Problem 65 - Differential Equations and Linear Algebra

Question

Use the determinant to show that $$A=\left[\begin{array}{llll} 1 & 2 & 3 & 4 \\ 2 & 1 & 2 & 3 \\ 3 & 2 & 1 & 2 \\ 4 & 3 & 2 & 1 \end{array}\right]$$ is invertible, and use \(A^{-1}\) to solve \(A \mathbf{x}=\mathbf{b}\) if \(\mathbf{b}=\) \([3,7,1,-4]^{T}\)

Step-by-Step Solution

Answer

To show matrix A is invertible, we find its determinant and get \(\det (A) = -16\), which is non-zero. Thus, A is invertible. We compute the inverse of A as: \( A^{-1} = \left[\begin{matrix} -1/8 & 1/4 & -1/8 & 1/4 \\ 1/4 & -1/2 & 1/4 & -1/2 \\ -1/8 & 1/4 & -1/8 & 1/4 \\ 1/4 & -1/2 & 1/4 & -1/2 \end{matrix}\right] \) To solve the system A**x**=**b**, we multiply the inverse of A with **b**: \( \mathbf{x}= A^{-1}\mathbf{b} = \left[\begin{matrix} -1 \\ 4 \\ -1 \\ 4 \end{matrix}\right] \)

Step 1: Find the determinant of A

To find the determinant of A, use the formula: \( \det(A)=\left|\begin{matrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 2 & 3 \\ 3 & 2 & 1 & 2 \\ 4 & 3 & 2 & 1 \end{matrix}\right| \) Finding the determinant of A, we get: \( \det(A) = -16 \) Since the determinant of A is non-zero, the matrix A is invertible.

Step 2: Find the inverse of A

To find the inverse of A, use the formula: \( A^{-1} = \frac{1}{\det(A)} \cdot adj(A) \) where adj(A) is the adjugate of A. Compute the adjugate of A and the inverse A^(-1). We get: \( A^{-1}= \left[\begin{matrix} -1/8 & 1/4 & -1/8 & 1/4 \\ 1/4 & -1/2 & 1/4 & -1/2 \\ -1/8 & 1/4 & -1/8 & 1/4 \\ 1/4 & -1/2 & 1/4 & -1/2 \end{matrix}\right] \)

Step 3: Solve A**x**=**b** for **x**

To solve the system of linear equations, use the formula: \( \mathbf{x}=A^{-1}\mathbf{b} \) Multiplying the inverse of A with the given vector **b**, we get: \( \mathbf{x}= \left[\begin{matrix} -1/8 & 1/4 & -1/8 & 1/4 \\ 1/4 & -1/2 & 1/4 & -1/2 \\ -1/8 & 1/4 & -1/8 & 1/4 \\ 1/4 & -1/2 & 1/4 & -1/2 \end{matrix}\right] \cdot \left[\begin{matrix} 3 \\ 7 \\ 1 \\ -4 \end{matrix}\right] \) After performing the matrix-vector multiplication, the solution **x** is: \( \mathbf{x}=\left[\begin{matrix} -1 \\ 4 \\ -1 \\ 4 \end{matrix}\right] \)

Key Concepts

Matrix Inversion

Matrix inversion is a key concept when dealing with square matrices like the one given in the exercise. The idea is to find a matrix called the inverse that "undoes" the effect of the original matrix when paired with it, similar to how multiplying a number by its reciprocal results in 1. For a matrix \( A \), the inverse is denoted \( A^{-1} \).

The existence of an inverse is tightly linked to the determinant of a matrix. A matrix is invertible if its determinant is non-zero. Thus, computing the determinant is a prerequisite for inversion; if the determinant is zero, the matrix lacks an inverse.

The formula for finding the inverse \( A^{-1} \) involves computing the adjugate (or adjoint) of the matrix and dividing it by the determinant:

• Calculate the determinant of \( A \).
• Compute the adjugate: this involves taking the transpose of the cofactor matrix.
• Calculate \( A^{-1} \) as \( \frac{1}{\det(A)}\cdot \text{adj}(A) \).

This process allows you to work through various applications, including solving systems of equations, by manipulating the matrix instead of working through individual equations.

System of Linear Equations

A system of linear equations is essentially a set of equations where each term is either a constant or the product of a constant and a single variable. These systems can be solved to find the values of variables that satisfy all the equations simultaneously.

Matrices provide a compact way to represent these systems. The equation \( A\mathbf{x}=\mathbf{b} \) describes the relationship where \( A \) is a known matrix, \( \mathbf{b} \) is the output vector, and \( \mathbf{x} \) is the vector of unknowns. Solving for \( \mathbf{x} \) means finding the values that satisfy this equation.

The usefulness of the matrix inverse \( A^{-1} \) comes into play here:

• If \( A \) is invertible, you can rewrite the system as \( \mathbf{x} = A^{-1}\mathbf{b} \).
• This method provides a straightforward technique to find solutions, as demonstrated in the exercise.

This approach is particularly powerful for complex systems where manual substitution isn't feasible. However, solving equations this way is only viable if \( A \) is invertible.

Matrix-Vector Multiplication

Matrix-vector multiplication is a fundamental operation in linear algebra. It involves multiplying a matrix by a vector, resulting in a new vector. This operation helps transform data, apply systems of equations, and handle multidimensional data.

When you multiply a matrix \( A \) with a vector \( \mathbf{b} \), each element of the resultant vector is derived from the dot product of corresponding rows of the matrix with the vector:

• Each entry in the resultant vector is computed by multiplying elements of a row in the matrix by corresponding elements in the vector and summing them up.
• This technique is used when solving \( A\mathbf{x} = \mathbf{b} \), as it allows for computing transformations of vectors.

In the exercise, solving \( \mathbf{x} = A^{-1}\mathbf{b} \) involves matrix-vector multiplication between the inverse of the matrix and the vector \( \mathbf{b} \). This approach simplifies the process of finding the solution to a system of equations represented in matrix form.