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Chapter 4: Problem 26

Construct an example of a \((2 \times 2)\) matrix function \(A(t)\) such that \(A^{2}(t)\) is a constant matrix but \(A(t)\) is not a constant matrix.

Short Answer

Expert verified
In summary, we have constructed a 2x2 matrix function A(t): $$ A(t) = \begin{pmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix} $$ We have shown that A^2(t) is constant, as it simplifies to the identity matrix, which is independent of t: $$ A^2(t) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$ We have also demonstrated that A(t) is not constant, as its elements are trigonometric functions of t, which vary with the input value of t.

Step by step solution

01

Find an appropriate form for A(t)

A possible form for A(t) is given by the following matrix: $$ A(t) = \begin{pmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix} $$ This is a rotation matrix, where t is the angle of rotation in radians.
02

Prove that A^2(t) is constant while A(t) is not

We will first show that A^2(t) is a constant matrix. To do that, we multiply A(t) by itself: $$ A^2(t) = A(t) \cdot A(t) = \begin{pmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix} \cdot \begin{pmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix} $$ Now, perform the matrix multiplication: $$ A^2(t) = \begin{pmatrix} \cos^2(t) + \sin^2(t) & \sin(t)\cos(t) - \sin(t)\cos(t) \\ -\sin(t)\cos(t) + \sin(t)\cos(t) & \sin^2(t) + \cos^2(t) \end{pmatrix} $$ Using the trigonometric identity \(\sin^2(t) + \cos^2(t) = 1\), we can simplify A^2(t) to: $$ A^2(t) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$ Since this matrix does not depend on t, A^2(t) is constant. Now, let's show that A(t) is not constant. If A(t) were constant, then its elements would not depend on t. However, the elements of A(t) are trigonometric functions of t, so A(t) cannot be constant. Therefore, A(t) is not a constant matrix.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotation Matrix
In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in 2-dimensional or 3-dimensional spaces. This kind of matrix is fundamental when it comes to transforming vectors through geometric rotations. In the context of our matrix function example, the specific form of the rotation matrix is: \[A(t) = \begin{pmatrix}\cos(t) & \sin(t) \-\sin(t) & \cos(t)\end{pmatrix}\] This matrix rotates any vector in the 2D plane by an angle \(t\) (in radians). Each element in this matrix depends on trigonometric functions which change with different values of \(t\). Hence, it allows for dynamic rotation, changing how vectors are oriented in the plane based on \(t\). The rotation matrix has important properties:
  • Its determinant is always 1, indicating a pure rotation without distortion or stretching.
  • When multiplied by its transpose, it results in the identity matrix, confirming it's an orthogonal matrix.
Trigonometric Identity
Trigonometric identities are equations that involve trigonometric functions and are true for every value of the occurring variables. They are indispensable in simplifying expressions and solving trigonometric equations. The most used trigonometric identity is: \[\sin^2(t) + \cos^2(t) = 1\] In our solution, this identity helps us simplify the squared matrix \(A^2(t)\). After performing matrix multiplication, this identity reduces the elements \(\cos^2(t) + \sin^2(t)\) to 1, which leads to: \[A^2(t) = \begin{pmatrix}1 & 0 \0 & 1\end{pmatrix}\] This simplification shows that \(A^2(t)\) is actually the identity matrix, which is crucial in proving that \(A^2(t)\) is constant regardless of \(t\). Understanding the role of trigonometric identities allows clearer insight into the simplification processes that make these cheeky square calculations work.
Matrix Multiplication
Matrix multiplication is a way to combine two matrices to produce a third matrix. It involves taking rows from the first matrix and columns from the second to form new components in the resultant matrix. In analyzing \(A^2(t)\), we multiply \(A(t)\) with itself. Here's how it works for our rotation matrix:
  • Row 1, Column 1: Combine and simplify \(\cos(t)\cdot\cos(t) + \sin(t)\cdot(-\sin(t))\)
  • Row 1, Column 2: Combine and simplify \(\cos(t)\cdot\sin(t) + \sin(t)\cdot\cos(t)\)
  • Row 2, Column 1: Combine and simplify \(-\sin(t)\cdot\cos(t) + \cos(t)\cdot\sin(t)\)
  • Row 2, Column 2: Combine and simplify \(-\sin(t)\cdot\sin(t) + \cos(t)\cdot\cos(t)\)
Applying the trigonometric identity \(\sin^2(t) + \cos^2(t) = 1\), these expressions simplify to form the identity matrix. Understanding this process helps recognize patterns in matrix calculations and utilize identities for simplification effectively.
Constant Matrix
A constant matrix is one whose elements remain unchanged. Regardless of the transformations or operations applied, the output does not vary; this matrix behaves like a steady fixture in your calculations. In our exercise, the matrix \(A^2(t)\), after multiplication and simplification, becomes: \[A^2(t) = \begin{pmatrix}1 & 0 \0 & 1\end{pmatrix}\] This result exhibits a perfect constant matrix, which is the identity matrix. Notably, it indicates that the outcome does not depend on \(t\); thus it remains constant.
  • This property confirms the uniform nature of this matrix in operations.
  • An identity matrix, often used as a multiplicative identity, does not alter other matrices it multiplies.
Being aware of how and why certain matrices remain constant across transformations is pivotal in matrix algebra, especially when dealing with varying components like time or angles, seen in matrix functions like \(A(t)\).

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Most popular questions from this chapter

Consider the homogeneous linear system \(\mathbf{y}^{\prime}=A \mathbf{y} .\) Recall that any associated fundamental matrix satisfies the matrix differential equation \(\Psi^{\prime}=A \Psi\). In each exercise, construct a fundamental matrix that solves the matrix initial value problem \(\Psi^{\prime}=A \Psi, \Psi\left(t_{0}\right)=\Psi_{0}\).\(\Psi^{\prime}=\left[\begin{array}{ll}3 & -4 \\\ 2 & -3\end{array}\right] \Psi, \quad \Psi(0)=\left[\begin{array}{ll}1 & 0 \\\ 0 & 1\end{array}\right]\)

Each initial value problem was obtained from an initial value problem for a higher order scalar differential equation. What is the corresponding scalar initial value problem? $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 0 & 1 \\ -3 & 2 \end{array}\right] \mathbf{y}+\left[\begin{array}{c} 0 \\ 2 \cos 2 t \end{array}\right], \quad \mathbf{y}(-1)=\left[\begin{array}{l} 1 \\ 4 \end{array}\right] $$

A Spring-Mass-Dashpot System with Variable Damping As we saw in Section \(3.6\), the differential equation modeling unforced damped motion of a mass suspended from a spring is \(m y^{\prime \prime}+\gamma y^{\prime}+k y=0\), where \(y(t)\) represents the downward displacement of the mass from its equilibrium position. Assume a mass \(m=1 \mathrm{~kg}\) and a spring constant \(k=4 \pi^{2} \mathrm{~N} / \mathrm{m}\). Also assume the damping coefficient \(\gamma\) is varying with time: $$ \gamma(t)=2 t e^{-t / 2} \mathrm{~kg} / \mathrm{sec} \text {. } $$ Assume, at time \(t=0\), the mass is pulled down \(20 \mathrm{~cm}\) and released. (a) Formulate the appropriate initial value problem for the second order scalar differential equation, and rewrite it as an equivalent initial value problem for a first order linear system. (b) Applying Euler's method, numerically solve this problem on the interval \(0 \leq t \leq 10 \mathrm{~min}\). Use a step size of \(h=0.005\). (c) Plot the numerical solution on the time interval \(0 \leq t \leq 10 \mathrm{~min}\). Explain, in qualitative terms, the effect of the variable damping upon the solution.

For each of the exercises, (a) Rewrite the equations from the given exercise in vector form as \(\mathbf{y}^{\prime}(t)=A \mathbf{y}(t)\), identifying the constant matrix \(A\). (b) Rewrite the solution of the equations in part (a) in vector form as \(\mathbf{y}(t)=c_{1} \mathbf{y}_{1}(t)+\) $$ \mathbf{y}^{\prime}=A \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r} -1 \\ 8 \end{array}\right], \text { where } A=\left[\begin{array}{ll} 3 & 2 \\ 4 & 1 \end{array}\right] \text { and } \mathbf{y}(t)=c_{1} e^{5 t}\left[\begin{array}{l} 1 \\ 1 \end{array}\right]+c_{2} e^{-t}\left[\begin{array}{r} -1 \\ 2 \end{array}\right] $$

Find the largest interval \(a
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