Question

T(1,1)=(5,0)
Let T:02 → ? be defined as, T(1,0) =(2,6) Then,
T(3,2)=(11,12)
(a) T(0,1)=(5,5)
(b) T(0,1)=(6,6
(c) T(0,1)=(7,7)
(d) T is not a linear transformation​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

Let T: R² → R² be defined as

T(1,1) = (5,0) , T(1,0) = (2,6) , T(3,2) = (11,12)

(a) T(0,1)=(5,5)

(b) T(0,1)=(6,6)

(c) T(0,1)=(7,7)

(d) T is not a linear transformation

EVALUATION

If possible let T: R² → R² be a linear transformation

Let (x,y) ∈ R²

Then there exists two non zero constants a and b such that

$\sf{(x,y) = a(1,1) + b(1,0)}$

$\sf{ \implies \: (x,y) = (a + b,b) }$

Comparing both sides we get

x = a + b and y = b

∴ a = x - y and b = y

Thus we get

$\sf{(x,y) =( x - y)(1,1) + y(1,0)}$

Since T is a linear transformation

$\sf{T(x,y) =( x - y)T(1,1) + yT(1,0)}$

$\sf{ \implies \: T(x,y) =( x - y)(5,0) + y(2,6)}$

$\sf{ \implies \: T(x,y) =(5 x -5 y,0) + (2y,6y)}$

$\sf{ \implies \: T(x,y) =(5 x -3 y,6y) }$

Now we find the value of T(3,2)

$\sf{ \implies \: T(3,2) =(5 \times 3 -3 \times 2,6 \times 2) }$

$\sf{ \implies \: T(3,2) =(15 -6,12) }$

$\sf{ \implies \: T(3,2) =(9,12) }$

But it is given that T(3,2) = (11,12)

So our assumption was wrong

So T is not a linear transformation

FINAL ANSWER

Hence the correct option is

(d) T is not a linear transformation

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. The eigen values of the matrix A are 2,3,5. Then the eigen values of adj A are

https://brainly.in/question/31051731

2. let A and B are square matrices such that AB=I then zero is an eigen value of

https://brainly.in/question/24255712

Answer 2

Answer:

please mark as best answer and thank me