Problem 18 - College Algebra

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 of the given matrix is itself.

Step 1: Recall the condition for invertibility

A matrix has an inverse if it is square and its determinant is not zero. The multiplicative inverse of a matrix \( A \) is the matrix \( A^{-1} \) such that \( AA^{-1} = I \), where \( I \) is the identity matrix.

Step 2: Calculate the determinant

Find the determinant of the given matrix by using the formula \( \ ext{det}(A) = ad - bc \) for a 2x2 matrix \( \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] \). Substitute the values: \( a = 0, b = 1, c = 1, \text{ and } d = 0 \). The determinant is \( 0 \times 0 - 1 \times 1 = -1 \). Since the determinant is not zero, the matrix is invertible.

Step 3: Find the Inverse Matrix Formula

For a 2x2 matrix \( \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] \), the inverse is given by \( \frac{1}{ad-bc} \left[ \begin{array}{cc} d & -b \ -c & a \end{array} \right] \). Plug in the values corresponding to the elements of the matrix.

Step 4: Substitute and Simplify

Using the inverse formula from Step 3: Substitute \( a = 0, b = 1, c = 1, \text{ and } d = 0 \) into the formula. After substitution, the inverse matrix becomes:\[ A^{-1} = \frac{1}{-1} \left[ \begin{array}{cc} 0 & -1 \ -1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & 1 \ 1 & 0 \end{array} \right] \].This confirms that the inverse matrix is exactly the original matrix.

Key Concepts

Determinant of a Matrix

The determinant of a matrix is an essential value that helps determine whether a matrix is invertible. It's a single number that can be calculated from the elements of a square matrix. For a 2x2 matrix, you can find the determinant using a straightforward formula: \[ \text{det}(A) = ad - bc \] where \( a \), \( b \), \( c \), and \( d \) are the elements of the matrix \( \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] \).

• If the determinant is zero, the matrix is not invertible, meaning it doesn't have a multiplicative inverse.
• If the determinant is non-zero, the matrix is invertible.

In the original exercise, the determinant was calculated as \(-1\), which is non-zero. This confirms that the matrix is invertible.

Invertible Matrix

An invertible matrix, also known as a non-singular or non-degenerate matrix, is a matrix that possesses a multiplicative inverse. In simpler terms, if a matrix \( A \) has an inverse, there exists another matrix \( A^{-1} \) such that the product of \( A \) and \( A^{-1} \) is the identity matrix \( I \). The identity matrix \( I \) for a 2x2 matrix looks like this:\[I = \left[\begin{array}{cc} 1 & 0 \ 0 & 1 \end{array}\right]\]The condition for a matrix to be invertible is that its determinant must not be zero. If the determinant is zero, the matrix is called singular, and no inverse exists. In our example, the determinant \(-1\) qualifies the matrix as invertible, allowing us to find its inverse successfully.

2x2 Matrix Inversion

Inversion of a 2x2 matrix is a fundamental process for solving linear equations and other mathematical problems. The formula to find the inverse is: \[A^{-1} = \frac{1}{ad-bc} \left[ \begin{array}{cc} d & -b \ -c & a \end{array} \right]\]Here, \( a \), \( b \), \( c \), and \( d \) are the elements of the original matrix, and \( ad-bc \) is the determinant. To find the inverse, compute the determinant first. If it's non-zero, use it in the formula.

• Swap the positions of \( a \) and \( d \).
• Change the signs of \( b \) and \( c \).
• Divide each term by the determinant \( ad-bc \).

In the example, the inverse of the matrix \( \left[ \begin{array}{cc} 0 & 1 \ 1 & 0 \end{array} \right] \) is again \( \left[ \begin{array}{cc} 0 & 1 \ 1 & 0 \end{array} \right] \), showing that it is symmetric enough to be its own inverse.