Find the value of T - linear transformation of matrix (incorrect solution)

[Eagerissac]

So I keep getting 2. ii) wrong below and I'm not sure what I'm doing incorrectly in my steps. I began by placing the values of the tuples in a 4x5 matrix consisting of:

1	0	-2	2	4
2	1	-7	3	3
1	1	-5	2	0
-3	-2	12	-5	-3

Using RREF, I got it simplified to:

1	0	-2	0	2
0	1	-3	0	-4
0	0	0	0	1
0	0	0	1	0

I'm getting stumped because I'm not exactly sure what I'm supposed to do with column 3 since it's a free variable. I assumed I would just ignore the 3rd column as well as the 3rd tuple given in the question. I did the following:

2(0 + 2x + 2x^2 + 1x^3) - 4(2 - x + 2x^2 + 2x^3) + 0(-1 + 0x + (-1)x^2 + 1x^3)

=0 + 4x + 4x^2 + 2x^3 - 8 + 4x - 8x^2 - 8x^3
=-8 + 8x - 4x^2 - 6x^3

but it keeps getting marked as wrong. I tried to just count the free variable/third column and tuple in anyways to see what happens but the new answer I got is also marked wrong. Are there steps that I'm missing or doing wrong completely? I would appreciate some help. I redid the RREF and substitutions multiple times but I keep getting the original answer that I got. I don't think I made a calculation error so I'm assuming there's something I'm not getting.

 

[HallsofIvy]

T is from \(\displaystyle R^4\) to \(\displaystyle R^4\) and we are given the results of T applied to four independent vectors so, yes, we can find the value of T applied to any vector in \(\displaystyle R^4\). One way to find T(4, 3, -3, 0) is to write (4, 3, -3, 0) as a linear combination of those four vectors: find a, b, c d such that
a(1, 2, 1, -3)+ b(0, 1, 1, -2)+ c(-2, -7, -5, 0)+ d(2, 3, 2, -5)= (a- 2c+ 2d, 2a+ b- 7c+ 3d, a+ b- 5c+ 2d, -3a- 2b- 5d)= (4, 3, 0, -3).

So we have four linear equation for a, b, c, and d:
a- 2c+ 2d= 4
2a+ b- 7c+ 3d= 3
a+ b- 5c+ 2d= 0
-3a- 2b- 5d= 0

I notice that there is no "b" in the first equation and no "c" in the fourth equation so I choose to eliminate b and c first.

Subtract the th

 

[Eagerissac]

sorry i'm still confused

 

[HallsofIvy]

Do you not know how to solve a system of equations?

What I was doing got cut off. I had
"So we have four linear equation for a, b, c, and d:
a- 2c+ 2d= 4
2a+ b- 7c+ 3d= 3
a+ b- 5c+ 2d= 0
-3a- 2b- 5d= 0

I notice that there is no "b" in the first equation and no "c" in the fourth equation so I choose to eliminate b and c first. Subtract the third equation from the second to eliminate b: a- 2c+ d= 3. Aha! Look at that first equation, a- 2c+ 2d= 4. Subtracting a- 2c+ d= 3 from that eliminates both a and c to give d= 1.

Putting d= 1 into the equations leaves
a- 2c= 2
2a+ b- 7c= 0
a+ b- 5c= -2
-3a- 2b= 5

If we subtract the third of those equations, a+ b- 5c= -2, from the second, 2a+ /b- 7c= 0, we get a- 2c= 2, the same as the first equation so we can drop one of them. From the first equation, we can write c= a/2- 1. from the last equation we can write b= -(3/2)a- 5/2. Then a+ b- 5c= a- (3/2)a- 5/2- (5/2)a+ 5= -3a+ 5/2= -2 so -3a= -9/2, a= 3/2. b= -(3/2)(3/2)- 5/2= -9/2- 5/2= -7. c= (3/2)/2- 1= 3/4- 1= -1/4. And, remember, d= 1.

 

[Dr.Peterson]

So I keep getting 2. ii) wrong below and I'm not sure what I'm doing incorrectly in my steps. I began by placing the values of the tuples in a 4x5 matrix consisting of:

1	0	-2	2	4
2	1	-7	3	3
1	1	-5	2	0
-3	-2	12	-5	-3

Using RREF, I got it simplified to:

1	0	-2	0	2
0	1	-3	0	-4
0	0	0	0	1
0	0	0	1	0

I'm getting stumped because I'm not exactly sure what I'm supposed to do with column 3 since it's a free variable. I assumed I would just ignore the 3rd column as well as the 3rd tuple given in the question. I did the following:

2(0 + 2x + 2x^2 + 1x^3) - 4(2 - x + 2x^2 + 2x^3) + 0(-1 + 0x + (-1)x^2 + 1x^3)

=0 + 4x + 4x^2 + 2x^3 - 8 + 4x - 8x^2 - 8x^3
=-8 + 8x - 4x^2 - 6x^3

but it keeps getting marked as wrong. I tried to just count the free variable/third column and tuple in anyways to see what happens but the new answer I got is also marked wrong. Are there steps that I'm missing or doing wrong completely? I would appreciate some help. I redid the RREF and substitutions multiple times but I keep getting the original answer that I got. I don't think I made a calculation error so I'm assuming there's something I'm not getting.

If your final matrix were correct, there would be no solution. Check your work.

(But note that Halls wrote the last equation incorrectly.)