Let
L: R2 �¨ R2
be a linear transformation for which we know that
L(1 1)= [-2 1]
L(-1 1)=[3 2]
a)what is L(-1 5)
b)what is L(u v)?
Quest qe jupoiur
Use the properties of linear transformations:
1. T(A+B)=T(A)+T(B)
2. T(kA)=kT(A)
then
let A=[1,1], B=[-1,1]
and
T(A+B)=T(0,2)=[-2,1]+[3,2]=[1,3]
=>
T(0,1)=[1/2, 3/2]=P
and
T(A-B)=T(2,0)=[-2,1]-[3,2]=[-5,-1]
=>
T(1,0)=[-5/2, -1/2]=Q
So
(a)
L(-1,5)=-P+5Q
=-[1/2,3/2]+5[-5/2,-1/2]
=[-13, -4]
(b)
L(u,v)=uP+vQ=...
To find the values of L(-1 5) and L(u v), we can first express these vectors in terms of the vectors (1 1) and (-1 1) for which we already know the images under the linear transformation L. We can then use the linearity property of L to find the corresponding images.
a) To find L(-1 5):
We can express (-1 5) as a linear combination of (1 1) and (-1 1) as follows:
(-1 5) = (-1)(1 1) + (6)(-1 1)
Since L is a linear transformation, we have:
L(-1 5) = L[(-1)(1 1) + (6)(-1 1)]
By using the linearity property:
L(-1 5) = (-1)L(1 1) + (6)L(-1 1)
Substituting the given values of L(1 1) and L(-1 1):
L(-1 5) = (-1)[-2 1] + (6)[3 2]
Performing the calculations:
L(-1 5) = [2 -1] + [18 12]
L(-1 5) = [20 11]
Therefore, L(-1 5) = [20 11]
b) To find L(u v):
We can express (u v) as a linear combination of (1 1) and (-1 1) as follows:
(u v) = (u)(1 1) + (v)(-1 1)
Again, using the linearity property of L:
L(u v) = L[(u)(1 1) + (v)(-1 1)]
Substituting the given values of L(1 1) and L(-1 1):
L(u v) = (u)L(1 1) + (v)L(-1 1)
L(u v) = (u)[-2 1] + (v)[3 2]
Performing the calculations:
L(u v) = [-2u u] + [3v 2v]
L(u v) = [-2u + 3v, u + 2v]
Therefore, L(u v) = (-2u + 3v, u + 2v)
To find the value of a linear transformation at a given input, we need to use the information about the transformation provided by the given inputs. In this case, we are given the values of L(1 1) and L(-1 1).
a) To find L(-1 5), we can use the information given and apply the linearity property of linear transformations. The linearity property states that for any scalars a and b and vectors u and v, L(a*u + b*v) = a*L(u) + b*L(v).
In this case, we can rewrite the vector -1 5 as (-1*(1 1) + 5*(1 1)). Using the linearity property, we can calculate L(-1 5) as follows:
L(-1 5) = -1 * L(1 1) + 5 * L(1 1)
Substituting the values given in the problem, we get:
L(-1 5) = -1 * [-2 1] + 5 * [-2 1]
Now, we can perform the scalar multiplication and addition:
L(-1 5) = [2 -1] + [-10 5]
L(-1 5) = [-8 4]
Therefore, L(-1 5) = [-8 4].
b) To find L(u v), we can use the same linearity property. We can express the given vector u v as (u*(1 1) + v*(1 1)). Applying the linearity property, we have:
L(u v) = u * L(1 1) + v * L(1 1)
Substituting the values given in the problem, we get:
L(u v) = u * [-2 1] + v * [-2 1]
Performing the scalar multiplication and addition:
L(u v) = [-2u u] + [-2v v]
L(u v) = [-2u - 2v u + v]
Therefore, L(u v) = [-2u - 2v u + v].