Question #122109
Qs: 04 Let T(x, y, z, w) = ( 3y -4z, x +2y+4z-w, 7z, -y-w)

Write the standard matrix for T. Check if T is one-to-one and onto.
1
Expert's answer
2020-06-15T15:30:37-0400

T(x,y,z,w)=(3y4z,x+2y+4zw,7z,yw)T(x, y, z, w) = ( 3y -4z, x +2y+4z-w, 7z, -y-w)

Write the standard matrix for T.


A=[0340124100700101]A=\begin{bmatrix} 0 & 3 & -4 & 0\\ 1 & 2 & 4 & -1 \\ 0 & 0 & 7 & 0\\ 0 & -1 & 0 & -1 \end{bmatrix}

One-to-one is the same as onto for square matrices.

In general, a transformation T is both one-to-one and onto if and only if T(x)=bT(\bf x )=\bf bhas exactly one solution for all b\bf b in Rm.\Bbb{R}^m.


[03400124100070001010]\begin{bmatrix} 0 & 3 & -4 & 0 & &0\\ 1 & 2 & 4 & -1 & &0 \\ 0 & 0 & 7 & 0 & &0\\ 0 & -1 & 0 & -1 & &0 \end{bmatrix}

Swap rows 1 and 2:


[12410034000070001010]\begin{bmatrix} 1 & 2 & 4 &-1& &0\\ 0 & 3 & -4 & 0 & &0 \\ 0 & 0 & 7 & 0 & &0\\ 0 & -1 & 0 & -1 & &0 \end{bmatrix}

R2=R2/3R_2={R_2/3}


[12410014/3000070001010]\begin{bmatrix} 1 & 2 & 4 &-1& &0\\ 0 & 1 & -4/3 & 0 & &0 \\ 0 & 0 & 7 & 0 & &0\\ 0 & -1 & 0 & -1 & &0 \end{bmatrix}

R1=R1(2)R2R_1=R_1-(2)R_2


[1020/310014/3000070001010]\begin{bmatrix} 1 & 0 & 20/3 &-1& &0\\ 0 & 1 & -4/3 & 0 & &0 \\ 0 & 0 & 7 & 0 & &0\\ 0 & -1 & 0 & -1 & &0 \end{bmatrix}

R4=R4+R2R_4=R_4+R_2


[1020/310014/30000700004/310]\begin{bmatrix} 1 & 0 & 20/3 &-1& &0\\ 0 & 1 & -4/3 & 0 & &0 \\ 0 & 0 & 7 & 0 & &0\\ 0 & 0 & -4/3 & -1 & &0 \end{bmatrix}

R3=R3/7R_3=R_3/7


[1020/310014/30000100004/310]\begin{bmatrix} 1 & 0 & 20/3 &-1& &0\\ 0 & 1 & -4/3 & 0 & &0 \\ 0 & 0 & 1 & 0 & &0\\ 0 & 0 & -4/3 & -1 & &0 \end{bmatrix}

R1=R1(20/3)R3R_1=R_1-(20/3)R_3


[10010014/30000100004/310]\begin{bmatrix} 1 & 0 & 0 &-1& &0\\ 0 & 1 & -4/3 & 0 & &0 \\ 0 & 0 & 1 & 0 & &0\\ 0 & 0 & -4/3 & -1 & &0 \end{bmatrix}

R2=R2+(4/3)R3R_2=R_2+(4/3)R_3


[100100100000100004/310]\begin{bmatrix} 1 & 0 & 0 &-1& &0\\ 0 & 1 & 0 & 0 & &0 \\ 0 & 0 & 1 & 0 & &0\\ 0 & 0 & -4/3 & -1 & &0 \end{bmatrix}

R4=R4+(4/3)R3R_4=R_4+(4/3)R_3


[10010010000010000010]\begin{bmatrix} 1 & 0 & 0 &-1& &0\\ 0 & 1 & 0 & 0 & &0 \\ 0 & 0 & 1 & 0 & &0\\ 0 & 0 & 0 & -1 & &0 \end{bmatrix}

R1=R1R4R_1=R_1-R_4


[10000010000010000010]\begin{bmatrix} 1 & 0 & 0 & 0& &0\\ 0 & 1 & 0 & 0 & &0 \\ 0 & 0 & 1 & 0 & &0\\ 0 & 0 & 0 & -1 & &0 \end{bmatrix}

R4=R4(1)R_4=R_4\cdot(-1)


[10000010000010000010]\begin{bmatrix} 1 & 0 & 0 & 0& &0\\ 0 & 1 & 0 & 0 & &0 \\ 0 & 0 & 1 & 0 & &0\\ 0 & 0 & 0 & 1 & &0 \end{bmatrix}

The columns of matrix are linearly independent, which happens precisely when the matrix has a pivot position in every column. 

Therefore TT is one-to-one.


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