Problem 14 - Algebra for College Students

Question

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} -2 & -5 \\ -3 & -6 \end{array}\right] $$

Step-by-Step Solution

Answer

The inverse is \(\begin{bmatrix} 2 & -\frac{5}{3} \\ -1 & \frac{2}{3} \end{bmatrix}\).

Step 1: Matrix Definition

We are given the matrix \(A = \begin{bmatrix} -2 & -5 \ -3 & -6 \end{bmatrix}\). Our task is to find the multiplicative inverse of this matrix if it exists.

Step 2: Check Determinant

The inverse of a matrix exists only if its determinant is not zero. To find the determinant of \(A\), use the formula \(det(A) = ad - bc\), where \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\). Here, \(a = -2\), \(b = -5\), \(c = -3\), and \(d = -6\). So, \(det(A) = (-2)(-6) - (-5)(-3) = 12 - 15 = -3\). Since the determinant is not zero, the inverse exists.

Step 3: Find the Adjugate Matrix

The adjugate of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is \(\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). For \(A\), this results in \(\begin{bmatrix} -6 & 5 \ 3 & -2 \end{bmatrix}\).

Step 4: Calculate the Inverse

The inverse of the matrix \(A\) is computed using the formula \(A^{-1} = \frac{1}{det(A)} \cdot adj(A)\). Here, \(det(A) = -3\) and \(adj(A) = \begin{bmatrix} -6 & 5 \ 3 & -2 \end{bmatrix}\). So, \(A^{-1} = \frac{1}{-3} \cdot \begin{bmatrix} -6 & 5 \ 3 & -2 \end{bmatrix} = \begin{bmatrix} 2 & -\frac{5}{3} \ -1 & \frac{2}{3} \end{bmatrix}\).

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that helps us understand certain properties of the matrix. For a 2x2 matrix, such as \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), its determinant is calculated using the formula:

• \( \text{det}(A) = ad - bc \)

This formula involves multiplying the diagonal elements (\( a \) and \( d \)) and then subtracting the product of the off-diagonal elements (\( b \) and \( c \)).
In our example, where \( A = \begin{bmatrix} -2 & -5 \ -3 & -6 \end{bmatrix} \), we substitute the values:

• \( a = -2 \), \( b = -5 \), \( c = -3 \), \( d = -6 \)

Thus, the determinant is calculated as:

• \( (-2)(-6) - (-5)(-3) = 12 - 15 = -3 \)

Since the determinant is \(-3\), which is not zero, this confirms that the matrix \( A \) has an inverse.

Adjugate Matrix

An adjugate matrix, also known as the adjoint matrix, is a key component in finding the inverse of a square matrix. It is basically the transpose of the cofactor matrix. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the adjugate matrix is:

• \( \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \)

To find the adjugate matrix for our example matrix \( A = \begin{bmatrix} -2 & -5 \ -3 & -6 \end{bmatrix} \):

• Swap \( a \) and \( d \), giving \( d = -6 \), and \( a = -2 \).
• Change the signs of \( b \) and \( c \), giving \( -b = 5 \), and \( -c = 3 \).

This results in the adjugate matrix:

• \( \begin{bmatrix} -6 & 5 \ 3 & -2 \end{bmatrix} \)

Calculating the adjugate matrix is a crucial step in determining the matrix inverse.

2x2 Matrix Inversion

To find the inverse of a 2x2 matrix \( A \), the formula is:

• \( A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \)

Here, \( \text{det}(A) \) is the determinant of the matrix, and \( \text{adj}(A) \) is the adjugate matrix.
Using our matrix example \( A = \begin{bmatrix} -2 & -5 \ -3 & -6 \end{bmatrix} \), we have already found:

• The determinant: \( \text{det}(A) = -3 \)
• The adjugate matrix: \( \begin{bmatrix} -6 & 5 \ 3 & -2 \end{bmatrix} \)

Plug these into the inversion formula:

• \( A^{-1} = \frac{1}{-3} \cdot \begin{bmatrix} -6 & 5 \ 3 & -2 \end{bmatrix} \)
• This results in an inverse of \( \begin{bmatrix} 2 & -\frac{5}{3} \ -1 & \frac{2}{3} \end{bmatrix} \)

The inverse provides us with a way to reverse the effects of the original matrix, which can be particularly useful in solving systems of equations.