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

Short Answer

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

Step by step solution

01

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.
02

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 \( \begin{bmatrix} 2 & -7 \ -9 & -3 \end{bmatrix} \).
03

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 \( \frac{1}{-69} \): \( A^{-1} = \begin{bmatrix} \frac{2}{-69} & \frac{-7}{-69} \ \frac{-9}{-69} & \frac{-3}{-69} \end{bmatrix} = \begin{bmatrix} \frac{-2}{69} & \frac{7}{69} \ \frac{9}{69} & \frac{-3}{69} \end{bmatrix} \).

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Key Concepts

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

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.

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Most popular questions from this chapter

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. You invested \(\$ 2,300\) into account \(1,\) and \(\$ 2,700\) into account 2 . If the total amount of interest after one year is \(\$ 254\), and account 2 has 1.5 times the interest rate of account 1 , what are the interest rates? Assume simple interest rates.

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. A bag of mixed nuts contains cashews, pistachios, and almonds. Originally there were 900 nuts in the bag. \(30 \%\) of the almonds, \(20 \%\) of the cashews, and \(10 \%\) of the pistachios were eaten, and now there are 770 nuts left in he bag. Originally, there were 100 more cashews than almonds. Figure out how many of each type of nut was in the bag to begin with.

For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramer’s Rule. Men aged \(20-29,30-39\) , and \(40-49\) made up 78\(\%\) of the population at a prison last year. This year, the same age groups made up 82.08\(\%\) of the population. The \(20-29\) age group increased by \(20 \%,\) the \(30-39\) age group increased by \(2 \%,\) and the \(40-49\) age group decreased to \(\frac{3}{4}\) of their previous population. Originally, the \(30-39\) age group had 2\(\%\) more prisoners than the \(20-29\) age group. Determine the prison population percentage for each age group last year.

For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{aligned} 8 x-2 y &=-3 \\ -4 x+6 y &=4 \end{aligned} $$

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. At an ice cream shop, three flavors are increasing in demand. Last year, banana, pumpkin, and rocky road ice cream made up \(12 \%\) of total ice cream sales. Th s year, the same three ice creams made up \(16.9 \%\) of ice cream sales. The rocky road sales doubled, the banana sales increased by \(50 \%,\) and the pumpkin sales increased by \(20 \%\). If the rocky road ice cream had one less percent of sales than the banana ice cream, fi \(\mathrm{d}\) out the percentage of ice cream sales each individual ice cream made last year.

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