Question

Find the inverse of the given matrix \(\mathbf{A}\) in each of Exercises 13-24. $$ \mathbf{A}=\left(\begin{array}{rr} 3 & -6 \\ -2 & 5 \end{array}\right) $$

Short answer

The inverse of matrix \(\mathbf{A}\) is: $$ \mathbf{A}^{-1}=\left(\begin{array}{rr} \frac{5}{3} & 2 \\ \frac{2}{3} & 1 \end{array}\right) $$

Step-by-step solution

Calculate the determinant of the matrix

To find the determinant of the \(2 \times 2\) matrix \(\mathbf{A}\), we will use the formula: \(\text{det}(\mathbf{A})=ad-bc\). In this case, \(\text{det}(\mathbf{A})=(3)(5)-(-6)(-2)\).

Check if the matrix is invertible

A matrix is invertible if and only if its determinant is not equal to 0. Calculate the determinant from the previous step: \(\text{det}(\mathbf{A})=15-12=3\). Since the determinant is not equal to 0, the matrix is invertible, and we can proceed to find the inverse.

Swap the elements on the main diagonal

Swap the elements \(a\) and \(d\) from matrix \(\mathbf{A}\): $$ \left(\begin{array}{rr} 5 & -6 \\ -2 & 3 \end{array}\right) $$

Change the sign of the elements on the other diagonal

Change the signs of elements \(b\) and \(c\) from matrix \(\mathbf{A}\): $$ \left(\begin{array}{rr} 5 & 6 \\ 2 & 3 \end{array}\right) $$

Divide by the determinant

Divide the resulting elements by the determinant of the original matrix, which is 3: $$ \mathbf{A}^{-1}=\frac{1}{3}\left(\begin{array}{rr} 5 & 6 \\\ 2 & 3 \end{array}\right) $$ Thus, the inverse of matrix \(\mathbf{A}\) is: $$ \mathbf{A}^{-1}=\left(\begin{array}{rr} \frac{5}{3} & 2 \\ \frac{2}{3} & 1 \end{array}\right) $$

Key concepts

Determinant of a Matrix

To begin with, the determinant of a matrix is a special number that can be calculated from a square matrix. For a 2x2 matrix, the formula to find the determinant is particularly straightforward:

• Let's denote the matrix as \( \mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix} \).
• The determinant, written as \( \text{det}(\mathbf{A}) \), is calculated using: \( ad - bc \).

This calculation is crucial because it tells us several things about the matrix. If the determinant equals zero, the matrix is not invertible. This means there's no possible way to "un-do" or reverse its effects, much like trying to divide by zero, which is undefined. In this exercise, the determinant of matrix \( \mathbf{A} \) was calculated as 3, reassuring us that an inverse does indeed exist.
Understanding determinants is fundamental in linear algebra, as they determine whether sets of linear equations have unique solutions.

Invertible Matrices

When discussing invertible matrices, what we're really talking about is whether a matrix can be reversed or inverted. This idea is similar to multiplying a number by its reciprocal to get one.

• A matrix is invertible when its determinant is not zero.
• An invertible matrix \( \mathbf{A} \) has an inverse, denoted \( \mathbf{A}^{-1} \), such that \( \mathbf{A} \cdot \mathbf{A}^{-1} = \mathbf{I} \), where \( \mathbf{I} \) represents the identity matrix.

The identity matrix is a special kind of matrix that serves as the "1" of matrix multiplication. For a 2x2 matrix, it looks like this: \(\mathbf{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \).If a matrix is invertible, any system of equations described by it will have a unique solution. Remember, we only find inverses for square matrices. In our topic, matrix \( \mathbf{A} \) proved invertible due to its non-zero determinant.

2x2 Matrix Inverse

Finding the inverse of a 2x2 matrix is a structured process with a standard formula. For our 2x2 matrix \( \mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the inverse is found by following these steps:

• First, check the determinant is not zero as discussed.
• Swap the elements of the main diagonal, \( a \) and \( d \).
• Change the signs of the other diagonal elements, \( b \) and \( c \).
• Finally, divide each element by the determinant.

This results in the inverse formula:\[\mathbf{A}^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}.\]Applying this to our matrix \( \mathbf{A} \), we exchanged and adjusted as needed, dividing everything by 3 to achieve \( \mathbf{A}^{-1} = \begin{pmatrix} \frac{5}{3} & 2 \ \frac{2}{3} & 1 \end{pmatrix} \). Inversions like this are incredibly useful for solving systems of linear equations and performing various operations in linear algebra.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, linear transformations, and systems of linear equations. Here are some key points about its role and importance:

• It provides the language and framework for expressing problems in mathematics and applied sciences.
• Matrices, determinants, and inverses are foundational elements used to solve linear systems.
• Applications of linear algebra abound in computer science, engineering, physics, economics, and more.

Concepts like invertible matrices and matrix inverses are everyday tools in linear algebra. They enable solutions to problems involving networks, modeling in economics, and rotations and transformations in computer graphics. Our discussion on matrix inversion is just one example of how linear algebra serves as a powerful tool for turning complex relationships into comprehensible and computable tasks.