Let T:R
3
→R
3
be a linear transformation such that T(1,0,0)=(2,−1,4),T(0,1,0)=(1,−2,3), and T(0,0,1)=(2,−2,0). Find the indicated image.
T(1,2,−4)
T(1,2,−4)=
To find T(1,2,-4), we can use the linearity of T. Since T is a linear transformation, we have:
T(a1,a2,a3) = a1T(1,0,0) + a2T(0,1,0) + a3T(0,0,1)
= a1(2,-1,4) + a2(1,-2,3) + a3(2,-2,0)
= (2a1 + a2 + 2a3, -a1 - 2a2 - 2a3, 4a1 + 3a2)
In this case, a1 = 1, a2 = 2, a3 = -4. Plugging these values into the expression above, we get:
T(1,2,−4) = (2(1) + 2(2) + 2(-4), -(1) - 2(2) - 2(-4), 4(1) + 3(2))
= (2 + 4 - 8, -1 - 4 + 8, 4 + 6)
= (-2, 3, 10)
Therefore, T(1,2,-4) = (-2, 3, 10).
To find the image of T(1,2,-4), we need to find the linear combination of the given basis vectors (1,0,0), (0,1,0), and (0,0,1) using the corresponding images.
The image of T(1,2,-4) can be found by multiplying the basis vectors by their respective images and summing the results.
T(1,2,-4) = (1)(T(1,0,0)) + (2)(T(0,1,0)) + (-4)(T(0,0,1))
Substituting the given images, we have:
T(1,2,-4) = (1)(2,-1,4) + (2)(1,-2,3) + (-4)(2,-2,0)
Simplifying the scalar multiplication and addition, we get:
T(1,2,-4) = (2,-1,4) + (2,-4,6) + (-8,8,0)
Adding the corresponding components, we have:
T(1,2,-4) = (2+2-8, -1-4+8, 4+6+0)
Simplifying further, we get:
T(1,2,-4) = (-4, 3, 10)
Therefore, T(1,2,-4) = (-4, 3, 10).