Answer
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Hint: Here we have to use the concept of one-one and onto function. Firstly, we will assume a matrix \[A\]. Then we will check the condition of one-one and then we will check the condition of onto. After this we will conclude whether the function is one-one or onto function.
Complete step-by-step answer:
Let matrix \[A\] be \[\left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\].
First, we will check the condition of the one- one or injectivity of the function \[f\]. We know that in one-one function the solution is never the same. Therefore, we get
\[f\left( {\left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right]} \right) = \left| {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right| = 0\], also
\[f\left( {\left[ {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right]} \right) = \left| {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right| = 0\]
\[\therefore f\left( {\left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]} \right) = f\left( {\left[ {\begin{array}{*{20}{c}}
1&0 \\
0&0
\end{array}} \right]} \right) = 0\]
So we can clearly say that the function \[f\] is not one-one function.
Now we will check the condition for onto or surjectivity of the function \[f\].
Let \[x\] be a number from the co-domain, such that \[f(A) = x\], we get
\[f(A) = f\left( {\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]} \right) = \left| {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right| = x\]
\[ \Rightarrow ad - bc = x\]
This means that \[a,b,c,d \in R\]. Therefore
\[A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right] \in R\]
So, the function \[f\] is onto.
Hence the function \[f\] is onto but not one-one function.
So, option B is the correct option.
Note: Here we should note the difference between the one-one and onto. One-one functions are the functions which have their unique image in the domain and onto functions are the functions which have multiple images in the domain. If the function is not one-one then the function is generally known as many one function.
One-one functions are generally known as injective functions and onto functions are generally known as the surjective functions.
Bijective Functions are the functions which are both one-one and onto.
Complete step-by-step answer:
Let matrix \[A\] be \[\left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\].
First, we will check the condition of the one- one or injectivity of the function \[f\]. We know that in one-one function the solution is never the same. Therefore, we get
\[f\left( {\left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right]} \right) = \left| {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right| = 0\], also
\[f\left( {\left[ {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right]} \right) = \left| {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right| = 0\]
\[\therefore f\left( {\left[ {\begin{array}{*{20}{c}}
0&0 \\
0&0
\end{array}} \right]} \right) = f\left( {\left[ {\begin{array}{*{20}{c}}
1&0 \\
0&0
\end{array}} \right]} \right) = 0\]
So we can clearly say that the function \[f\] is not one-one function.
Now we will check the condition for onto or surjectivity of the function \[f\].
Let \[x\] be a number from the co-domain, such that \[f(A) = x\], we get
\[f(A) = f\left( {\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]} \right) = \left| {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right| = x\]
\[ \Rightarrow ad - bc = x\]
This means that \[a,b,c,d \in R\]. Therefore
\[A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right] \in R\]
So, the function \[f\] is onto.
Hence the function \[f\] is onto but not one-one function.
So, option B is the correct option.
Note: Here we should note the difference between the one-one and onto. One-one functions are the functions which have their unique image in the domain and onto functions are the functions which have multiple images in the domain. If the function is not one-one then the function is generally known as many one function.
One-one functions are generally known as injective functions and onto functions are generally known as the surjective functions.
Bijective Functions are the functions which are both one-one and onto.
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