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Chapter 7: Problem 14

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

Short Answer

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

Step by step solution

01

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:\[ A^{-1} = \frac{1}{\text{det}(A)} \left[\begin{array}{cc} d & -b \ -c & a \end{array}\right] \]
02

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.
03

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:\[ A^{-1} = \left[\begin{array}{cc} -\frac{1}{8} & \frac{1}{4} \ \frac{3}{8} & \frac{1}{4} \end{array}\right] \]
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. Three roommates shared a package of 12 ice cream bars, but no one remembers who ate how many. If Tom ate twice as many ice cream bars as Joe, and Albert ate three less than Tom, how many ice cream bars did each roommate eat?

For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{array}{r} 5 x-4 y=2 \\ -4 x+7 y=6 \end{array} $$

For the following exercises, solve a system using the inverse of a \(3 \times 3\) matrix. $$ \begin{array}{l} 4 x-2 y+3 z=-12 \\ 2 x+2 y-9 z=33 \\ 6 y-4 z=1 \end{array} $$

For the following exercises, solve a system using the inverse of a \(3 \times 3\) matrix. $$ \begin{array}{l} 6 x-5 y+2 z=-4 \\ 2 x+5 y-z=12 \\ 2 x+5 y+z=12 \end{array} $$

For the following exercises, solve the system using the inverse of a \(2 \times 2\) matrix. $$ \begin{array}{l} \frac{8}{5} x-\frac{4}{5} y=\frac{2}{5} \\ -\frac{8}{5} x+\frac{1}{5} y=\frac{7}{10} \end{array} $$
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