T(x) = A(x), find x so T(x) = b; determine if x is unique

[warwick]

(Column vectors)

With T defined as T(x) = Ax, find a vector x whose image under T is b, and determine whether x is unique.

1 0 -2
-2 1 6 = A
3 -2 -5

-1
7 = b
-3

1 0 -2 -1
-2 1 6 7
3 -2 -5 -3

I'm getting x1 = 17/5, x2 = 3/5, x3 = 11/5

 

[mmm4444bot]

Almost ...

Hi Warwick:

\(\displaystyle \left[ \begin{array}{ccc} 1 & 0 & -2 & -2 & 1 & 6 & 3 & -2 & -5 \end{array} \right] \cdot \left[ \begin{array}{c} 17/5 & 3/5 & 11/5 \end{array} \right] \;=\; \left[ \begin{array}{c} -1 & 7 & -2 \end{array} \right]\)

\(\displaystyle \left[ \begin{array}{ccc} 1 & 0 & -2 & -2 & 1 & 6 & 3 & -2 & -5 \end{array} \right] \cdot \left[ \begin{array}{c} 3 & 1 & 2 \end{array} \right] \;=\; \left[ \begin{array}{c} -1 & 7 & -3 \end{array} \right]\)

Your augmented matrix is a good start; since using row operations to obtain a reduced matrix is fairly mechanical, many false results come from arithmetic mistakes. Check over your row operations and arithmetic. If you cannot locate your mistake, then please show your steps so that we can see what you're doing.

Cheers,

~ Mark

 

[warwick]

Re: Not getting the right vector

Well, you did get the right answer. I made several other arithmetical mistakes last night, too. I guess it doesn't help doing homework with only four hours of sleep the morning of. lol. I'll go over it again.