Problem 61 - Calculus for Biology and Medicine

Question

Write down the inverse of \(A\). $$ A=\left[\begin{array}{rr} -1 & 4 \\ 5 & 0 \end{array}\right] $$

Step-by-Step Solution

Answer

The inverse of matrix \(A\) is \(\begin{bmatrix} 0 & \frac{1}{5} \\ \frac{1}{4} & \frac{1}{20} \end{bmatrix}\).

Step 1: Determine the Formula for Inverse

To find the inverse of a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), use the formula: \(A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). Here, \(ad - bc\) is the determinant of the matrix.

Step 2: Calculate the Determinant

For the given matrix \(A = \begin{bmatrix} -1 & 4 \ 5 & 0 \end{bmatrix}\), the determinant is calculated as follows: \(det(A) = (-1)(0) - (4)(5) = 0 - 20 = -20\).

Step 3: Verify the Determinant is Non-zero

An inverse exists only if the determinant is non-zero. Here, the determinant is \(-20\), which is non-zero, so the inverse exists.

Step 4: Apply the Formula for the Inverse

Substitute the values into the inverse formula: \(A^{-1} = \frac{1}{-20} \begin{bmatrix} 0 & -4 \ -5 & -1 \end{bmatrix}\).

Step 5: Simplify the Inverse Matrix

Simplify the matrix by multiplying \(\frac{1}{-20}\) with each element: \(A^{-1} = \begin{bmatrix} 0 \times \frac{1}{-20} & -4 \times \frac{1}{-20} \ -5 \times \frac{1}{-20} & -1 \times \frac{1}{-20} \end{bmatrix} = \begin{bmatrix} 0 & \frac{1}{5} \ \frac{1}{4} & \frac{1}{20} \end{bmatrix}\).

Key Concepts

Determinant Calculation

The determinant of a matrix is a special number that can be calculated from its elements, and it's crucial for finding the inverse of the matrix. For a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is calculated using the formula: \(det(A) = ad - bc\). This formula involves multiplying the elements on the main diagonal \(a\) and \(d\), and subtracting the product of the elements on the other diagonal \(b\) and \(c\).
In the case of our specific matrix \(A = \begin{bmatrix} -1 & 4 \ 5 & 0 \end{bmatrix}\), the calculation is straightforward:

• Multiply the diagonal elements: \((-1) \cdot 0 = 0\).
• Subtract the product of the off-diagonal elements: \(4 \cdot 5 = 20\).
• So, \(det(A) = 0 - 20 = -20\).

This calculation shows that the determinant is \(-20\), a non-zero number, indicating that the matrix is invertible.

2x2 Matrix

A 2x2 matrix is one of the simplest types of matrices, consisting of two rows and two columns. It's represented in the form \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\). These small matrices are often used as examples in teaching because they are easier to handle and understand compared to larger matrices.
A 2x2 matrix is significant as it lays the foundation for understanding more complex matrix operations. In terms of the matrix operations, each element in the matrix plays a role in determining both the determinant and the inverse.

• The diagonal elements, \(a\) and \(d\), are crucial for calculating the determinant, as their product needs to be considered first.
• The off-diagonal elements \(b\) and \(c\), although not on the main diagonal, are equally important in both determinant calculation and in forming the inverse matrix.

This basic understanding is essential for anyone looking to grasp matrix algebra.

Inverse Formula

Finding the inverse of a 2x2 matrix is made simple by using the inverse formula. For matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the inverse is given by: \(A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). It is important to note that this formula is applicable only when the determinant \(ad-bc\) is non-zero.
Let's break this formula down further:

• The term \(\frac{1}{ad-bc}\) is a scalar that adjusts each element of the matrix to find the inverse.
• The positions of \(a\) and \(d\) are swapped in the matrix.
• The signs of \(b\) and \(c\) are flipped, making them \(-b\) and \(-c\).

For our specific matrix \(A = \begin{bmatrix} -1 & 4 \ 5 & 0 \end{bmatrix}\), applying the formula:

• Substitute the values: \(A^{-1} = \frac{1}{-20} \begin{bmatrix} 0 & -4 \ -5 & -1 \end{bmatrix}\).
• Simplify each element by multiplying it with \(\frac{1}{-20}\), resulting in \(A^{-1} = \begin{bmatrix} 0 & \frac{1}{5} \ \frac{1}{4} & \frac{1}{20} \end{bmatrix}\).

This comprehensive approach ensures that students understand how each part of the matrix contributes to its inverse.