Tool to invert a matrix. The inverse of a square matrix M is a matrix denoted M^-1 such as que M.M^-1=I where I is the identity matrix.

Inverse of a Matrix - dCode

Tag(s) : Matrix

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The inverse of a square and invertible matrix $ A $ is a matrix $ A^{-1} $ such that $ A \times A^{-1} = A^{-1} \times A = I $, where $ I $ is the identity matrix.

Inverting a matrix undoes the initial transformation, returning each point to its original position.

The inverse $ M^{-1} $ of a square matrix $ M $ can be calculated using several methods that dCode applies for all square matrix sizes.

— Calculation via transpose of the cofactor matrix:

$$ M^{-1} = \frac{1}{\det M} \left( \operatorname{cof}(M) \right)^\mathsf{T} = \frac{1}{\ det M} \operatorname{comp}(M) $$

This formula requires calculating the determinant of the matrix $ \det M $ as well as the transpose of the cofactor matrix (also called the complementary matrix $ \operatorname{comp}(M) $).

— Calculation using the Gauss pivot method:

The method requires carrying out elementary operations on the rows of the matrix M in order to bring it back to the identity matrix. To obtain the inverse matrix, perform the same operations but this time from the identity matrix.

If the matrix is small, the cofactor method does not require too many calculations and gives a general formula:

— For a matrix of order 1 (1x1):

$$ M^{-1} = \begin{bmatrix} a \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{a} \end{bmatrix} $$

— For a matrix of order 2 (2x2):

$$ M = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \\ \det(M) = ad - bc \\ \operatorname{cof}(M) = \begin{bmatrix} d & -c \\ -b & a \end{bmatrix} \\ \operatorname{comp}(M) = \left( \operatorname{cof}(M) \right)^\mathsf{T} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ M^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \\ \end{bmatrix} $$

__Example:__ $$ M = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} \Rightarrow M^{-1} = \frac{1}{\det(M)} \begin{bmatrix} 4 & -2 \\ -3 & 1 \\ \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \\ \end{bmatrix} $$

— For a matrix of order 3 (3x3):

$$ M^{-1} = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}^{-1} = \frac{1}{-c e g+b f g+c d h-a f h-b d i+a e i} \begin{bmatrix} e i-f h & c h-b i & b f-c e \\ f g-d i & a i-c g & c d-a f \\ d h-e g & b g-a h & a e-b d \end{bmatrix} $$

Form an augmented matrix by concatenating the matrix A with the corresponding identity matrix, i.e. a matrix of the form [A | I].

Apply the Gaussian pivot method to reduce the left part (matrix A) to a reduced echelon form, by performing elementary row operations.

If the reduced form is the identity matrix, then the right part is the inverse of the A matrix.

This method may take longer in calculation time but can be used to check the inverse of a matrix.

A matrix is invertible if its determinant is non-zero (different from 0). So to prove that a matrix has an inverse, calculate the determinant of the matrix, if it is different from 0, then the matrix is invertible.

The matrix must be square (same number of rows and columns) and not rectangular to be invertible.

A non invertible matrix is called *singular* (inversion is not possible).

Avoid the term *inversible* which is wrong.

A diagonal matrix is invertible if all elements on its diagonal are non-zero.

The inverse of a diagonal matrix is also a diagonal matrix.

$$ A = \left( \begin{array}{cccc} a_{1,1} & 0 & \dots & 0 \\ 0 & a_{2,2} & \dots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \dots & a_{n,n} \end{array} \right) \quad A^{-1} = \left( \begin{array}{cccc} \frac{1}{a_{1,1}} & 0 & \dots & 0 \\ 0 & \frac{1}{a_{2,2}} & \dots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \dots & \frac{1}{a_{n,n}} \end{array} \right) $$

A triangular matrix is invertible if all elements on its diagonal are non-zero.

The inverse of a triangular matrix is also a triangular matrix.

A matrix with a determinant equal to 0 is not invertible. It does not have an inverse, it is not possible to calculate its inverse.

The inverse of an orthogonal matrix $ Q $ is its matrix transpose: $ Q^{-1} = Q^T $

The multiplication of the matrix by its inverse must give the identity matrix. So the computation of $ M . M^{-1} = I $.

The principle is the same, but instead of calculating the determinant, calculate the modular inverse of the matrix determinant. See Hill cipher.

Solving a system of linear equations $ A \times X = B $ where $ A $ is an invertible matrix, involves calculating the inverse of $ A $, the matrix $ A^{-1} $ obtained can then be multiplied on both sides of the equation to obtain $ X = A^{-1} \times B $.

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Cite as source (bibliography):

*Inverse of a Matrix* on dCode.fr [online website], retrieved on 2024-09-10,

- Square Matrix Inverse Calculator NxN
- Matrix Modular Inverse Calculator
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- How to calculate the inverse of an invertible matrix?
- How to calculate the inverse of a matrix using the minors?
- How to calculate the inverse of a matrix using Gauss's pivot?
- How to prove that a matrix is invertible?
- How to prove that a diagonal matrix is invertible?
- How to prove that a triangular matrix is invertible?
- How to inverse a matrix with zero determinant?
- How to inverse an orthoginal matrix?
- How to check that a matrix is the inverse of another?
- How to calculate the modular inverse of a matrix?
- How to solve a system of linear equations with an inverse matrix?

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