Problem 45 - Calculus for Biology and Medicine

Question

$$ A=\left[\begin{array}{rr} -1 & 1 \\ 2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & -2 \\ 3 & 2 \end{array}\right] $$ Find the inverse (if it exists) of \(A\).

Step-by-Step Solution

Answer

The inverse of matrix \(A\) is \(\begin{bmatrix} -\frac{3}{5} & \frac{1}{5} \\ \frac{2}{5} & -\frac{1}{5} \end{bmatrix}.\)

Step 1: Determine If the Matrix Is Invertible

To find if a square matrix is invertible, calculate its determinant. If the determinant is not zero, the matrix is invertible. The determinant of a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is \(ad - bc\). For matrix \(A\), the determinant is calculated as follows:\[\text{det}(A) = (-1)(3) - (1)(2) = -3 - 2 = -5.\]Since \(\text{det}(A) eq 0\), matrix \(A\) is invertible.

Step 2: Calculate the Inverse of the Matrix

The inverse of a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by:\[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}.\]For matrix \(A\), this is:\[A^{-1} = \frac{1}{-5} \begin{bmatrix} 3 & -1 \ -2 & -1 \end{bmatrix} = \begin{bmatrix} -\frac{3}{5} & \frac{1}{5} \ \frac{2}{5} & -\frac{1}{5} \end{bmatrix}.\]Thus, \(A^{-1}\) is as calculated above.

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from its elements. For a square matrix, the determinant provides essential information about the matrix's properties, especially regarding invertibility. In a 2x2 matrix, the formula for the determinant is simple and elegant. Given a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant \( \text{det}(A) \) is calculated as

• \(ad - bc\)

This formula reveals the product of the diagonal elements (\(a\) multiplied by \(d\)) minus the product of the off-diagonal elements (\(b\) times \(c\)).
The importance of the determinant cannot be overstated. If \( \text{det}(A) \) is zero, the matrix is singular and does not have an inverse. But if it is non-zero, as is the case with our example where \( \text{det}(A) = -5 \), it signals that the matrix is invertible, opening the door to finding its inverse.

Invertible Matrix

An invertible matrix, also known as a non-singular or nondegenerate matrix, has a unique property: it can be multiplied by another matrix (its inverse) to yield the identity matrix. This means:

• For a matrix \(A\), if an inverse \(A^{-1}\) exists, then \( A \times A^{-1} = A^{-1} \times A = I \).
• Here, \(I\) is the identity matrix, which acts like the number 1 in matrix arithmetic.

The existence of an inverse is crucial in solving a variety of mathematical and practical problems, from systems of linear equations to 3D graphics transformations. Recognizing an invertible matrix involves checking that its determinant is not zero.
Our exercise showed that matrix \(A\) is invertible because its determinant, \( \text{det}(A) = -5 \), is not zero. This enables us to confidently proceed to calculate the inverse using known formulas.

2x2 Matrix

The 2x2 matrix is one of the simplest, yet most fundamental types of matrices. It comprises two rows and two columns, making it ideal for introducing key linear algebra concepts. Whenever dealing with a 2x2 matrix, it's essential to become familiar with its structure and how operations like inversion are performed.

• Structure: A typical 2x2 matrix \(A\) is arranged as \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \).
• Operations: Calculations on this matrix type, like finding the determinant or the inverse, use simple formulas.

In our original exercise, the step-by-step approach was applied to matrix \(A\), leading to the solution of its inverse. It's remarkable how straightforward arithmetic on a 2x2 matrix can solve what's theoretically a complex problem, illustrating the power and simplicity of algebraic methods.