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Chapter 2: Problem 33

Let $$ A=\left[\begin{array}{rr} 2 & 3 \\ -4 & -5 \end{array}\right] $$ a. Find \(A^{-1}\). b. Show that \(\left(A^{-1}\right)^{-1}=A\).

Short Answer

Expert verified
\( A^{-1} = \begin{bmatrix} -\frac{5}{2} & -\frac{3}{2} \\ 2 & 1 \end{bmatrix} \) and \(\left(A^{-1}\right)^{-1} = \begin{bmatrix} 2 & 3 \\ -4 & -5 \end{bmatrix} = A\).

Step by step solution

01

Find the inverse of matrix A

To find the inverse of a 2x2 matrix, we use the formula: \[ A^{-1} = \frac{1}{\text{det} (A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] where \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] and \[ \text{det} (A) = ad - bc \] In our case, a = 2, b = 3, c = -4 and d = -5. First, we find the determinant of A: \[ \text{det} (A) = ad - bc = (2)(-5) - (3)(-4) = -10 + 12 = 2 \] Now, we can find A inverse using the formula: \[ A^{-1} = \frac{1}{\text{det} (A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -5 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{5}{2} & -\frac{3}{2} \\ 2 & 1 \end{bmatrix} \] So, the inverse of matrix A is: \[ A^{-1} = \begin{bmatrix} -\frac{5}{2} & -\frac{3}{2} \\ 2 & 1 \end{bmatrix} \]
02

Find the inverse of the inverse of matrix A (i.e., \(\left(A^{-1}\right)^{-1}\))

Now, we must find the inverse of the matrix \(A^{-1}\) which is denoted as \(\left(A^{-1}\right)^{-1}\). We will use the same method as in step 1. \[ A^{-1} = \begin{bmatrix} -\frac{5}{2} & -\frac{3}{2} \\ 2 & 1 \end{bmatrix} \] In this case, a = -5/2, b = -3/2, c = 2, d = 1. Calculating the determinant of \(A^{-1}\): \[ \text{det} \left( A^{-1} \right) = ad - bc = \left(-\frac{5}{2}\right)(1) - \left(-\frac{3}{2}\right)(2) = -\frac{5}{2} + 3 = \frac{1}{2} \] Now, we can find \(\left(A^{-1}\right)^{-1}\) using the formula: \[ \left(A^{-1}\right)^{-1} = \frac{1}{\text{det} \left( A^{-1} \right)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = 2 \begin{bmatrix} 1 & \frac{3}{2} \\ -2 & -\frac{5}{2} \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ -4 & -5 \end{bmatrix} \] So, the inverse of the inverse of matrix A, \(\left(A^{-1}\right)^{-1}\) is: \[ \left(A^{-1}\right)^{-1} = \begin{bmatrix} 2 & 3 \\ -4 & -5 \end{bmatrix} \]
03

Verify that \(\left(A^{-1}\right)^{-1} = A\)

Now that we have found \(\left(A^{-1}\right)^{-1}\), we can compare it to the given matrix A: \[ A = \begin{bmatrix} 2 & 3 \\ -4 & -5 \end{bmatrix} \] Since both matrices are equal, we have verified that \(\left(A^{-1}\right)^{-1} = A\).

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Key Concepts

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

Determinant Calculation
Understanding how to calculate the determinant of a matrix is key in finding its inverse. The determinant of a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\)is calculated using the formula:
  • \(\text{det}(A) = ad - bc\)
For the given matrix, \(A = \begin{bmatrix} 2 & 3 \ -4 & -5 \end{bmatrix}\),we substitute the values into the formula:
  • \(\text{det}(A) = (2)(-5) - (3)(-4) = -10 + 12 = 2\)
A non-zero determinant means that the matrix is invertible. If the determinant were zero, the matrix wouldn't have an inverse. This step ensures we can move forward in calculating the inverse of the matrix.
Inverse of a Matrix
The inverse of a matrix is a fundamental concept in linear algebra, represented for a matrix \(A\) by \(A^{-1}\). For a 2x2 matrix, the inverse is found using the formula:
  • \[ A^{-1} = \frac{1}{\text{det} (A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]
Here, for \(A = \begin{bmatrix} 2 & 3 \ -4 & -5 \end{bmatrix}\),with a determinant of 2, substitute the values:
  • \[ A^{-1} = \frac{1}{2} \begin{bmatrix} -5 & -3 \ 4 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{5}{2} & -\frac{3}{2} \ 2 & 1 \end{bmatrix} \]
This inverse matrix undoes the effect of matrix \(A\), meaning if you multiply \(A^{-1}\) by \(A\), you get the identity matrix. Calculating the inverse requires careful attention to the details and understanding how the determinant's role in this formula affects the process.
Matrix Algebra
Matrix Algebra is the base of many mathematical operations, where understanding functions like inversion and multiplication is crucial. The problem demonstrated a key property:
  • The inverse of the inverse of a matrix returns the original matrix: \((A^{-1})^{-1} = A\).
We start with \(A^{-1} = \begin{bmatrix} -\frac{5}{2} & -\frac{3}{2} \ 2 & 1 \end{bmatrix}\)and find its inverse using similar steps:
  • The determinant of \(A^{-1}\)is \(\frac{1}{2}\).We use:\[ (A^{-1})^{-1} = 2 \begin{bmatrix} 1 & \frac{3}{2} \ -2 & -\frac{5}{2} \end{bmatrix} = \begin{bmatrix} 2 & 3 \ -4 & -5 \end{bmatrix} \].
It verifies that \((A^{-1})^{-1} = A\), showcasing a fundamental aspect of matrix algebra that links inverse operations back to the original. This consistency is pivotal in solving many algebraic problems efficiently.

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Most popular questions from this chapter

Hartman Lumber Company has two branches in the city. The sales of four of its products for the last year (in thousands of dollars) are represented by the matrix $$ B=\begin{array}{l} \text { Branch I } \\ \text { Branch II } \end{array} \quad\left[\begin{array}{rrrr} 5 & 3 & C & D \\ 3 & 4 & 6 & 8 \end{array}\right] $$ For the present year, management has projected that the sales of the four products in branch I will be \(10 \%\) more than the corresponding sales for last year and the sales of the four products in branch II will be \(15 \%\) more than the corresponding sales for last year. a. Show that the sales of the four products in the two branches for the current year are given by the matrix \(A B\), where $$ A=\left[\begin{array}{ll} 1.1 & 0 \\ 0 & 1.15 \end{array}\right] $$ Compute \(A B\). b. Hartman has \(m\) branches nationwide, and the sales of \(n\) of its products (in thousands of dollars) last year are represented by the matrix Product $$ \left.B=\begin{array}{c} 1 & 2 & 3 & \cdots & n \\ \text { Branch } 1 \\ \text { Branch 2 } \\ \vdots & a_{11} & a_{12} & a_{13} & \cdots & a_{1 n} \\ a_{21} & a_{22} & a_{23} & \cdots & a_{2 n} \\ \vdots & \vdots & \vdots & & \vdots \\ a_{m 1} & a_{m 2} & a_{m 3} & \cdots & a_{m n} \end{array}\right] $$ Also, management has projected that the sales of the \(n\) products in branch 1 , branch \(2, \ldots\), branch \(m\) will be \(r_{1} \%, r_{2} \%, \ldots, r_{m} \%\), respectively, more than the corresponding sales for last year. Write the matrix \(A\) such that \(A B\) gives the sales of the \(n\) products in the \(m\) branches for the current year.

The problems in exercise correspond to those in exercises 15-27, Section 2.1. Use the results of your previous work to help you solve these problems. The annual returns on Sid Carrington's three investments amounted to $$\$ 21,600$$: \(6 \%\) on a savings account, \(8 \%\) on mutual funds, and \(12 \%\) on bonds. The amount of Sid's investment in bonds was twice the amount of his investment in the savings account, and the interest earned from his investment in bonds was equal to the dividends he received from his investment in mutual funds. Find how much money he placed in each type of investment.

Let $$ A=\left[\begin{array}{rr} 2 & 4 \\ 5 & -6 \end{array}\right] \text { and } B=\left[\begin{array}{rr} 4 & 8 \\ -7 & 3 \end{array}\right] $$ a. Find \(A^{T}\) and show that \(\left(A^{T}\right)^{T}=A\). b. Show that \((A+B)^{T}=A^{T}+B^{T}\). c. Show that \((A B)^{T}=B^{T} A^{T}\).

Let $$ A=\left[\begin{array}{rr} 3 & 1 \\ 2 & 4 \\ -4 & 0 \end{array}\right] \text { and } B=\left[\begin{array}{rr} 1 & 2 \\ -1 & 0 \\ 3 & 2 \end{array}\right] $$ $$ (3+5) A=3 A+5 A $$

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A^{-1}\) does not exist, then the system \(A X=B\) of \(n\) linear equations in \(n\) unknowns does not have a unique solution.
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