/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 Find \(\left(A^{2}\right)^{-1}\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(\left(A^{2}\right)^{-1}\) and \(\left(A^{-1}\right)^{2},\) where \(A=\left[\begin{array}{rr}1 & -2 \\ -1 & 3\end{array}\right] .\) What do you observe? Use the definition of the inverse of a matrix, together with the fact that \((A B)^{-1}=A^{-1} B^{-1},\) to show that \(\left(A^{2}\right)^{-1}=\left(A^{-1}\right)^{2}\) for every square matrix \(A\)

Short Answer

Expert verified
The inverse of \(A^2\) and the square of \(A^{-1}\) for the given matrix \(A\) are both the identity matrix, thus \(\left(A^2\right)^{-1} = \left(A^{-1}\right)^{2}\). This result aligns with the given property that \((AB)^{-1} = B^{-1}A^{-1}\), used here with \(AB = A^2\).

Step by step solution

01

Find The Square Of Matrix A

First, to find \(A^{2}\), you multiply matrix \(A\) by itself. Matrix \(A\) is \(\left[\begin{array}{rr}1 & -2 \ -1 & 3\end{array}\right]\). When you multiply this matrix by itself, you receive the matrix \(A^{2} = \left[\begin{array}{rr}1 & 0 \ 0 & 1\end{array}\right]\).
02

Find The Inverse Of \(A^2\)

The inverse of a 2x2 matrix can be found using the formula \(\frac{1}{ad-bc} \left[\begin{array}{rr}d & -b \ -c & a\end{array}\right]\), where \(a, b, c, d\) are the entries of the matrix. Since \(A^2\) is the identity matrix, its inverse is also the identity matrix, i.e., \((A^{2})^{-1} = \left[\begin{array}{rr}1 & 0 \ 0 & 1\end{array}\right]\).
03

Find The Inverse Of Matrix A

Using the same formula, the inverse of matrix \(A\), denoted \(A^{-1}\), is \(\left[\begin{array}{rr}3 & 2 \ 1 & 1\end{array}\right]\).
04

Find The Square Of \(A^{-1}\)

To find \(\left(A^{-1}\right)^{2}\), you multiply matrix \(A^{-1}\) by itself. By performing this operation, you find \(\left(A^{-1}\right)^{2} = \left[\begin{array}{rr}1 & 0 \ 0 & 1\end{array}\right]\).
05

Observations And Conclusion

By comparing the results from steps 2 and 4, you can see that both \(\left(A^{2}\right)^{-1}\) and \(\left(A^{-1}\right)^{2}\) equal the identity matrix. This demonstrates that \(\left(A^{2}\right)^{-1} = \left(A^{-1}\right)^{2}\), consistent with the given rule that \((AB)^{-1} = B^{-1}A^{-1}\) for all square matrices \(A\) and \(B\) when \(AB = A^2\).

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Key Concepts

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

Inverse of a Matrix
The inverse of a matrix is akin to the reciprocal of a number. It's a concept that helps us "undo" a matrix operation. For a square matrix \(A\), its inverse \(A^{-1}\) is defined such that when you multiply \(A\) by \(A^{-1}\), you get the identity matrix: \(AA^{-1} = A^{-1}A = I\). The identity matrix acts like the number 1 in matrix multiplication, where multiplying any matrix by the identity matrix leaves it unchanged.
To find the inverse of a 2x2 matrix, there's a handy formula:
  • First, check if the matrix is invertible by ensuring its determinant is not zero. The determinant for a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is calculated as \( ad - bc \).
  • If the determinant is non-zero, the inverse is given by: \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).
It's important to grasp this concept as it forms the foundation for working with more complex operations like solving linear equations, transformations, and eventually understanding deeper matrix properties.
Matrix Multiplication
Matrix multiplication is a process of multiplying two matrices to produce another matrix. This operation is not as straightforward as regular number multiplication. It's essential to grasp how it works:
  • Ensure the matrices are compatible for multiplication. If matrix \( A \) has dimensions \( m \times n \) and matrix \( B \) has dimensions \( n \times p \), then the result, \( AB \), will have dimensions \( m \times p \).
  • The entry in the \( i \)-th row and \( j \)-th column of the product matrix \( AB \) is found by multiplying each element of the \( i \)-th row of \( A \) with the corresponding element of the \( j \)-th column of \( B \), and adding the results.
Understanding matrix multiplication is pivotal because it underpins operations like finding the power of a matrix and composing transformations. For instance, when you found \( A^2 \) in the original exercise, you were performing matrix multiplication of matrix \( A \) with itself. This process helps us in scenarios like applying multiple linear transformations sequentially.
Properties of Matrices
Matrices exhibit several interesting properties, and understanding them can simplify complex problems. These properties are often used in practically everything related to matrices.
  • Associativity: Matrix multiplication is associative. That means for matrices \( A \), \( B \), and \( C \), the equation \( (AB)C = A(BC) \) holds true.
  • Distributivity: Matrix operations are distributive over addition, i.e., \( A(B + C) = AB + AC \).
  • Transpose of a Product: The transpose of a product of matrices is the product of their transposes in reverse order, \( (AB)^T = B^TA^T \).
  • Inverse Property: The inverse of a product of matrices is the product of their inverses in reverse order: \( (AB)^{-1} = B^{-1}A^{-1} \).
Recognizing these properties empowers you in matrix algebra. For instance, in the exercise, the property \( (A^2)^{-1} = (A^{-1})^2 \) is a result of applying these properties correctly, showcasing the elegant symmetry in matrix operations and their inverses.

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Most popular questions from this chapter

A chemist wishes to make 10 gallons of a \(15 \%\) acid solution by mixing a \(10 \%\) acid solution with a \(25 \%\) acid solution. (a) Let \(x\) and \(y\) denote the total volumes (in gallons) of the \(10 \%\) and \(25 \%\) solutions, respectively. Using the variables \(x\) and \(y,\) write an equation for the total volume of the \(15 \%\) solution (the mixture). (b) Using the variables \(x\) and \(y,\) write an equation for the total volume of acid in the mixture by noting that Volume of acid in \(15 \%\) solution \(=\) volume of acid in \(10 \%\) solution \(+\) volume of acid in \(25 \%\) solution. (c) Solve the system of equations from parts (a) and (b), and interpret your solution. (d) Is it possible to obtain a \(5 \%\) acid solution by mixing a \(10 \%\) solution with a \(25 \%\) solution? Explain without solving any equations.

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. The athletic director of a local high school is ordering equipment for spring sports. He needs to order twice as many baseballs as softballs. The total number of balls he must order is \(300 .\) How many of each type should he order?

A financial advisor offers three specific investment instruments: a stock- based mutual fund, a high-yield bond, and a certificate of deposit (CD). Risk factors for individual instruments can be quantified on a scale of 1 to \(5,\) with 1 being the most risky. The risk factors associated with these particular instruments are summarized in the following table.$$\begin{array}{lc} \text { Type of Investment } & \text { Risk Factor } \\ \text { Stock-based mutual fund } & 3 \\\\\text { High-yield bond } & 1 \\\\\text { CD } & 5\end{array}$$.One of the advisor's clients can tolerate an overall risk level of \(3.5 .\) In addition, the client stipulates that the amount of money invested in the mutual fund must equal the sum of the amounts invested in the high-yield bond and the CD. To satisfy the client's requirements, what percentage of the total investment should be allocated to each instrument?

A family owns and operates three businesses. On their income-tax return, they have to report the depreciation deductions for the three businesses separately. In \(2004,\) their depreciation deductions consisted of use of a car, plus depreciation on 5 -year equipment (on which onc-fifth of the original value is deductible per year) and 10-year equipment (on which one- tenth of the original valuc is deductible per year). The car use (in miles) for cach business in 2004 is given in the following table, along with the original value of the depreciable 5 - and 10 year equipment used in each business that year. $$\begin{array}{|c|c|c|c|}\hline & \begin{array}{c}\text { Car } \\\\\text { Use }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 5-Year }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 10-Year }\end{array} \\\\\text { Business } & \text { (miles) } & \text { Equipment (S) } & \text { Equipment (S) } \\\\\hline 1 & 3200 & 9850 & 435 \\\2 & 8800 & 12,730 & 980 \\\3 & 6880 & 2240 & 615\\\\\hline\end{array}$$ The depreciation deduction for car use in 2004 was 37.5 cents per mile. Use matrix multiplication to determine the total depreciation deduction for each business in 2004.

Find the decoding matrix for each encoding matrix. $$\left[\begin{array}{ll}1 & -3 \\\1 & -2\end{array}\right]$$

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