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(D) The function $f: S ightarrow R$ is defined by $f(A) = \det(A)$, where $A \in S$ and $S$ is the set of all non-singular matrices of order $2 	im

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$f$ is neither one-one nor onto

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(D) The function $f: S ightarrow R$ is defined by $f(A) = \det(A)$, where $A \in S$ and $S$ is the set of all non-singular matrices of order $2 	imes 2$. $A$ matrix is non-singular if its determinant is non-zero. Thus, the range of $f$ is $R \setminus \{0\}$.$1$. Check for one-one: Consider two matrices $A_1 = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}$ and $A_2 = \begin{bmatrix} 4 & 0 \ 0 & 1 \end{bmatrix}$. Both $A_1, A_2 \in S$ because $\det(A_1) = 4 
eq 0$ and $\det(A_2) = 4 
eq 0$. Here, $f(A_1) = 2(2) - 0(0) = 4$ and $f(A_2) = 4(1) - 0(0) = 4$. Since $f(A_1) = f(A_2)$ but $A_1 
eq A_2$, the function $f$ is not one-one.$2$. Check for onto: The codomain is $R$. For any $y = 0 \in R$, there exists no matrix $A \in S$ such that $f(A) = 0$, because $S$ only contains non-singular matrices (where $\det(A) 
eq 0$). Thus, $f$ is not onto.Therefore, $f$ is neither one-one nor onto.

If $f: S \rightarrow R$ where $S$ is the set of all non-singular matrices of order $2$ over $R$ and $f\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = ad - bc$, then:

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