GATE CSE 1996 | Question: 2.1

Question

Let $R$ denote the set of real numbers. Let $f:R\times R \rightarrow R \times R$ be a bijective function defined by $f(x,y) = (x+y, x-y)$. The inverse function of $f$ is given by

• A. $f^{-1} (x,y) = \left( \frac {1}{x+y}, \frac{1}{x-y}\right)$
  
  
• B. $f^{ -1} (x,y) = (x-y , x+y)$
  
  
• C. $f^{-1} (x,y) = \left( \frac {x+y}{2}, \frac{x-y}{2}\right)$
  
  
• D. $f^{-1}(x,y)=\left [ 2\left(x-y\right),2\left(x+y\right) \right ]$

Comments

just draw mapping and whith help of e.g find out inverse function

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c

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The faster way to do it is to plug in a number. 

Pick any point, say (2, 3). Then f(2,3) = (5, −1). So f⁻¹(5, −1) must give back (2, 3). Test each option with x = 5, y = −1:

• A: (1/4, 1/6) ✗
• B: (6, 4) ✗
• C: ((5−1)/2, (5+1)/2) = (2, 3) ✓
• D: (12, 8) ✗

Answer 1

to find inverse of the function take

$z_1=x+y \text{          } \to(1)$

$z_2=x-y \text{          } \to (2)$

Adding (1) and (2) we get,

$x = \frac{z_1+z_2}{2}$ and $y = \frac{z_1-z_2}{2}$

So, $f \left(\frac{z_1}{2},\frac{z_2}{2}\right) = \left(\frac{z_1+z_2}{2},\frac{z_1-z_2} {2}\right) = (x, y) \\ \implies f^{-1}(x, y) = \left(\frac{z_1}{2},\frac{z_2}{2}\right) \\=  \left\{\frac{x+y}{2},\frac{x-y}{2} \right\}$

Correct Answer: $C$

Comments

yes sir

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Hello Sir,

Can you please explain why we have taken f1(z1/2, z2 /2)

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@JPranavc

Can you please explain why we have taken f1(z1/2, z2 /2)

Please check the image below, hope that will help.

Answer 2

Taking an example:

$f(2,3)=(2+3,2-3)=(5,-1)$

$f^{-1}(5,-1)$ should be $(2,3).$

Substituting the values we get (C) as answer.

Comments

amazing!

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did it like this only!

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chamka bhai

 

Answer 3

these kind of questions are always very very interesting

 

Comments

Great Solution Should be at the top

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Thank you Vandit.

~Shashank

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best answer !!

Answer 4

Answer : C

f(x,y) = ( x+y , x−y ) . for invertible function you have to find that there should be bijection (one to one correspondence) . 

 if f(a) = b then  a = $f^{-1}$(b)

apply this concept here , f(x,y) = f(x+y , x-y) , so (x,y) = $f^{-1}$(x+y , x-y) -----------(1)

lets assume

p1 = x+y ------------(i)

p2 = x-y ------------(ii)

By Adding           (i)+(ii)                    $\frac{(p1+ p2) }{2}$ = x   

By Subtracting    (i)-(ii)                    $\frac{(p1- p2) }{2}$ = y

put value in--(1) 

$\left ( \frac{( p1+p2 )}{2} , \frac{( p1-p2 )}{2} \right )$ = $f^{-1}$( p1 , p2 ) 

$f^{-1}\left ( x,y \right )$ =  $\left ( \frac{( x+y )}{2} , \frac{( x-y )}{2} \right )$

 

Comments

but u assumed p1=x+y and p2=x-y .. then how you substituted p1=x and p2=y in the last step??       pls explain?

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Given: f(x,y) = (x+y, x-y)
Let f(x,y) = (u,v)
=> u = x+y, v = x-y
=> x = (u+v)/2, y = (u-v)/2
Hence, f^-1(u,v) = ((u+v)/2, (u-v)/2)
Verification: f(f^-1(u,v)) = (u,v)
 

Answer 5

Another procedure to solve the inverse of a function using Matrices.

 

 

 

 

So, Option C  is the answer.

Comments

Woah! Nice Solution. Is this method applicable to some other problem too, then pls post it. Really short and easy to digest

Answer 6

Given $$f(x,y)=(x+y,x-y)$$

Now inverse is such a function, that if you plug in the output of $f$ [here $(x+y,x-y)$] as input to $f^{-1}$ then you shall get back the input to $f$ [i.e. $(x,y)$] which gave the output [$(x+y,x-y)$].

So our input to $f^{-1}$ should be $(x+y,x-y)$ to get output as $(x,y)$. This gives the hint that we need to choose such an inverse function which adds/subtracts the two components of the input [since the two components of the input are present in sum/subtraction form, and adding or subtracting shall result in only one particular term of each input component.]

Using option $(C)$ we have:

$f^{-1}(x+y,x-y)= \left( \frac{(x+y)+(x-y)}{2},\frac{(x+y)+(x-y)}{2}\right)=(x,y)$

So option (C) is the required answer.

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Formal method of solving it.

$$f(x,y)=(x+y,x-y)$$

Let $X=x+y$ and $Y=x-y$. We need to change the equation from $x,y$ to $X,Y$. So

$$X+Y=2x \implies x=\frac{X+Y}{2}$$

$$X-Y=2y \implies y=\frac{X-Y}{2}$$

Substituting these in our function $f$,

$$f\left(\frac{X+Y}{2},\frac{X-Y}{2}\right)=(X,Y)$$

$$\implies \left(\frac{X+Y}{2},\frac{X-Y}{2}\right)=f^{-1}(X,Y)$$

so option (C) is correct.

Comments

a very good answer