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$\to$ R defined by f(A) = |A| for every A ∈ M, is

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M = \left\{ A = \begin{bmatrix}a & b \ c & d\end{bmatrix}: a, b, c, d \in R ight\} f: M 	o 	ext{R is given by}f\left( A ight)=\left| A ight| Injectivity: f\left( \begin{bmatrix}0 & 0 \ 0 & 0\end{bmatrix} ight) = \begin{vmatrix}0 & 0 \ 0 & 0\end{vmatrix} = 0 	ext{and} f\left( \begin{bmatrix}1 & 0 \ 0 & 0\end{bmatrix} ight) = \begin{vmatrix}1 & 0 \ 0 & 0\end{vmatrix} = 0 \Rightarrow f\left( \begin{bmatrix}0 & 0 \ 0 & 0\end{bmatrix} ight) = f\left( \begin{bmatrix}1 & 0 \ 0 & 0\end{bmatrix} ight) = 0 So, f is not one-one. Surjectivity : Let y be an element of the co-domain, such that f\left( A ight) = - y, A = \begin{bmatrix}a & b \ c & d\end{bmatrix} \Rightarrow \begin{vmatrix}a & b \ c & d\end{vmatrix} = y \Rightarrow ad - bc = y \Rightarrow a, b, c, d \in R \Rightarrow A = \begin{bmatrix}a & b \ c & d\end{bmatrix} \in M ⇒ f is onto. So, the answer is (d).

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