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Chapter 12: Problem 43

Use your calculator to find the multiplicative inverse of \(\left[\begin{array}{ll}\frac{1}{2} & \frac{2}{5} \\ \frac{3}{4} & \frac{1}{4}\end{array}\right]\) What difficulty did you encounter?

Short Answer

Expert verified
The inverse matrix is \( \begin{bmatrix} -\frac{10}{7} & \frac{16}{7} \\ \frac{30}{7} & -\frac{20}{7} \end{bmatrix} \). Finding fractions by hand can be challenging.

Step by step solution

01

Identify Matrix Elements

The given matrix is \( A = \begin{bmatrix} \frac{1}{2} & \frac{2}{5} \ \frac{3}{4} & \frac{1}{4} \end{bmatrix} \). Let's identify its elements: \( a = \frac{1}{2} \), \( b = \frac{2}{5} \), \( c = \frac{3}{4} \), \( d = \frac{1}{4} \).
02

Calculate the Determinant

The determinant of the 2x2 matrix \( A \) is calculated as follows:\[ \text{det}(A) = ad - bc = \left(\frac{1}{2} \times \frac{1}{4}\right) - \left(\frac{2}{5} \times \frac{3}{4}\right)\]\[ = \frac{1}{8} - \frac{6}{20} = \frac{1}{8} - \frac{3}{10} = \frac{5}{40} - \frac{12}{40} = -\frac{7}{40} \]Ensure that the determinant is non-zero; the matrix is invertible.
03

Calculate the Inverse Matrix

The formula for the inverse of a 2x2 matrix \( A \) is:\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]Substitute the known values:\[ A^{-1} = \frac{1}{-\frac{7}{40}} \begin{bmatrix} \frac{1}{4} & -\frac{2}{5} \ -\frac{3}{4} & \frac{1}{2} \end{bmatrix} \]\[ = -\frac{40}{7} \begin{bmatrix} \frac{1}{4} & -\frac{2}{5} \ -\frac{3}{4} & \frac{1}{2} \end{bmatrix} \]Multiply through by \(-\frac{40}{7}\):\[ A^{-1} = \begin{bmatrix} -\frac{40}{28} & \frac{80}{35} \ \frac{120}{28} & -\frac{40}{14} \end{bmatrix} \]Simplify:\[ A^{-1} = \begin{bmatrix} -\frac{10}{7} & \frac{16}{7} \ \frac{30}{7} & -\frac{20}{7} \end{bmatrix} \].
04

Use a Calculator

Use a calculator or matrix algebra software to verify your calculations of the inverse matrix. Ensure that the multiplication of \( A \) and \( A^{-1} \) gives the identity matrix \( I \).
05

Address Any Difficulties

In manually calculating the determinants and matrix inverses, maintaining accuracy in fractions can be challenging without a calculator. Ensure precision in simplifying fractions and arithmetic.

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Key Concepts

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

Determinant Calculation
The determinant is a special number that can be calculated from a square matrix. It is essential for finding the inverse of a matrix. For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant \( \text{det}(A) \) is computed as:
  • \( \text{det}(A) = ad - bc \)
To calculate \( \text{det}(A) \) from our example matrix, we first identify the matrix's elements:
  • \( a = \frac{1}{2} \), \( b = \frac{2}{5} \)
  • \( c = \frac{3}{4} \), \( d = \frac{1}{4} \)
Then, substitute these elements into the determinant formula:
  • \( \text{det}(A) = \left(\frac{1}{2} \cdot \frac{1}{4}\right) - \left(\frac{2}{5} \cdot \frac{3}{4}\right) = \frac{1}{8} - \frac{3}{10} \)
  • Convert fractions to a common denominator: \( \frac{5}{40} - \frac{12}{40} \)
  • So, \( \text{det}(A) = -\frac{7}{40} \)
Ensure that the determinant is not equal to zero, which confirms that the matrix is invertible.
Matrix Algebra
Matrix algebra involves operations with matrices, such as addition, multiplication, and finding inverses. To find the inverse of a matrix is a crucial operation. If a 2x2 matrix has a non-zero determinant, it is invertible. The inverse is found using the formula:
  • For a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), its inverse is \( A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \)
Using this formula for our example:
  • Substitute \( d = \frac{1}{4} \), \( -b = -\frac{2}{5} \), \( -c = -\frac{3}{4} \), and \( a = \frac{1}{2} \)
  • The inverse matrix calculation involves multiplying these values by \( \frac{1}{-\frac{7}{40}} \) to balance the scale imposed by the determinant's reciprocal
  • The result is \( A^{-1} = \begin{bmatrix} -\frac{10}{7} & \frac{16}{7} \ \frac{30}{7} & -\frac{20}{7} \end{bmatrix} \)
Make sure to verify this computation by multiplying the original matrix by its inverse, which should yield the identity matrix \( I \). This verification step checks for any errors in calculation.
Fraction Arithmetic
Fraction arithmetic, crucial in matrix operations, involves addition, subtraction, multiplication, and division of fractions. When dealing with matrix elements in fractional form:
  • Always find a common denominator when adding or subtracting: For \( \frac{1}{8} - \frac{3}{10} \), convert to \( \frac{5}{40} - \frac{12}{40} \)
  • Multiply fractions across numerators and denominators: For example, \( \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \).
  • Division of fractions is just as important, particularly when dealing with determinants: Multiply by the reciprocal of the divisor, for example, divide by \( -\frac{7}{40} \) is to multiply by its reciprocal \( -\frac{40}{7} \).
Accurate fraction arithmetic ensures that all matrix operations, including determinant and inverse calculations, are precise. When manually dealing with fractions, careful steps are essential to avoid errors, especially in simplification.

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Most popular questions from this chapter

For Problems \(13-26\), compute \(A B\) and \(B A\). $$ A=\left[\begin{array}{rr} -8 & -5 \\ 3 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} -2 & -5 \\ 3 & 8 \end{array}\right] $$

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} 1 & -1 \\ 1 & 1 \end{array}\right] $$

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} 5 & -1 \\ 3 & 4 \end{array}\right] $$

For Problems 27-40, use the method of matrix inverses to solve each system. $$ \left(\begin{array}{l} 3 x-5 y=2 \\ 4 x-3 y=-1 \end{array}\right) $$

For Problems \(9-20\), find \(A B\) and \(B A\), whenever they exist. $$ A=\left[\begin{array}{rrr} -2 & 3 & -1 \\ 7 & -4 & 5 \end{array}\right], \quad B=\left[\begin{array}{rr} 1 & -1 \\ -2 & 3 \\ -5 & -6 \end{array}\right] $$
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