Problem 15 - College Algebra

Question

For the following exercises, find the multiplicative inverse of each matrix, if it exists. $$\left[\begin{array}{rr}-3 & 7 \\ 9 & 2\end{array}\right]$$

Step-by-Step Solution

Answer

The multiplicative inverse is \( \begin{bmatrix} \frac{-2}{69} & \frac{7}{69} \\ \frac{9}{69} & \frac{-3}{69} \end{bmatrix} \).

Step 1: Verify the Determinant

To find the multiplicative inverse of a matrix, first ensure the matrix is invertible by checking if its determinant is non-zero. Calculate the determinant of matrix \( A = \begin{bmatrix}-3 & 7 \ 9 & 2\end{bmatrix} \). Use the formula for the determinant of a \( 2 \times 2 \) matrix: \( \text{det}(A) = ad - bc \). Here, \( a = -3 \), \( b = 7 \), \( c = 9 \), and \( d = 2 \). Thus, \( \text{det}(A) = (-3)(2) - (7)(9) = -6 - 63 = -69 \). Since the determinant is not zero, the matrix is invertible.

Step 2: Calculate the Adjugate Matrix

The adjugate, or adjoint, of a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is \( \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \). For our matrix, this is \( \begin{bmatrix} 2 & -7 \ -9 & -3 \end{bmatrix} \).

Step 3: Compute the Inverse Matrix

Use the formula for the inverse of a \( 2 \times 2 \) matrix: \( A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A) \). Substitute the determinant and adjugate values: \(A^{-1} = \frac{1}{-69} \begin{bmatrix}2 & -7 \ -9 & -3\end{bmatrix} \). Simplify by multiplying each entry of the adjugate matrix by \( \frac{1}{-69} \): \( A^{-1} = \begin{bmatrix} \frac{2}{-69} & \frac{-7}{-69} \ \frac{-9}{-69} & \frac{-3}{-69} \end{bmatrix} = \begin{bmatrix} \frac{-2}{69} & \frac{7}{69} \ \frac{9}{69} & \frac{-3}{69} \end{bmatrix} \).

Key Concepts

Determinant

The determinant is a special number that can be calculated from a square matrix. It helps in determining whether a matrix is invertible. If the determinant of a matrix is zero, the matrix is not invertible. Conversely, if the determinant is non-zero, then the matrix is invertible.

To calculate the determinant of a 2x2 matrix, use the formula \( \text{det}(A) = ad - bc \). Here, \( a, b, c, \text{ and } d \) are the elements of the matrix located respectively in the first row first column, first row second column, second row first column, and second row second column. For example:

\[ A = \begin{bmatrix} -3 & 7 \ 9 & 2 \end{bmatrix} \]

In this case:

• \( a = -3 \)
• \( b = 7 \)
• \( c = 9 \)
• \( d = 2 \)

Substituting these values into the formula gives:

\[ \text{det}(A) = (-3)(2) - (7)(9) = -6 - 63 = -69 \]

Since the determinant is not zero, matrix \( A \) is invertible.

Adjugate Matrix

The adjugate matrix, also known as the adjoint, is critical in finding the inverse of a matrix. For a 2x2 matrix, the adjugate is created by swapping the elements on the main diagonal and changing the signs of the off-diagonal elements.

Given the matrix:

\[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \]

The adjugate matrix is:

\[ \text{adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]

For our example matrix, where \( a = -3 \), \( b = 7 \), \( c = 9 \), and \( d = 2 \), the adjugate becomes:

\[ \text{adj}(A) = \begin{bmatrix} 2 & -7 \ -9 & -3 \end{bmatrix} \]

Having the adjugate is vital for calculating the inverse of the matrix, as it directly features in the formula for the inverse.

Invertible Matrix

An invertible matrix is one that has an inverse, meaning you can find another matrix that when multiplied with the original gives the identity matrix. For a matrix to be invertible, its determinant must not be zero.

The process of finding an inverse involves calculating the determinant and the adjugate of the matrix. Next, use the formula for the inverse of a 2x2 matrix:

\[ A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A) \]

In the given exercise, the determinant \( \text{det}(A) = -69 \), which is not zero. Thus, matrix \( A \) is invertible. Using the previously calculated adjugate, the inverse is:

\[ A^{-1} = \frac{1}{-69} \begin{bmatrix} 2 & -7 \ -9 & -3 \end{bmatrix} \]

By simplifying each element in the matrix we get:

\[ A^{-1} = \begin{bmatrix} \frac{-2}{69} & \frac{7}{69} \ \frac{9}{69} & \frac{-3}{69} \end{bmatrix} \]

Each step ensures that \( A^{-1} \) correctly satisfies the condition \( A \times A^{-1} = I \), where \( I \) is the identity matrix.