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Inverse of a Matrix

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.

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Inverse of a Matrix -

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Inverse of a Matrix

Square Matrix Inverse Calculator NxN

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Matrix Modular Inverse Calculator

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See also: Hill Cipher

Answers to Questions (FAQ)

How to calculate the inverse of an invertible matrix?

The inverse of a square matrix $ M $ is noted $ M^{-1} $ and can be calculated in several ways. The most suitable for 2x2 or 3x3 matrix sizes is the cofactor method which necessitate to calculate the determinant of the matrix $ \det M $ and the transposed cofactor matrix (also called adjugate matrix $ \operatorname{adj}(M) $):

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

The dCode calculator works for any size of square matrix.

For a 2x2 matrix (order 2):

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

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 3x3 matrix (order 3):

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

It is essential that the determinant of the matrix to be inverted is not equal to zero for the matrix to be invertible.

How to prove that a matrix is invertible?

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.

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

Avoid the term inversible which is wrong.

How to inverse a matrix with zero determinant?

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.

How to check that a matrix is the inverse of another?

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

How to calculate the modular inverse of a matrix?

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

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