Let X = left{begin{bmatrix} a & b c & d end{ | vedclass

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(B) We have $X = \left\{\begin{bmatrix} a & b \ c & d \end{bmatrix} : a, b, c, d \in \mathbb{R} ight\}$ and $f(A) = \operatorname{det}(A) = ad - bc

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onto but not one-one

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(B) We have $X = \left\{\begin{bmatrix} a & b \ c & d \end{bmatrix} : a, b, c, d \in \mathbb{R} ight\}$ and $f(A) = \operatorname{det}(A) = ad - bc$.For the function to be onto,for every $y \in \mathbb{R}$,there must exist a matrix $A \in X$ such that $f(A) = y$.Consider the matrix $A = \begin{bmatrix} y & 0 \ 0 & 1 \end{bmatrix}$. Then $\operatorname{det}(A) = y(1) - 0(0) = y$. Since such a matrix exists for any $y \in \mathbb{R}$,the function is onto.For the function to be one-one,$f(A_1) = f(A_2)$ must imply $A_1 = A_2$.Consider $A_1 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ and $A_2 = \begin{bmatrix} 2 & 1 \ 1 & 1 \end{bmatrix}$.$f(A_1) = (1)(1) - (0)(0) = 1$ and $f(A_2) = (2)(1) - (1)(1) = 1$.Since $f(A_1) = f(A_2)$ but $A_1 
eq A_2$,the function is not one-one.Therefore,$f$ is onto but not one-one.

Let $X = \left\{\begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \mathbb{R} \right\}$. Define $f: X \rightarrow \mathbb{R}$ by $f(A) = \operatorname{det}(A), \forall A \in X$. Then,$f$ is

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