Problem 20 - College Algebra with Modeling and Visualization

Question

( Refer to Examples 3-5.) LetA be the given matrix. Find \(A^{-1}\) without a calculator. $$ \left[\begin{array}{ll} -2 & 4 \\ -5 & 9 \end{array}\right] $$

Step-by-Step Solution

Answer

The inverse of the matrix is \( A^{-1} = \begin{bmatrix} \frac{9}{2} & -2 \\ \frac{5}{2} & -1 \end{bmatrix} \).

Step 1: Understand the Matrix

The given matrix \( A \) is a \( 2 \times 2 \) matrix: \( A = \begin{bmatrix} -2 & 4 \ -5 & 9 \end{bmatrix} \). To find the inverse, we need to calculate the determinant and apply the inverse formula for \( 2 \times 2 \) matrices.

Step 2: Calculate the Determinant

The determinant of matrix \( A \), denoted as \( \text{det}(A) \), is calculated using the formula: \( \text{det}(A) = ad - bc \) for a matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \). Here, \( a = -2, b = 4, c = -5, \) and \( d = 9 \). Thus, \( \text{det}(A) = (-2)(9) - (4)(-5) = -18 + 20 = 2 \).

Step 3: Check if Determinant is Non-Zero

To have an inverse, the determinant must not be zero. Since \( \text{det}(A) = 2 \), which is non-zero, matrix \( A \) is invertible.

Step 4: Apply the Inverse Formula

For a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse \( A^{-1} \) is given by \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]. Substitute the values: \[ A^{-1} = \frac{1}{2} \begin{bmatrix} 9 & -4 \ 5 & -2 \end{bmatrix} \].

Step 5: Simplify the Inverse Matrix

Multiply each element of the matrix \( \begin{bmatrix} 9 & -4 \ 5 & -2 \end{bmatrix} \) by \( \frac{1}{2} \): \[ A^{-1} = \begin{bmatrix} \frac{9}{2} & -2 \ \frac{5}{2} & -1 \end{bmatrix} \]. This is the inverse of matrix \( A \).

Key Concepts

Determinant Calculation

The determinant is crucial for finding the inverse of a matrix. For a \( 2 \times 2 \) matrix, the determinant helps us determine if an inverse exists. The determinant of a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is calculated using the formula:

• \( \text{det}(A) = ad - bc \)

To apply this formula, substitute the values from the matrix into the formula. In our case, with \( a = -2, b = 4, c = -5, \) and \( d = 9 \), it looks like this:

• \( \text{det}(A) = (-2)(9) - (4)(-5) \)
• \( \text{det}(A) = -18 + 20 \)
• \( \text{det}(A) = 2 \)

The determinant \( 2 \) is non-zero, proving that the matrix has an inverse. Without a non-zero determinant, the matrix would not be invertible.

2x2 Matrix

A \( 2 \times 2 \) matrix is a simple structure often used in various calculations in linear algebra. It consists of two rows and two columns. Matrices are toolkits for solving system of equations, transformations, and more. Our matrix is:

• \( \begin{bmatrix} -2 & 4 \ -5 & 9 \end{bmatrix} \)

Each element in the matrix is referenced by its position:

• The top-left is \( a \), top-right is \( b \)
• The bottom-left is \( c \), the bottom-right is \( d \)

Understanding this basic structure is essential for further operations like calculating the inverse or the determinant of the matrix.

Inverse Matrix Formula

Finding the inverse of a \( 2 \times 2 \) matrix depends on the determinant. Once you've determined that the determinant is non-zero, you can apply the inverse formula. The formula for finding the inverse \( A^{-1} \) of a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is:\[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]Let's break down how we applied this to our example:

• The determinant \( ad-bc \) was calculated as 2.
• Substitute the values: \[ A^{-1} = \frac{1}{2} \begin{bmatrix} 9 & -4 \ 5 & -2 \end{bmatrix} \]
• Now distribute \( \frac{1}{2} \) across each element: \[ A^{-1} = \begin{bmatrix} \frac{9}{2} & -2 \ \frac{5}{2} & -1 \end{bmatrix} \]

That resulting matrix is the inverse of our original matrix. Understanding each step in obtaining the inverse matrix formula ensures accuracy in calculations for more complex problems.