14 pts Let T R3 R3 be a linear transformation such that 711

[14 pts] Let T : R3 R3 be a linear transformation such that 7(1,1,1)=(2,0-1), T(1,1,2)-(-3,2-1), and T(L0,1)-(1,1,0) . Find T(1.35) 1,0). Find T(1.3.5)

Solution

Here , first we try to see whether (1,1,1) , (1,-1,2) and (1,0,1) are independent.

a(1,1,1) + b(1,-1,2) + c(1,0,1) = (0,0,0)

=> (a + b + c , a - b , a + 2b + c ) = (0,0,0)

=> a + b+ c =0 (1), a - b = 0 (2), a + 2b +c = 0 (3)

=> a = b , (3) - (1) gives b = 0 => a = 0

implies c = 0 .

So, a = b = c = 0 .

So, these three vectors are independent and forms a basis of R³ .

Therefore, there exists one and only one transformation staed above.

We first represent (1,3,5) = e(1,1,1) + f(1,-1,2) + g(1,0,1) = (e + f + g , e - f , e + 2f + g )

e + f+g = 1 ..(4) , e - f = 3 ...(5) , e +2f +g = 5 ...(6)

(6) - (4) = f = 4, putting f = 4 in (5) , we get e = 3 + f = 3 + 4 = 7

e = 7 and f = 4 in (4) we get g = 1 - 7 - 4 = - 10 .

So, (1,3,5) = 7*(1,1,1) + 4*(1,-1,2) + (-10)*(1,0,1)

So, T(1,3,5) = 7*T(1,1,1) + 4*T(1,-1,2) + (-10)*T(1,0,1) = 7*(2,0,-1) + 4*(-3,2,-1) + (-10)*(1,1,0)

=> T(1,3,5) = ( 14 -12 - 10 , 0 + 8 -10 , -7 - 4 ) = ( - 8 , -2 , -11 )