Problem 62 - A Graphical Approach to Precalculus with Limits

Question

Given a square matrix \(A^{-1}\), find matrix \(A\). $$A^{-1}=\left[\begin{array}{ll} \frac{3}{20} & \frac{1}{4} \\ -\frac{1}{20} & \frac{1}{4} \end{array}\right]$$

Step-by-Step Solution

Answer

Matrix \( A \) is \( \left[ \begin{array}{cc} 5 & -5 \\ \frac{4}{7} & \frac{24}{7} \end{array} \right] \).

Step 1: Understanding the Problem

The problem provides the inverse matrix \( A^{-1} \) and asks us to find the original matrix \( A \). For a square matrix \( A \), the relationship between \( A \) and its inverse is given by the equation \( A \times A^{-1} = I \), where \( I \) is the identity matrix.

Step 2: Set Up the Equation Using the Identity Matrix

For a 2x2 matrix, the identity matrix \( I \) is \( \left[\begin{array}{cc} 1 & 0 \ 0 & 1 \end{array}\right] \). Thus, we need to solve \( A \times A^{-1} = I \). Given \( A^{-1}\), express \( A \) in general form as \( A = \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \) and set up the multiplication \( A \times A^{-1} = I \).

Step 3: Perform Matrix Multiplication

Multiply the general matrix \( A = \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \) by \( A^{-1} = \left[\begin{array}{cc} \frac{3}{20} & \frac{1}{4} \ -\frac{1}{20} & \frac{1}{4} \end{array}\right] \). The result should yield the identity matrix:\[\begin{pmatrix}a \cdot \frac{3}{20} + b \cdot -\frac{1}{20} & a \cdot \frac{1}{4} + b \cdot \frac{1}{4} \c \cdot \frac{3}{20} + d \cdot -\frac{1}{20} & c \cdot \frac{1}{4} + d \cdot \frac{1}{4}\end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\]

Step 4: Solve for Entries of Matrix A

From the equality of matrices, solve the following system of equations:1. \( a \cdot \frac{3}{20} - b \cdot \frac{1}{20} = 1 \)2. \( a \cdot \frac{1}{4} + b \cdot \frac{1}{4} = 0 \)3. \( c \cdot \frac{3}{20} - d \cdot \frac{1}{20} = 0 \)4. \( c \cdot \frac{1}{4} + d \cdot \frac{1}{4} = 1 \)

Step 5: Solve the System for Entries a and b

From equation 2:\( \frac{a + b}{4} = 0 \) implies \( a + b = 0 \). Thus, \( b = -a \).Substitute \( b = -a \) into equation 1:\[ a \cdot \frac{3}{20} - (-a) \cdot \frac{1}{20} = 1 \]\[ \frac{4a}{20} = 1 \rightarrow a = 5 \]\( b = -5 \).Hence, \( a = 5 \) and \( b = -5 \).

Step 6: Solve the System for Entries c and d

From equation 4:\( \frac{c + d}{4} = 1 \) implies \( c + d = 4 \). Thus, \( d = 4 - c \).Substitute \( d = 4 - c \) into equation 3:\[ \frac{3c}{20} - \frac{4-c}{20} = 0 \]\[ \frac{7c}{20} = \frac{4}{20} \rightarrow c = \frac{4}{7} \]\( d = 4 - \frac{4}{7} = \frac{24}{7} \).Hence, \( c = \frac{4}{7} \) and \( d = \frac{24}{7} \).

Step 7: Write the Matrix A

Now write the entries of \( A \) as follows:\[A = \begin{pmatrix} 5 & -5 \ \frac{4}{7} & \frac{24}{7} \end{pmatrix}\] This is the original matrix \( A \) such that \( A \times A^{-1} = I \).

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, often necessary when working with transformations, solving equations, and more. Here's a simple breakdown of how matrix multiplication works. To multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. In our scenario, we are multiplying a 2x2 matrix by another 2x2 matrix, which fits this requirement.

Each element in the resulting matrix is created by taking the dot product of the corresponding row from the first matrix and the column from the second matrix. For example, suppose you have matrices:

• Matrix A: \[\begin{pmatrix} a & b \ c & d\end{pmatrix}\]
• Matrix B: \[\begin{pmatrix} x & y \ z & w\end{pmatrix}\]

To find the element at the first row and first column of the product matrix, you calculate:

• \( ax + bz \)

This process is repeated for every element in the resulting matrix. Matrix multiplication is essential because it provides a method to understand composite transformations when these matrices represent transformations.

System of Equations

A system of equations involves finding unknown variables that satisfy a set of equations simultaneously. In the context of matrices, solving a system can often involve using what's known as an augmented matrix, which combines the coefficients and constants of the equations into a larger matrix.

In our exercise, we needed to solve for four variables—namely, the entries of matrix \( A \):

• \( a \), \( b \)
• \( c \), \( d \)

We started by setting matrix \( A \times A^{-1} = I \), breaking it down into individual equations:

• \( a \cdot \frac{3}{20} - b \cdot \frac{1}{20} = 1 \)
• \( a \cdot \frac{1}{4} + b \cdot \frac{1}{4} = 0 \)
• \( c \cdot \frac{3}{20} - d \cdot \frac{1}{20} = 0 \)
• \( c \cdot \frac{1}{4} + d \cdot \frac{1}{4} = 1 \)

By solving these equations step-by-step, we find the values of \( a, b, c, \) and \( d \). Solving such systems is crucial for determining matrix characteristics and relationships, as demonstrated in obtaining matrix \( A \).

Identity Matrix

The identity matrix is a special type of matrix that acts like the number 1 in matrix multiplication. For any matrix \( A \, (A \times I = A) \), where \( I \) is the identity matrix. In this exercise, the identity matrix was critical because we used the relationship \( A \times A^{-1} = I \) to find the original matrix \( A \).

The identity matrix is typically a diagonal matrix where all diagonal elements are 1, and all others are 0, like so for a 2x2 matrix:

• \[\begin{pmatrix}1 & 0 \0 & 1\end{pmatrix}\]

Using the identity matrix in the context of inverse matrices is important because it confirms that multiplying a matrix by its inverse truly returns the identity matrix—proof that the calculations are correct and the matrix has been inverted accurately. Understanding the role of the identity matrix allows us to validate our solution work and ensure results align with mathematical principles.