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Consider two linear spaces V and W. A function T is said to be linear transformation if the following holds.

$$\begin{gathered} \mathbf{T}\left(f+g\right)\mathbf{=}\mathbf{T}\left(f\right)\mathbf{+}\mathbf{T}\left(g\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{T}\left(\mathrm{kf}\right)\mathbf{=}\mathbf{kT}\left(f\right) \end{gathered}$$

For all elements f,g of V and is scalar.

An invertible linear transformation is called an isomorphism.

The given transformation as follows.

$\mathrm{T}\left(\mathrm{M}\right)=\det \left(\mathrm{M}\right)\mathrm{from}{\mathrm{R}}^{2\mathrm{x}2}\mathrm{to}\mathrm{R}$

By using the definition of linear transformation as follows.

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

Consider 2X2 the matrix for A and B as follows.

$$\begin{gathered} \mathrm{A}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \mathrm{B}=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right] \end{gathered}$$

Simplify the matrix $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for A and $\mathrm{A}=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$for B in $\mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)$as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\det \left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\right) \\ =\det \left[\begin{array}{cc}1-1& 0\\ 0& 1-1\end{array}\right] \\ =\det \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Now, simplify for T (A ) = A and T (B) = B as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}\right)=\mathrm{A} \\ =\det \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \mathrm{T}\left(\mathrm{A}\right)=1 \end{gathered}$$

Similarly, simplify for $\mathrm{T}\left(\mathrm{B}\right)=\mathrm{B}$as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{B}\right)=\mathrm{B} \\ =\det \left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right] \\ \mathrm{T}\left(\mathrm{B}\right)=1 \end{gathered}$$

Then adding the transformations as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)=\mathrm{A}+\mathrm{B} \\ \mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)=1+1 \\ \mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)=2 \end{gathered}$$

Now, substitute the value 0 for T ( A+B ) and 2 for $\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)$in the transformation $\mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)$as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right) \\ 0=2 \\ 0鈮�2 \end{gathered}$$

Therefore, the transformation is $\mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)鈮�\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)$.

Hence, the transformation is not linear.