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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
Options
one-one and onto
neither one-one nor onto
one-one but-not onto
onto but not one-one
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Solution
\[M = \left\{ A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}: a, b, c, d \in R \right\}\]
\[f: M \to \text{R is given by}f\left( A \right)=\left| A \right|\]
Injectivity:
\[f\left( \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} \right) = \begin{vmatrix}0 & 0 \\ 0 & 0\end{vmatrix} = 0\]
\[\text{and} f\left( \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} \right) = \begin{vmatrix}1 & 0 \\ 0 & 0\end{vmatrix} = 0\]
\[ \Rightarrow f\left( \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} \right) = f\left( \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} \right) = 0\]
So, f is not one-one.
Surjectivity :
Let y be an element of the co-domain, such that
\[f\left( A \right) = - y, A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}\]
\[ \Rightarrow \begin{vmatrix}a & b \\ c & d\end{vmatrix} = y\]
\[ \Rightarrow ad - bc = y\]
\[ \Rightarrow a, b, c, d \in R\]
\[ \Rightarrow A = \begin{bmatrix}a & b \\ c & d\end{bmatrix} \in M\]
⇒ f is onto.
So, the answer is (d).
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