# Quick Math Inverse Matrix

## Quick Math Inverse Matrix

1
##### Enter expression in this form

Example (2x2 Matrix):  Enter expression as [2, 5], [11, 7]

### Inverse Matrix: Explanation and Example

In linear algebra, the inverse of a matrix is a matrix that, when multiplied by the original matrix, yields the identity matrix. The concept of an inverse is fundamental because it allows for solving systems of linear equations, among other applications.

#### Definition

For a square matrix $$A$$, its inverse is denoted as $$A^{-1}$$. The matrix $$A$$ must be non-singular (i.e., it has a non-zero determinant) to have an inverse. The relationship is given by:

$A \cdot A^{-1} = A^{-1} \cdot A = I$

where $$I$$ is the identity matrix of the same dimension as $$A$$.

#### Conditions for Inverses

1. Square Matrix: The matrix must be square (same number of rows and columns).
2. Non-Singular Matrix: The determinant of the matrix must be non-zero ($$\text{det}(A) \neq 0$$).

#### Finding the Inverse

To find the inverse of a 2×2 matrix $$A$$:

$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

The inverse $$A^{-1}$$ is calculated as:

$A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

where $$\text{det}(A) = ad – bc$$.

#### Example

Let’s find the inverse of the matrix:

$A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$

1. Calculate the Determinant:
$\text{det}(A) = (2 \cdot 4) – (3 \cdot 1) = 8 – 3 = 5$

Swap the elements on the main diagonal and change the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$

3. Calculate the Inverse:
Multiply the adjugate matrix by the reciprocal of the determinant:
$A^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{4}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix}$

#### Verification

To verify that $$A^{-1}$$ is correct, multiply $$A$$ by $$A^{-1}$$ and check if the result is the identity matrix:

$A \cdot A^{-1} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} \frac{4}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix}$

#### Perform the multiplication:

$= \begin{pmatrix} 2 \cdot \frac{4}{5} + 3 \cdot -\frac{1}{5} & 2 \cdot -\frac{3}{5} + 3 \cdot \frac{2}{5} \\ 1 \cdot \frac{4}{5} + 4 \cdot -\frac{1}{5} & 1 \cdot -\frac{3}{5} + 4 \cdot \frac{2}{5} \end{pmatrix} = \begin{pmatrix} \frac{8}{5} – \frac{3}{5} & -\frac{6}{5} + \frac{6}{5} \\ \frac{4}{5} – \frac{4}{5} & -\frac{3}{5} + \frac{8}{5} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Since the product is the identity matrix, $$A^{-1}$$ is correctly calculated. This example illustrates the steps to find and verify the inverse of a matrix.