit more complex than addition. You need to be cautious about the dimensions of the matrices involved. A matrix \( A \) with dimensions \( m \times n \) can only be multiplied by another matrix \( B \) if \( B \) has dimensions \( n \times p \). The result will be a new matrix with dimensions \( m \times p \).
- The element at position \((i, j)\) in the resulting matrix is the dot product of the \(i\)th row of the first matrix and the \(j\)th column of the second matrix.
- Matrix multiplication is associative but not commutative. This means that while the grouping of matrices in a multiplication expression can change without affecting the outcome, the order cannot.
Trigonometric Functions
Trigonometric functions are used extensively in mathematics to relate angles to ratios of sides in right-angled triangles. The main functions are sine, cosine, and tangent. In linear algebra, specifically in rotational transformations, these functions are crucial.
- The cosine of an angle is the adjacent side divided by the hypotenuse in a right triangle, and sine is the opposite side divided by the hypotenuse.
- When an angle is used to rotate vectors in a two-dimensional space, matrices that incorporate sine and cosine values can represent these rotations.
Other exercises in this chapter
Problem 166
If \(A+B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\) and \(A-2 B=\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]\), then find \(A\).
View solution Problem 167
Find \(A\) and \(B\) if \(2 A-B=\left[\begin{array}{ccc}3 & -3 & 0 \\ 3 & 3 & 2\end{array}\right]\) and \(2 B+A=\left[\begin{array}{ccc}4 & 1 & 5 \\ -1 & 4 & -4
View solution Problem 169
What is the order of \(\left[\begin{array}{lll}x & y & z\end{array}\right]\left[\begin{array}{lll}a & h & g \\ h & b & f \\ g & f & c\end{array}\right]\left[\be
View solution Problem 170
Given \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 1 & 2 & -1\end{array}\right]\) and \(B=\left[\begin{array}{lll}1 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0\end{array}\ri
View solution