Problem 14 - College Algebra with Corequisite Support

Question

For the following exercises, find the multiplicative inverse of each matrix, if it exists. $$ \left[\begin{array}{cc} -2 & 2 \\ 3 & 1 \end{array}\right] $$

Step-by-Step Solution

Answer

The multiplicative inverse is \( \left[\begin{array}{cc} -\frac{1}{8} & \frac{1}{4} \\ \frac{3}{8} & \frac{1}{4} \end{array}\right] \).

Step 1: Understanding the Multiplicative Inverse

The multiplicative inverse of a matrix \( A \) is another matrix \( A^{-1} \) such that when they are multiplied together they yield the identity matrix. For a \( 2 \times 2 \) matrix, \( A = \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \), the inverse \( A^{-1} \) exists if and only if the determinant \( \text{det}(A) eq 0 \). The formula for the inverse is given by:\[ A^{-1} = \frac{1}{\text{det}(A)} \left[\begin{array}{cc} d & -b \ -c & a \end{array}\right] \]

Step 2: Calculate the Determinant

Compute the determinant of the matrix \( A = \left[\begin{array}{cc} -2 & 2 \ 3 & 1 \end{array}\right] \):\[ \text{det}(A) = (-2)(1) - (2)(3) = -2 - 6 = -8 \] Since the determinant \( \text{det}(A) = -8 eq 0 \), the matrix is invertible.

Step 3: Use the Formula for the Inverse

Apply the formula for the inverse using the calculated determinant:\[ A^{-1} = \frac{1}{-8} \left[\begin{array}{cc} 1 & -2 \ -3 & -2 \end{array}\right] \]This simplifies to:\[ A^{-1} = \left[\begin{array}{cc} -\frac{1}{8} & \frac{1}{4} \ \frac{3}{8} & \frac{1}{4} \end{array}\right] \]

Step 4: Verify the Result

To confirm the accuracy, multiply \( A \) and \( A^{-1} \):\[ \left[\begin{array}{cc} -2 & 2 \ 3 & 1 \end{array}\right] \times \left[\begin{array}{cc} -\frac{1}{8} & \frac{1}{4} \ \frac{3}{8} & \frac{1}{4} \end{array}\right] = \left[\begin{array}{cc} 1 & 0 \ 0 & 1 \end{array}\right] \]Both the product of \( A \) and \( A^{-1} \) and \( A^{-1} \) and \( A \) give the identity matrix, confirming the solution is correct.

Key Concepts

Determinant Calculation

When dealing with matrices, the determinant is a crucial value that we must calculate in order to determine if a matrix is invertible. For a \( 2 \times 2 \) matrix, \( A = \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \), the determinant is found using the formula:\[ \text{det}(A) = ad - bc \]
In this specific exercise, the matrix is \( \left[\begin{array}{cc} -2 & 2 \ 3 & 1 \end{array}\right] \). To find the determinant, we can apply the formula as follows:

• Multiply \(-2\) (the top left number) by \(1\) (the bottom right number) to get -2.
• Multiply \(2\) (the top right number) by \(3\) (the bottom left number) to get 6.
• Subtract the second product from the first, thus: \(-2 - 6 = -8\).

The determinant is calculated to be \(-8\). Since it's not zero, the matrix is invertible.

Matrix Inversion Steps

Finding the inverse of a matrix involves several steps following a specific formula. For a \( 2 \times 2 \) matrix \( A = \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \), the inverse \( A^{-1} \) can be determined if the determinant is not zero.
Here's the general formula used for inversion:\[ A^{-1} = \frac{1}{\text{det}(A)} \left[\begin{array}{cc} d & -b \ -c & a \end{array}\right] \]Steps involved are:

• Calculate the determinant, \( \text{det}(A) \), as shown before.
• Create the adjugate of matrix \( A \) by swapping \( a \) and \( d \), and changing the signs of \( b \) and \( c \).
• Divide each element of this new matrix by the determinant.

Applying these steps to our example gives:

• The adjugate matrix: \( \left[\begin{array}{cc} 1 & -2 \ -3 & -2 \end{array}\right] \).
• Dividing each element by \(-8\), we get: \( \left[\begin{array}{cc} -\frac{1}{8} & \frac{1}{4} \ \frac{3}{8} & \frac{1}{4} \end{array}\right] \).

Thus, we have successfully found the inverse matrix.

Identity Matrix Verification

Once we have found a matrix's inverse, it's essential to confirm its accuracy. This is done by verifying if the product of the original matrix and its inverse yields the identity matrix. For any matrix \( A \) and its inverse \( A^{-1} \), their product should result in:\[ A \times A^{-1} = I \]where \( I \) is the identity matrix, defined as \( \left[\begin{array}{cc} 1 & 0 \ 0 & 1 \end{array}\right] \).
The same result must occur when the inverse is multiplied by the original:\[ A^{-1} \times A = I \]For our example:

• Multiply the given matrix \( \left[\begin{array}{cc} -2 & 2 \ 3 & 1 \end{array}\right] \) by its calculated inverse \( \left[\begin{array}{cc} -\frac{1}{8} & \frac{1}{4} \ \frac{3}{8} & \frac{1}{4} \end{array}\right] \).
• If both products equal the identity matrix \( \left[\begin{array}{cc} 1 & 0 \ 0 & 1 \end{array}\right] \), the original inverse calculation is confirmed as correct.

This process solidifies the correctness of our inverse matrix finding.