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(B) The set $X$ consists of all $2 	imes 2$ matrices with real entries. The function $f: X ightarrow \mathbb{R}$ is defined as $f(A) = \det(A) = ad

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onto but not one-one

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(B) The set $X$ consists of all $2 	imes 2$ matrices with real entries. The function $f: X ightarrow \mathbb{R}$ is defined as $f(A) = \det(A) = ad - bc$.To check if $f$ is one-one, consider two different matrices $A_1 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ and $A_2 = \begin{bmatrix} 2 & 0 \ 0 & 0.5 \end{bmatrix}$.Here, $f(A_1) = (1)(1) - (0)(0) = 1$ and $f(A_2) = (2)(0.5) - (0)(0) = 1$.Since $f(A_1) = f(A_2)$ but $A_1 
eq A_2$, the function $f$ is not one-one.To check if $f$ is onto, for any real number $k \in \mathbb{R}$, we can find a matrix $A = \begin{bmatrix} k & 0 \ 0 & 1 \end{bmatrix} \in X$ such that $f(A) = (k)(1) - (0)(0) = k$.Since every element in the codomain $\mathbb{R}$ has a pre-image in $X$, the function $f$ is onto.Therefore, $f$ is onto but not one-one.

Let $X = \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \mathbb{R} \right\}$. If $f: X \rightarrow \mathbb{R}$ is defined by $f(A) = \det(A)$ for all $A \in X$, then $f$ is

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