Quick Math Inverse Matrix
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:
\[ \text{det}(A) = (2 \cdot 4) – (3 \cdot 1) = 8 – 3 = 5 \]
2. Form the Adjugate Matrix:
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.
