Let M be the set of all 2 × 2 matrices with entries from the set R of real numbers.

Solution:

$M=\left\{A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]: a, b, c, d \in R\right\}$

$f: M \rightarrow R$ is given by $f(A)=|A|$

Injectivity:

$f\left(\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\right)=\left|\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right|=0$

and $f\left(\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\right)=\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|=0$

$\Rightarrow f\left(\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\right)=f\left(\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\right)=0$

So, f is not one-one.

Surjectivity:
Let y be an element of the co-domain, such that

$f(A)=-y, A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$\Rightarrow\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|=y$

$\Rightarrow a d-b c=y$

$\Rightarrow a, b, c, d \in R$

$\Rightarrow A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \in M$

$\Rightarrow f$ is onto.

So, the answer is (d).