I agree with Omerโs solution. A few additional remarks could be of interest.
First, the function \(T: M_{22}->C^1\) defined by \[T \left( \left[ \begin{array}{rr} a & b \\ c & d \\ \end{array} \right] \right) = a + b + c - d \]
is a linear transformation. This can be verified by noting that
\[T(A + B) = T(A) + T(B)\] for matrices \(A\) and \(B\) and that \(\alpha T(A) = T( \alpha \cdot A)\).
\[T\left( \begin{array}{rr} a_1 + a_2 & b_1 + b_2 \\ c_1 + c_2 & d_1 + d_2 \\ \end{array} \right) = a_1 + a_2 + b_1 + b_2 + c_1 + c_2 - (d_1 + d_2) = (a_1 + b_1 + c_1 - d_1) + (a_2 + b_2 + c_2 - d_2) = T\left( \begin{array}{rr} a_1 & b_1 \\ c_1 & d_1 \\ \end{array} \right) + T\left( \begin{array}{rr} a_2 & b_2 \\ c_2 & d_2 \\ \end{array} \right) \]
\[ \alpha T\left( \begin{array}{rr} a & b \\ c & d \\ \end{array} \right) = \alpha \cdot (a + b + c - d) = \alpha a + \alpha b + \alpha c - \alpha d = T\left( \begin{array}{rr} \alpha a & \alpha b \\ \alpha c & \alpha d \end{array} \right) \]
A second observation is that the pre-image \(T^{-1}(3)\) is a 3-dimensional hyperplane in a 4-dimensional space. If we view the space \(M_{2,2}\) as being equivalent to \(R^4\), then the comparison is even clearer.