Mathematics: In this exercise, t: r2 → r2 is a function. for each of the following parts, state why t is not linear. (a) t(a1, a2)= (1, a2) (b) t(a, a2) (a1,ai) (c) t(a1,a2)= (sin a1,0) (d) t(al, aa) = (lal, a2) (e) t(a1, d12) = (a1 + 1, a2)

169.15 is the correct answer...

In this exercise, t: r2 → r2 is a function. for

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Demondevilg

13.01.2022

Mathematics

a) T is not homogeneousb) T is not additivec) T is not homogeneousd) T is not additivee) T is not additive

Step-by-step explanation:

In each case there is an example where one property of linear functions fails.

T(2(1,0))=T((2,0))=(1,0); 2T((1,0))=2(1,0)=(2,0)

. These vectors are not equal, then T doesn't satisfy the condition of scalar multiplication (homogeneity).b) T((1,2)+(2,3))=T(3,5)=(3,9); T((1,2))+T((2,3))=(1,1)+(2,4)=(3,5)

. Because these vectors are not equal, T doesn't satisfy the property of vector addition (additivity).c) $T(\frac{1}{2}(\pi,0))=T((\frac{\pi}{2},0))=(\sin(\frac{\pi}{2}),0)=(1,0); \frac{1}{2}T((\pi,0))=\frac{1}{2}(0,0)=(0,0)$ so T is not homogeneous.d) T((-1,0)+(1,0))=T(0,0)=(0,0); T((-1,0))+T((1,0))=(1,0)+(1,0)=(2,0)

T((-1,0)+(1,0))=T(0,0)=(0,0); T((-1,0))+T((1,0))=(1,0)+(1,0)=(2,0)

then T is not additive.e) T((0,0)+(1,0))=T(1,0)=(2,0); T((0,0))+T((1,0))=(1,0)+(2,0)=(3,0)

then T is not additive.

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In this exercise, t: r2 → r2 is a function. for each of the following parts, state why t is not linear. (a) t(a1, a2)= (1, a2) (b) t(a, a2) (a1,ai) (c) t(a1,a2)= (sin a1,0) (d) t(al, aa) = (lal, a2) (e) t(a1, d12) = (a1 + 1, a2)

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