Question

verify that the given matrix satisfies the given differential equation. \(\mathbf{\Psi}^{\prime}=\left(\begin{array}{rrr}{1} & {-1} & {4} \\ {3} & {2} & {-1} \\ {2} & {1} & {-1}\end{array}\right) \mathbf{\Psi}, \quad \mathbf{\Psi}(t)=\left(\begin{array}{rrr}{e^{t}} & {e^{-2 t}} & {e^{3 t}} \\\ {-4 e^{t}} & {-e^{-2 t}} & {2 e^{3 t}} \\ {-e^{t}} & {-e^{-2 t}} & {e^{3 t}}\end{array}\right)\)

Short answer

Answer: Yes, the given matrix \(\mathbf{\Psi}(t)\) satisfies the differential equation, as we have shown that \(\mathbf{\Psi}'(t) = A\mathbf{\Psi}(t)\) in our step-by-step solution.

Step-by-step solution

Compute the derivative of \(\mathbf{\Psi}(t)\) with respect to time \(t\)

To compute the derivative of the matrix \(\mathbf{\Psi}(t)\), we need to compute the derivative of each entry of the matrix with respect to \(t\). The derivative of \(\mathbf{\Psi}(t)\) is given by: $\mathbf{\Psi}'(t) = \left(\begin{array}{rrr} {\frac{d(e^{t})}{dt}} & {\frac{d(e^{-2t})}{dt}} & {\frac{d(e^{3t})}{dt}} \\ {\frac{d(-4e^{t})}{dt}} & {\frac{d(-e^{-2t})}{dt}} & {\frac{d(2e^{3t})}{dt}} \\ {\frac{d(-e^{t})}{dt}} & {\frac{d(-e^{-2t})}{dt}} & {\frac{d(e^{3t})}{dt}} \end{array}\right)$ Calculating the derivatives, we get: $\mathbf{\Psi}'(t) = \left(\begin{array}{rrr} {e^{t}} & {-2e^{-2t}} & {3e^{3t}} \\ {-4e^{t}} & {2e^{-2t}} & {6e^{3t}} \\ {-e^{t}} & {2e^{-2t}} & {3e^{3t}} \end{array}\right)$

Multiply the given matrix by \(\mathbf{\Psi}(t)\)

Now, we need to multiply the given matrix \(A = \left(\begin{array}{rrr}{1} & {-1} & {4} \\\ {3} & {2} & {-1} \\\ {2} & {1} & {-1}\end{array}\right)\) by \(\mathbf{\Psi}(t)\): \(A\mathbf{\Psi}(t) = \left(\begin{array}{rrr}{1} & {-1} & {4} \\\ {3} & {2} & {-1} \\\ {2} & {1} & {-1}\end{array}\right) \left(\begin{array}{rrr}{e^{t}} & {e^{-2 t}} & {e^{3 t}} \\\ {-4 e^{t}} & {-e^{-2 t}} & {2 e^{3 t}} \\\ {-e^{t}} & {-e^{-2 t}} & {e^{3 t}}\end{array}\right)\) After performing the matrix multiplication, we obtain: $A\mathbf{\Psi}(t) = \left(\begin{array}{rrr} {e^{t}} & {-2e^{-2t}} & {3e^{3t}} \\ {-4e^{t}} & {2e^{-2t}} & {6e^{3t}} \\ {-e^{t}} & {2e^{-2t}} & {3e^{3t}} \end{array}\right)$

Check that \(\mathbf{\Psi}'(t)\) is equal to \(A\mathbf{\Psi}(t)\)

Now, we can compare the results of step 1 and step 2. We find that: $\mathbf{\Psi}'(t) = \left(\begin{array}{rrr} {e^{t}} & {-2e^{-2t}} & {3e^{3t}} \\ {-4e^{t}} & {2e^{-2t}} & {6e^{3t}} \\ {-e^{t}} & {2e^{-2t}} & {3e^{3t}} \end{array}\right)$ $A\mathbf{\Psi}(t) = \left(\begin{array}{rrr} {e^{t}} & {-2e^{-2t}} & {3e^{3t}} \\ {-4e^{t}} & {2e^{-2t}} & {6e^{3t}} \\ {-e^{t}} & {2e^{-2t}} & {3e^{3t}} \end{array}\right)$ Since \(\mathbf{\Psi}'(t) = A\mathbf{\Psi}(t)\), we can conclude that the given matrix \(\mathbf{\Psi}(t)\) indeed satisfies the given differential equation.

Key concepts

Understanding Differential Equations

Differential equations are a cornerstone of understanding change and dynamics in various physical systems. In essence, they are mathematical equations that relate a function with its derivatives. When we work with matrix differential equations, we deal with equations that set a matrix-valued function's derivative equal to the function itself multiplied by another matrix.

In our exercise, the matrix differential equation's solution, \( \mathbf{\Psi}(t) \), must satisfy the equation \( \mathbf{\Psi}^{\textprime} = A \mathbf{\Psi} \), where A is a constant matrix and \( \mathbf{\Psi}^{\textprime} \) is the derivative of \( \mathbf{\Psi}(t) \). Each entry in the derivative matrix \( \mathbf{\Psi}^{\textprime}(t) \) represents the rate of change of the respective entry in \( \mathbf{\Psi}(t) \).

Matrix Exponentiation

Matrix exponentiation is a powerful tool for solving systems of linear differential equations. This concept becomes crucial when dealing with time evolution in systems, especially in physics and engineering. Exponentiating a matrix involves creating a series of terms wherein the matrix is raised to an incrementing power and multiplied by a scalar derived from the exponential power series formula.

The matrices given in our exercise contain exponential functions of 't' as their entries, such as \( e^{t} \), \( e^{-2t} \) and \( e^{3t} \). These functions describe how the system evolves over time. To verify that \( \mathbf{\Psi}(t) \) satisfies the differential equation, we use techniques of differentiating these matrix entries, which inherently relies on understanding matrix exponentiation.

Matrix Multiplication

In the realm of matrices, multiplication is not simply an element-wise operation; it's a process where the row elements of one matrix are multiplied with the corresponding column elements of another to yield a new matrix.

In the given problem, we multiply matrix A by \( \mathbf{\Psi}(t) \), and this step is crucial for checking if \( \mathbf{\Psi}(t) \) truly satisfies the differential equation presented. This multiplication is not commutative, meaning that \( A \mathbf{\Psi} \) is not necessarily equal to \( \mathbf{\Psi} A \), which underscores the importance of the order in which we multiply these matrices.

Derivatives of Matrix Entries

When working with matrices in a context that involves calculus, we encounter scenarios where we need to take derivatives of matrices themselves. This operation is performed entry-wise, meaning we differentiate each entry of the matrix independently with respect to the variable of interest.

In our example, the differentiation of \( \mathbf{\Psi}(t) \) with respect to 't' required us to compute the derivative of an exponential function in each entry. Since each function is an expression involving 't', we apply the standard rules of differentiation to obtain \( \mathbf{\Psi}^{\textprime}(t) \) as needed for our comparison with the product \( A \mathbf{\Psi}(t) \) in the final verification step.