Problem 18 - College Algebra with Corequisite Support

Question

For the following exercises, find the multiplicative inverse of each matrix, if it exists. $$ \left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] $$

Step-by-Step Solution

Answer

The multiplicative inverse is the matrix itself: \( \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \).

Step 1: Identify the matrix

We are given the matrix \( A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \). Our task is to find its multiplicative inverse if it exists.

Step 2: Check if the matrix is invertible

A 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is invertible if its determinant \( ad - bc \) is not zero. For matrix \( A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \), the determinant is \( 0 \times 0 - 1 \times 1 = -1 \). Since the determinant is not zero, the matrix is invertible.

Step 3: Use the inverse formula for a 2x2 matrix

The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \). For our matrix \( A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \), \( a=0, b=1, c=1, d=0 \) and the determinant is \(-1\). Thus, the inverse is \( \begin{bmatrix} 0 & -1 \ -1 & 0 \end{bmatrix} \) divided by \(-1\).

Step 4: Simplify the inverse matrix

Dividing each element of \( \begin{bmatrix} 0 & -1 \ -1 & 0 \end{bmatrix} \) by \(-1\), we get \( \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \). Therefore, the multiplicative inverse of matrix \( A \) is \( \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \).

Key Concepts

Determinant

The determinant is a special value calculated from a square matrix. It is an essential part of finding out whether a matrix is invertible or not. Essentially, for a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated by the formula \( ad - bc \).

In simple terms:

• Multiply the top left and bottom right entries together (\( a \) and \( d \)).
• Multiply the top right and bottom left entries together (\( b \) and \( c \)).
• Subtract the second product from the first (\( ad - bc \)).

If the determinant is zero, the matrix cannot have an inverse. If it's not zero, you're in luck—the matrix is invertible. For example, in the given matrix \( A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \), its determinant is \(-1\). Since \(-1\) is not zero, it confirms that the matrix can be inverted.

Invertible Matrix

An invertible matrix is one that has an inverse, meaning another matrix can multiply with it to produce the identity matrix. For a 2x2 matrix, being invertible is synonymous with having a non-zero determinant.

The identity matrix in this case is \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \). When you multiply an invertible matrix by its inverse, the result is always this identity matrix.

• If \( A \times A^{-1} = I \) (where \( I \) is the identity matrix), then \( A \) is invertible.
• If the calculation results in something other than the identity matrix, then \( A \) lacks an inverse.

Knowing that a matrix is invertible is crucial for solving various mathematical problems, such as systems of equations and transforming data sets. As established, since the determinant of matrix \( A \) is \(-1\), it confirms \( A \) has an inverse and is indeed invertible.

2x2 Matrix Inverse

Finding the inverse of a 2x2 matrix involves a straightforward formula, particularly useful for small matrices. For a matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by: \[\frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]

Here's the step-by-step process:

• First, find the determinant \( ad-bc \).
• Next, flip the entries \( a \) and \( d \), and change the signs of \( b \) and \( c \).
• Finally, multiply each element by \( 1/(ad-bc) \).

For the given matrix \( A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \), the steps yield its inverse as \( \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \) after simplifying \( \begin{bmatrix} 0 & -1 \ -1 & 0 \end{bmatrix} \) by dividing each element by \(-1\). Having this inverse is crucial for many algebraic applications.