M.4 Matrix Inverse

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Inverse of a Matrix

The matrix B is the inverse of matrix A if \(AB = BA = I\). This is often denoted as \(B = A^{-1}\) or \(A = B^{-1}\). When taking the inverse of the product of two matrices A and B,

\[(AB)^{-1} = B^{-1} A^{-1}\]

 

When taking the determinate of the inverse of the matrix A,

\[ det(A^{-1}) = \frac{1}{det(A)} = det(A)^{-1}\]

Note that not all matrices have inverses. For a matrix A to have an inverse, that is to say for A to be invertible, A must be a square matrix and \(det(A) \neq 0\). For that reason, invertible matrices are also called nonsingular matrices.

Two examples are shown below

\[ det(A) = \begin{vmatrix} 4 & 5 \\ -2 & 1 \end{vmatrix} = 4*1-5*-2 = 14 \neq 0 \]

\[ det(C) = \begin{vmatrix} 1 & 2 & -1\\ 5 & 3 & 2 \\ 6 & 0 & 6 \end{vmatrix} = -2 \begin{vmatrix} 5 & 2 \\ 6 & 6 \end{vmatrix} + 3 \begin{vmatrix} 1 & -1\\ 6 & 6 \end{vmatrix} + 0 \begin{vmatrix} 1 & -1\\ 5 & 2 \end{vmatrix}\]

\[ det(C)= - 2(5*6-2*6) + 3(1*6-(-1)*6) - 0(1*2-(-1)*5) = 0 \]

So C is not invertible, because its determinate is zero. However, A is an invertible matrix, because its determinate is nonzero. To calculate that matrix inverse of a 2 × 2 matrix, use the below formula.

\[ A^{-1} = \begin{pmatrix} a_{1,1} & a_{1,2}\\ a_{2,1} & a_{2,2} \end{pmatrix} ^{-1} = \frac{1}{det(A)} \begin{pmatrix} a_{2,2} & -a_{1,2} \\  -a_{2,1} & a_{1,1} \end{pmatrix} = \frac{1}{a_{1,1} * a_{2,2} - a_{1,2}*a_{2,1}} \begin{pmatrix} a_{2,2} & -a_{1,2} \\  -a_{2,1} & a_{1,1} \end{pmatrix}\]

For example

\[ A^{-1} = \begin{pmatrix} 4 & 5 \\ -2 & 1 \end{pmatrix} ^{-1} = \frac{1}{det(A)} \begin{pmatrix} 1 & -5 \\ 2 & 4 \end{pmatrix} = \frac{1}{4*1 - 5*(-2)} \begin{pmatrix} 1 & -5 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} \frac{1}{14} & \frac{-5}{14} \\ \frac{2}{14} & \frac{4}{14} \end{pmatrix}\]

 

For finding the matrix inverse in general, you can use Gauss-Jordan Algorithm. However, this is a rather complicated algorithm, so usually one relies upon the computer or calculator to find the matrix inverse.