Today, we’re diving into the world of linear algebra to explore how to find the inverse of a 2x2 matrix. Specifically, we’ll demonstrate the formula for the inverse and prove that it’s correct. The key here is to use matrix notation and perform some basic matrix multiplication.

Understanding the Problem

A matrix, A, has an inverse, A⁻¹, if when multiplied by A in either order (i.e., AA⁻¹ or A⁻¹A), the result is the identity matrix, denoted as I. For a 2x2 matrix, the identity matrix is:

\(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

We are given a 2x2 matrix A:

\(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)

And we are told that if ad - bc ≠ 0, then the inverse of A, denoted as A⁻¹, is:

\(A^{-1} = \begin{bmatrix} \frac{d}{ad-bc} & \frac{-b}{ad-bc} \\ \frac{-c}{ad-bc} & \frac{a}{ad-bc} \end{bmatrix}\)

Our goal is to prove that A⁻¹ as defined above is indeed the inverse of A by showing that AA⁻¹ = I. We will be using the standard matrix multiplication rules.

Proof of the Inverse Formula

To prove this, we need to perform the matrix multiplication of A and A⁻¹ and show that the resulting matrix is the identity matrix I:

\(AA^{-1} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \frac{d}{ad-bc} & \frac{-b}{ad-bc} \\ \frac{-c}{ad-bc} & \frac{a}{ad-bc} \end{bmatrix}\)

Performing matrix multiplication, we get:

\(AA^{-1} = \begin{bmatrix} a \cdot \frac{d}{ad-bc} + b \cdot \frac{-c}{ad-bc} & a \cdot \frac{-b}{ad-bc} + b \cdot \frac{a}{ad-bc} \\ c \cdot \frac{d}{ad-bc} + d \cdot \frac{-c}{ad-bc} & c \cdot \frac{-b}{ad-bc} + d \cdot \frac{a}{ad-bc} \end{bmatrix}\)

Simplifying each term:

\(AA^{-1} = \begin{bmatrix} \frac{ad-bc}{ad-bc} & \frac{-ab+ba}{ad-bc} \\ \frac{cd-dc}{ad-bc} & \frac{-cb+da}{ad-bc} \end{bmatrix}\)

Further simplification yields:

\(AA^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

Which is the identity matrix, I.

The Determinant

The quantity ad - bc is known as the determinant of the matrix A. It is often denoted as det(A). The condition that ad - bc ≠ 0 is crucial because if the determinant is zero, the matrix has no inverse (division by zero). A matrix with a determinant of zero is called a singular matrix.

Conclusion

We have successfully demonstrated, by direct matrix multiplication, that:

\(AA^{-1} = I\)

when:

\(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)

and

\(A^{-1} = \begin{bmatrix} \frac{d}{ad-bc} & \frac{-b}{ad-bc} \\ \frac{-c}{ad-bc} & \frac{a}{ad-bc} \end{bmatrix}\)

provided that ad - bc ≠ 0.

This proof confirms the formula for the inverse of a 2x2 matrix and highlights the importance of the determinant in determining whether a matrix has an inverse. This result is a fundamental concept in linear algebra and finds wide applications in various fields, from computer graphics to solving systems of linear equations.