Let $M$ be the set of all $2 \times 2$ matrices with entries from the set R of real numbers. Then the function $f: M \rightarrow R$ defined by $f(A)=|A|$ for every $A \in M$ is

The function $f: M \rightarrow R$ is defined as $f(A) = |A|$, where $M$ is the set of all $2 \times 2$ matrices with real entries and $|A|$ denotes the determinant of matrix $A$.

To check if $f$ is one-one, consider two distinct matrices $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 0 & 0.5 \end{pmatrix}$. Both matrices are in $M$ and $A \neq B$. The determinant of $A$ is $|A| = (1)(1) - (0)(0) = 1$. The determinant of $B$ is $|B| = (2)(0.5) - (0)(0) = 1$. Since $f(A) = f(B)$ but $A \neq B$, the function $f$ is not one-one.

To check if $f$ is onto, consider any real number $k \in R$. We need to determine if there exists a matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M$ such that $|A| = ad - bc = k$. For any $k \in R$, we can choose $A = \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}$. Then $|A| = k(1) - 0(0) = k$. Since for every $k \in R$ there exists at least one matrix $A \in M$ such that $f(A) = k$, the function $f$ is onto.

Therefore, the function $f$ is onto but not one-one.

Answer: onto but not one-one