Game of Mappings

Algebra Level pending

Consider a function f : Z 4 → Z f: \mathbb{Z^4} \to \mathbb{Z} such that

f ( a , b , c , d ) f(a,b,c,d) is equal to the determinant of the matrix [ a b c d ] \begin{bmatrix} {a} && {b} \\ {c} && {d} \end{bmatrix} .

Which of the following is the most accurate option regarding the property of f f ?


Notation: Z \mathbb Z denotes the set of integers .

Many-to-one Onto One-to-one Bijective Onto and Many-to-one

Kushal Bose by Kushal Bose , 28, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tom Engelsman
Jan 27, 2017

The require function in question is f ( a , b , c , d ) = a d − b c f(a,b,c,d) = ad - bc . It cannot be one-to-one (or injective) by the example:

f ( 1 , 2 , 3 , 4 ) = f ( − 1 , − 2 , − 3 , − 4 ) = − 2 f(1,2,3,4) = f(-1,-2,-3,-4) = -2

There is no unique quadruple to yield one unique value of f f . Also, it cannot be bijective because it's not injective. This leaves the function as onto (or surjective) AND many-to-one, which if we expand on the above example:

f ( 1 , 2 , 3 , 4 ) = f ( − 1 , − 2 , − 3 , − 4 ) = − 2 f(1,2,3,4) = f(-1,-2,-3,-4) = -2 AND f ( 1 , 2 , 3 , 4 ) = f ( 1 , 2 , 6 , 10 ) = − 2 f(1,2,3,4) = f(1,2,6,10) = -2 .

there are many quadruples that will yield the entire range of integers Z \mathbb{Z} , as well as yield the same value of f ⇒ f \Rightarrow choice B is correct.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...