Question #271400

Given the linear transformation below:

T (x1, x2, x3) → (x1-x2+2x3, 2x1-2x3, -x1-x2+4x3, 3x1-x2)

T = R3 → R4

1. Determine the transformation matrix of the linear transformation above

2. Determine Ker(T) and Range(T)


1
Expert's answer
2021-11-29T13:21:35-0500

1.

Ax=(112202114310)(x1x2x3)=(x1x2+2x32x12x3x1x2+4x33x1x2)Ax=\begin{pmatrix} 1 & -1&2 \\ 2&0 & -2\\ -1&-1&4\\ 3&-1&0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3\\ \end{pmatrix}=\begin{pmatrix} x_1-x_2+2x_3 \\ 2x_1-2x_3 \\ -x_1-x_2+4x_3\\ 3x_1-x_2 \end{pmatrix}


transformation matrix:

A=(112202114310)A=\begin{pmatrix} 1 & -1&2 \\ 2&0 & -2\\ -1&-1&4\\ 3&-1&0 \end{pmatrix}


2.

kernel:

x1x2+2x3=0x_1-x_2+2x_3=0

2x12x3=02x_1-2x_3=0

x1x2+4x3=0-x_1-x_2+4x_3=0

3x1x2=03x_1-x_2=0


x1=x3,x2=3x1x_1=x_3,x_2=3x_1

ker T=span(1,3,1)ker\ T=span(1,3,1)


range:

range T=(x1x2+2x32x12x3x1x2+4x33x1x2)=x1(1213)+x2(1011)+x3(2240)=range\ T=\begin{pmatrix} x_1-x_2+2x_3 \\ 2x_1-2x_3 \\ -x_1-x_2+4x_3\\ 3x_1-x_2 \end{pmatrix}=x_1\begin{pmatrix} 1 \\ 2 \\ -1\\ 3 \end{pmatrix}+x_2\begin{pmatrix} -1 \\ 0 \\ -1\\ -1 \end{pmatrix}+x_3\begin{pmatrix} 2 \\ -2 \\ 4\\ 0 \end{pmatrix}=


=span((1,2,1,3),(1,0,1,1),(2,2,4,0))=span((1,2,-1,3),(-1,0,-1,-1),(2,-2,4,0))


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