Problem 53 - Precalculus

Question

find \(A^{-1},\) if possible. $$A=\left[\begin{array}{ll}\frac{1}{3} & 2 \\\5 & \frac{3}{4} \end{array}\right]$$

Step-by-Step Solution

Answer

The inverse of matrix \(A\) is \(A^{-1} = \begin{bmatrix} -\frac{1}{13} & \frac{8}{39} \\ \frac{20}{39} & -\frac{4}{117} \end{bmatrix}\).

Step 1: Check Invertibility

A matrix is invertible if and only if its determinant is non-zero. To determine if matrix \(A\) is invertible, we first calculate its determinant. Matrix \(A\) is given by:\[A=\begin{bmatrix} \frac{1}{3} & 2 \ 5 & \frac{3}{4} \end{bmatrix}\]The determinant of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is computed as \(ad - bc\). For matrix \(A\), this is:\[\det(A) = \left(\frac{1}{3} \times \frac{3}{4}\right) - (2 \times 5)\]Calculate:\[\det(A) = \frac{1}{4} - 10 = \frac{1}{4} - \frac{40}{4} = -\frac{39}{4}\]Since \(\det(A) eq 0\), the matrix is invertible.

Step 2: Calculate the Inverse Matrix

For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), if the determinant is non-zero, the inverse is given by:\[A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]Using this formula, substitute the elements of matrix \(A\) and its determinant:\[A^{-1} = \frac{1}{-\frac{39}{4}} \begin{bmatrix} \frac{3}{4} & -2 \ -5 & \frac{1}{3} \end{bmatrix}\]Simplify by multiplying through by the reciprocal of the determinant:\[A^{-1} = -\frac{4}{39} \begin{bmatrix} \frac{3}{4} & -2 \ -5 & \frac{1}{3} \end{bmatrix}\]Calculate each entry:\[A^{-1} = \begin{bmatrix} -\frac{4}{39} \times \frac{3}{4} & -\frac{4}{39} \times -2 \ -\frac{4}{39} \times -5 & -\frac{4}{39} \times \frac{1}{3} \end{bmatrix}\]This simplifies to:\[A^{-1} = \begin{bmatrix} -\frac{1}{13} & \frac{8}{39} \ \frac{20}{39} & -\frac{4}{117} \end{bmatrix}\]

Step 3: Verify the Inverse by Multiplication

As a final verification step, multiply matrix \(A\) by its inverse \(A^{-1}\) and check if the result is the identity matrix \(I\).Compute:\[A \times A^{-1} = \begin{bmatrix} \frac{1}{3} & 2 \ 5 & \frac{3}{4} \end{bmatrix} \times \begin{bmatrix} -\frac{1}{13} & \frac{8}{39} \ \frac{20}{39} & -\frac{4}{117} \end{bmatrix}\]Perform the matrix multiplication:1. First row, first column: \(\frac{1}{3} \times -\frac{1}{13} + 2 \times \frac{20}{39} = 1\)2. First row, second column: \(\frac{1}{3} \times \frac{8}{39} + 2 \times -\frac{4}{117} = 0\)3. Second row, first column: \(5 \times -\frac{1}{13} + \frac{3}{4} \times \frac{20}{39} = 0\)4. Second row, second column: \(5 \times \frac{8}{39} + \frac{3}{4} \times -\frac{4}{117} = 1\)The product is the identity matrix:\[I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]This verifies our inverse was computed correctly.

Key Concepts

Determining Invertibility

To determine if a matrix is invertible, we focus on its determinant. The determinant is a special number calculated from the elements of the square matrix, and it provides key information about the matrix's properties. For a matrix to have an inverse, its determinant must be non-zero.
If the determinant is zero, then the matrix is called singular, and it does not have an inverse.

In this exercise, we start by calculating the determinant of the 2x2 matrix \(A\):

• Matrix \(A\) is \(\begin{bmatrix} \frac{1}{3} & 2 \ 5 & \frac{3}{4} \end{bmatrix}\).
• The formula for the determinant of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is \(ad - bc\).
• Plugging in the elements, we calculate \(\det(A) = (\frac{1}{3} \times \frac{3}{4}) - (2 \times 5)\).

This results in \(\det(A) = \frac{1}{4} - 10 = -\frac{39}{4}\). Since \(\det(A) eq 0\), the matrix is confirmed as invertible.

2x2 Matrix

The 2x2 matrix is one of the simplest forms of a square matrix and serves as a fundamental building block in linear algebra. Understanding these matrices is crucial as they provide an introductory framework for more complex matrices used in advanced mathematics.
A 2x2 matrix is represented as:

• Two rows and two columns.
• Each position filled with a specific element (numbers in our case).

The square shape (same number of rows and columns) allows us to compute operations like determinants and inverses. This forms the basis for many applications involving linear transformations, geometry, and more.

For instance, in matrix \(A\) given by \(\begin{bmatrix} \frac{1}{3} & 2 \ 5 & \frac{3}{4} \end{bmatrix}\), each position has its distinct role in defining the properties of the matrix. The simplicity of a 2x2 matrix also makes it a powerful illustration tool for mathematical concepts like invertibility, as seen from earlier calculations.

Matrix Multiplication Verification

Once an inverse is calculated, it is beneficial to verify the result. Matrix multiplication is used here to check the correctness of the inverse matrix.
The goal is to confirm that multiplying the original matrix by its inverse yields the identity matrix, a special matrix with 1s on the diagonal and 0s elsewhere. For our 2x2 case, the identity matrix is \(I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\). This is a critical step in confirming that the inverse computation is accurate.

Let's go through verification for our exercise:

• Original matrix \(A\): \(\begin{bmatrix} \frac{1}{3} & 2 \ 5 & \frac{3}{4} \end{bmatrix}\)
• Computed inverse \(A^{-1}\): \(\begin{bmatrix} -\frac{1}{13} & \frac{8}{39} \ \frac{20}{39} & -\frac{4}{117} \end{bmatrix}\)
• Multiply \(A\) and \(A^{-1}\):
• First row, first column: \(\frac{1}{3} \times -\frac{1}{13} + 2 \times \frac{20}{39} = 1\).
• First row, second column: \(\frac{1}{3} \times \frac{8}{39} + 2 \times -\frac{4}{117} = 0\).
• Second row, first column: \(5 \times -\frac{1}{13} + \frac{3}{4} \times \frac{20}{39} = 0\).
• Second row, second column: \(5 \times \frac{8}{39} + \frac{3}{4} \times -\frac{4}{117} = 1\).

The resulting matrix is \(I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\), proving that the inverse was calculated correctly. This verification step assures us of the accuracy and correctness of the inverse processing.