Ikunupe6v

Answered

2021-12-20

How to determine which of the following transformations are linear transformations?

Determine which of the following transformations are linear transformations

A. The transformation $T}_{1$ defined by ${T}_{1}({x}_{1},{x}_{2},{x}_{3})=({x}_{1},0,{x}_{3})$

B. The transformation $T}_{2$ defined by ${T}_{2}({x}_{1},{x}_{2})=(2{x}_{1}-3{x}_{2},{x}_{1}+4,5{x}_{2})$.

C. The transformation $T}_{3$ defined by ${T}_{3}({x}_{1},{x}_{2},{x}_{3})=({x}_{1},{x}_{2},-{x}_{3})$

D. The transformation $T}_{4$ defined by ${T}_{4}({x}_{1},{x}_{2},{x}_{3})=(1,{x}_{2},{x}_{3})$

E. The transformation $T}_{5$ defined by ${T}_{5}({x}_{1},{x}_{2})=(4{x}_{1}-2{x}_{2},3\left|{x}_{2}\right|)$.

Answer & Explanation

Anzante2m

Expert

2021-12-21Added 34 answers

To test whether T is a linear transformation, you must verify that. for some vectors a and b and some constant c

$T(a+b)=T\left(a\right)+T\left(b\right)$

$T\left(ca\right)=cT\left(a\right)$

$T\left(0\right)=0$

Example:

A. $T({x}_{1},{x}_{2},{x}_{3})=({x}_{1},0.{x}_{3})$

$T({x}_{1}+{y}_{1},{x}_{2}+{y}_{2},{x}_{3}+{y}_{3})=({x}_{1}+{y}_{1},0({x}_{2}+{y}_{2}),{x}_{3}+{y}_{3})=T({x}_{1},0,{x}_{3})+T({y}_{1},0,{y}_{2})$

$T(c{x}_{1},c{x}_{2},c{x}_{3})=T(c{x}_{1},\left(c\right)0,c{x}_{3})=cT({x}_{1},0,{x}_{3})$

$T(0,0,0)=0$

B. $T({x}_{1},{x}_{2})=(2{x}_{1}-3{x}_{2},{x}_{1}+4,5{x}_{2})$

$T({x}_{1}+{y}_{1},{x}_{2}+{y}_{2})=(2({x}_{1}+{y}_{1})-3({x}_{2}+{y}_{2}),({x}_{1}+{y}_{1})+4,5({x}_{2}+{y}_{2}))=(2{x}_{1}+2{y}_{1}-3{x}_{2}-3{y}_{2},{x}_{1}+{y}_{1}+4,5{x}_{2}+5{y}_{2})$

Pademagk71

Expert

2021-12-22Added 34 answers

For T to be a linear transformation, two criteria need to be satisfied:

1.$T(x+y)=T\left(x\right)+T\left(y\right)$

$T\left(ax\right)=aT\left(x\right)$ for a a scalar/constant.

As an example, suppose$x=({x}_{1},{x}_{2},{x}_{3})\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}y=({y}_{1},{y}_{2},{y}_{3})$ . Lets

1.

As an example, suppose

nick1337

Expert

2021-12-28Added 573 answers

Therefore the first transformation is linear as you correctly guessed. Just repeat the same procedure for B-E and see if it works or not.

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