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Let $A$ be the set of all $3 \times 3$ scalar matrices with real entries. If $f: A \rightarrow R$ is defined by $f(m)=\operatorname{det}(m) \forall ; m \in A$, then $f$ is
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The correct answer is:
bijective
$A$ be a scalar matrix
$$
\begin{aligned}
A & =\left[\begin{array}{ccc}
m & 0 & 0 \\
0 & m & 0 \\
0 & 0 & m
\end{array}\right] \\
|A| & =m^3 \Rightarrow f(m)=m^3
\end{aligned}
$$
$f(m)$ is increasing function for all values of $m$.
$\therefore f(m)$ is injective.
Range of $f(m)=R=$ codomain
$\therefore f(m)$ is surjective.
Hence, $f(m)$ is bijective.
$$
\begin{aligned}
A & =\left[\begin{array}{ccc}
m & 0 & 0 \\
0 & m & 0 \\
0 & 0 & m
\end{array}\right] \\
|A| & =m^3 \Rightarrow f(m)=m^3
\end{aligned}
$$
$f(m)$ is increasing function for all values of $m$.
$\therefore f(m)$ is injective.
Range of $f(m)=R=$ codomain
$\therefore f(m)$ is surjective.
Hence, $f(m)$ is bijective.
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