Question

Let M be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then, the function f : M → R defined by f(A) = |A| for every A ∈ M, is
(a) one-one and onto
(b) neither one-one nor onto
(c) one-one but-not onto
(d) onto but not one-one

Answer 1

$$\begin{gathered} M=\left\{A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]:a,b,c,d\in R\right\} \\ f:M\to R\text{is given by}\mathit{\text{f}}\left(A\right)\mathit{\text{=}}\left|A\right| \end{gathered}$$

Injectivity:
$$\begin{gathered} f\left(\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right)=\left|\begin{array}{cc}0& 0\\ 0& 0\end{array}\right|=0 \\ \mathrm{and}f\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=\left|\begin{array}{cc}1& 0\\ 0& 0\end{array}\right|=0 \\ \Rightarrow f\left(\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right)=f\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=0 \end{gathered}$$
So, f is not one-one.

Surjectivity:
Let y be an element of the co-domain, such that
$$\begin{gathered} f\left(A\right)=-y,A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right] \\ \Rightarrow \left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=y \\ \Rightarrow ad-bc=y \\ \Rightarrow a,b,c,d\in R \\ \Rightarrow A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\in M \end{gathered}$$
$\Rightarrow$f is onto.
So, the answer is (d).