Problem 39 - Precalculus: Mathematical for Calculus

Question

Find the inverse of the matrix. $$\begin{aligned} &\left[\begin{array}{rr} a & -a \\ a & a \end{array}\right]\\\ &(a \neq 0) \end{aligned}$$

Step-by-Step Solution

Answer

The inverse is \( \begin{bmatrix} \frac{1}{2a} & \frac{1}{2a} \\ \frac{-1}{2a} & \frac{1}{2a} \end{bmatrix} \).

Step 1: Identify the matrix

The matrix given is a 2x2 matrix: \[ A = \begin{bmatrix} a & -a \ a & a \end{bmatrix} \] where \( a eq 0 \).

Step 2: Use the inverse formula for a 2x2 matrix

For a 2x2 matrix \( \begin{bmatrix} e & f \ g & h \end{bmatrix} \), the inverse is calculated as \[ \frac{1}{(eh - fg)} \begin{bmatrix} h & -f \ -g & e \end{bmatrix} \]. Apply this formula to matrix \( A \).

Step 3: Calculate the determinant of the matrix

Compute the determinant of \( A \): \[ \det(A) = (a \cdot a) - (-a \cdot a) = a^2 + a^2 = 2a^2 \]. Since \( a eq 0 \), \( \det(A) = 2a^2 eq 0 \) indicating that the matrix has an inverse.

Step 4: Apply the inverse formula

Using the inverse formula, plug in the values: \[ A^{-1} = \frac{1}{2a^2} \begin{bmatrix} a & a \ -a & a \end{bmatrix} \].

Step 5: Simplify the inverse matrix

Multiply each term in the matrix by \( \frac{1}{2a^2} \): \[ A^{-1} = \begin{bmatrix} \frac{a}{2a^2} & \frac{a}{2a^2} \ \frac{-a}{2a^2} & \frac{a}{2a^2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2a} & \frac{1}{2a} \ \frac{-1}{2a} & \frac{1}{2a} \end{bmatrix} \].

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that is derived from a square matrix. It is a critical component in determining if a matrix is invertible, meaning if it has an inverse. For a 2x2 matrix, the formula to find the determinant is simple and follows a straightforward pattern. You calculate the determinant of a 2x2 matrix \( \begin{bmatrix} e & f \ g & h \end{bmatrix} \) using the equation \( \det(A) = eh - fg \). This calculation involves multiplying elements from the diagonals and subtracting the off-diagonal product.
This concept plays a pivotal role in linear algebra concepts, allowing us to understand transformations represented by matrices. If the determinant is zero, the matrix is singular, which means it has no inverse. In our example, the determinant is \( 2a^2 \), and since \( a eq 0 \), the determinant is non-zero, thus the matrix is invertible.

2x2 Matrix Inverse

Inverting a 2x2 matrix is a fundamental operation in linear algebra, often used in solving linear systems of equations. The inverse of a matrix \( A \) is another matrix, denoted as \( A^{-1} \), which, when multiplied with the original matrix, results in an identity matrix. For a 2x2 matrix, finding the inverse involves a particular formula.

• First, compute the determinant of the matrix.
• Then, apply the inverse formula: For a 2x2 matrix \( \begin{bmatrix} e & f \ g & h \end{bmatrix} \), its inverse is \( \frac{1}{eh - fg} \begin{bmatrix} h & -f \ -g & e \end{bmatrix} \).

Here, the inverse relies on multiplying the reciprocal of the determinant with a matrix formed by swapping and changing the signs of certain elements in the original matrix.
In our specific problem, after computing the determinant as \( 2a^2 \), the inverse was calculated as \[ A^{-1} = \begin{bmatrix} \frac{1}{2a} & \frac{1}{2a} \ \frac{-1}{2a} & \frac{1}{2a} \end{bmatrix} \]. This result is achieved by dividing each element of the matrix by \( 2a^2 \).

Linear Algebra Concepts

Linear algebra is a branch of mathematics focusing on vector spaces and linear mappings between these spaces. It encompasses various concepts such as matrices, vectors, determinants, and linear transformations. Understanding how to compute the inverse of a matrix, for example, is crucial in many applications like solving systems of linear equations and computer graphics.
Matrices serve as representations of linear transformations, which could be rotations, reflections, or scaling of vector spaces. When we look at the inverse of a matrix, it essentially represents the transformation that undoes the effect of the original matrix. This undoing is what makes the inverse so significant; it's like having the key to unlock the function represented by the matrix.
The importance of matrices in linear algebra cannot be overstated, as they serve in defining systems of equations, they are pivotal in engineering, physics, computer science, and more.