Let A be the set of all 3 times 3 scalar matr | vedclass

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(C) scalar matrix $M$ of order $3 	imes 3$ is of the form $M = \begin{bmatrix} m & 0 & 0 \ 0 & m & 0 \ 0 & 0 & m \end{bmatrix}$ for some $m \in R$.

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bijective

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(C) scalar matrix $M$ of order $3 	imes 3$ is of the form $M = \begin{bmatrix} m & 0 & 0 \ 0 & m & 0 \ 0 & 0 & m \end{bmatrix}$ for some $m \in R$. The determinant of $M$ is given by $\operatorname{det}(M) = m^3$. Thus,the function $f: A ightarrow R$ is defined as $f(M) = m^3$. Since $f(m) = m^3$ is a strictly increasing function,for any $m_1, m_2 \in R$,$f(m_1) = f(m_2) \implies m_1^3 = m_2^3 \implies m_1 = m_2$. Therefore,$f$ is one-one (injective). For any real number $y \in R$,there exists a real number $m = \sqrt[3]{y}$ such that $f(m) = (\sqrt[3]{y})^3 = y$. Thus,the range of $f$ is $R$,which is equal to the codomain. Therefore,$f$ is onto (surjective). Since $f$ is both one-one and onto,it is bijective.

Let $A$ be the set of all $3 \times 3$ scalar matrices with real entries. If $f: A \rightarrow R$ is defined by $f(M) = \operatorname{det}(M)$ for all $M \in A$,then $f$ is

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