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Chapter 8: Problem 58

Explain how to find the multiplicative inverse for a \(2 \times 2\) invertible matrix.

Short Answer

Expert verified
The inverse of a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by \(A^{-1} = \frac{1}{ad - bc} \times \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\)

Step by step solution

01

- Identify the Elements of the Given Matrix

Given a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), identify the elements a, b, c, and d at their respective positions in the matrix.
02

- Calculate the Determinant

The determinant of the matrix \(A\) is calculated by evaluating \(ad - bc\). It is important to note that for a matrix to be invertible, the determinant must not be zero.
03

- Switch a and d

Swap the positions of a and d in the matrix. So your matrix now looks like this \(\begin{bmatrix} d & b \ c & a \end{bmatrix}\).
04

- Negate b and c

Then, switch the signs of b and c so that your matrix looks according to the formula \(\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\).
05

- Divide by the Determinant

Divide each element in the new matrix by the calculated determinant. This results in the inverse of the matrix, i.e., \(A^{-1} = \frac{1}{ad - bc} \times \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\)

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Key Concepts

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

Understanding Invertible Matrices
An invertible matrix, also known as a non-singular or non-degenerate matrix, is a square matrix that has an inverse. This means that for a matrix \(A\), there exists another matrix \(A^{-1}\) such that when they are multiplied together, the result is the identity matrix \(I\). Put simply, this can be written as:
  • \(AA^{-1} = A^{-1}A = I\)
An essential condition for a matrix to be invertible is that it must have a non-zero determinant. In the case of a \(2 \times 2\) matrix, this concept is fairly straightforward to apply, making it easier to check and calculate the inverse. This necessity of a non-zero determinant ensures that the operations of the matrix preserve the space's volume, so to say, which is key for the invertibility.
Introduction to 2x2 Matrices
A \(2 \times 2\) matrix is a simple and small square matrix made up of two rows and two columns. This format makes it relatively easy to handle in terms of mathematical operations, including finding an inverse. The general representation of a \(2 \times 2\) matrix looks like this:
  • \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\)
Each letter (a, b, c, and d) represents an element of the matrix which has a specific position. These matrices are often the building blocks for understanding larger matrices and many fundamental concepts in linear algebra, such as transformations, are well illustrated using a \(2 \times 2\) matrix.
Calculating the Determinant of a Matrix
The determinant is a special number that can be calculated from a matrix. For a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant helps determine whether the matrix is invertible or not. The formula for the determinant of a \(2 \times 2\) matrix is:
  • \(det(A) = ad - bc\)
This determinant is significant because if \(ad - bc = 0\), the matrix does not have an inverse, meaning it is singular. However, if \(ad - bc eq 0\), then the matrix is invertible, and you can proceed with finding its inverse by using the determinant in further calculations, such as dividing the modified elements of the matrix by this determinant.

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Most popular questions from this chapter

Use a graphing utility to find the multiplicative inverse of each matrix. Check that the displayed inverse is correct. $$\left[\begin{array}{llll}1 & 2 & 0 & 0 \\\0 & 0 & 1 & 0 \\\1 & 3 & 0 & 1 \\\4 & 0 & 0 & 2 \end{array}\right]$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I used Gauss-Jordan elimination to find the multiplicative inverse of a \(3 \times 3\) matrix.

Will help you prepare for the material covered in the next section. Simplify the expression in each exercise. $$2(-30-(-3))-3(6-9)+(-1)(1-15)$$

Prove the following statement: If \(A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right], a \neq 0, b \neq 0, c \neq 0\) then \(A^{-1}=\left[\begin{array}{ccc}\frac{1}{a} & 0 & 0 \\ 0 & \frac{1}{b} & 0 \\ 0 & 0 & \frac{1}{c}\end{array}\right]\)

Find \(A^{-1}\) by forming \([A | I]\) and then using row operations to obtain \([I | B],\) where \(A^{-1}=[B] .\) Check that \(A A^{-1}=I\) and \(A^{-1} A=I\) $$A=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{array}\right]$$
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