91Ó°ÊÓ

The inverse of a square matrix Ais the matrix ${\mathbf{A}}^{\mathbf{-}\mathbf{1}}$so that when multiplied by that matrix gives a unit matrix l .

${\mathbf{AA}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{l}$

The inverse that operates on square matrices needs to be determined.

Check if the determinant of a matrix is a linear operator.

A linear operator satisfies the condition.

$$\begin{gathered} \mathrm{O}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{O}\left(\mathrm{A}\right)+\mathrm{O}\left(\mathrm{B}\right) \\ \mathrm{O}\left(\mathrm{kA}\right)=\mathrm{kO}\left(\mathrm{A}\right) \end{gathered}$$

The inverse of a square matrix is given.

${\mathrm{M}}^{-1}=\frac{1}{\mathrm{detM}}{\mathrm{C}}^{\mathrm{T}}$

Take the sum of two square matrices A and B,

The determinant D=A+B is${\mathrm{D}}^{-1}=\frac{1}{\det \left(\mathrm{A}+\mathrm{B}\right)}{\mathrm{C}}^{\mathrm{T}}$

If the inverse were a linear operator, this would be equal to$\frac{1}{detA}{C}_a^T+\frac{1}{detA}{C}_b^T+$which cannot be true generally.

Also, the result was obtained from the second condition.

$$\begin{gathered} {\left(\mathrm{k}\mathrm{A}\right)}^{-1}=\frac{1}{\det \left(\mathrm{k}\mathrm{A}\right)}{\mathrm{C}}^{\mathrm{T}} \\ =\frac{1}{\mathrm{kdetA}}{\mathrm{C}}^{\mathrm{T}} \\ =\frac{1}{\mathrm{k}}{\mathrm{A}}^{-1} \end{gathered}$$

Due to the property $\det \left(\mathrm{k}\mathrm{A}\right)={\mathrm{k}}^{\mathrm{n}}\det \mathrm{A}$, where is the dimension of the matrix (the elements of C are determinants with one row and column of deleted so that it have one less power of k ). This property is the consequence of the property of the determinant which states that if one row of A is multiplied with k , then the determinant is k det A. Therefore, the inverse is not a linear operator.