Question

1)find matrix such that:

Answer 1

work the systems of equations that are produced ....

Answer 2

-1a -2b = 2
-1c -2d =16


3a +4b = -12
4c +4d = -36


you have 2 sets of systems of equations to play with ... solve each system to determine the values of abcd

Answer 3

or we can combine them into one big 4x4 matrix

Answer 4

a b c d
----------------
-1 0 -2 0 -2
3 0 4 0 12
0 -1 0 -2 -16
0 4 0 4 36


ok, a 4x5 and rref

Answer 5

a=-8
b=3
c=-100
d=42 but how do i put it in the from asked in the question T(v)=[ ________ ](v)?

Answer 6

I'd say \(\left(\begin{matrix}-8&3\\-100&42\end{matrix}\right)\)

Answer 7

i put that answer on webwork but its wrong

Answer 8

@satellite73 any idea? I'm not so familiarized with matrices

Answer 9

answer is \[\left[\begin{matrix}-8 & 3 \\ -4& -6\end{matrix}\right] \] you can solve it just like amistre64 said but just fix the mistake 4c+4d=-36 is suppose to be 3c+4d=-36 =]

Answer 10

Check the answer if you're unsure
\[\left(\begin{matrix}-8 & 3 \\ -100 & 42\end{matrix}\right)\left(\begin{matrix}-1 \\ -2\end{matrix}\right) = \left(\begin{matrix}2 \\ 16\end{matrix}\right)\]\[\left(\begin{matrix}-8 & 3 \\ -100 & 42\end{matrix}\right)\left(\begin{matrix}3 \\ 4\end{matrix}\right) = \left(\begin{matrix}-12 \\ -132\end{matrix}\right)\]The second one isn't correct.

Let's redo this from the beginning.
We have the following problem:
\[T(\left(\begin{matrix}-1 \\ -2\end{matrix}\right)) = \left(\begin{matrix}2 \\ 16\end{matrix}\right)\]\[T(\left(\begin{matrix}3 \\ 4\end{matrix}\right)) = \left(\begin{matrix}-12 \\ -36\end{matrix}\right)\]using some matrix \[T = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)\]
This can be translated into the following problem:
\[\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)\left(\begin{matrix}-1 \\ -2\end{matrix}\right) = \left(\begin{matrix}2 \\ 16\end{matrix}\right) \rightarrow \left(\begin{matrix} -1a -2b \\ -1c-2d\end{matrix}\right) = \left(\begin{matrix}2 \\ 16\end{matrix}\right)\]\[\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)\left(\begin{matrix}3 \\ 4\end{matrix}\right) = \left(\begin{matrix}-12 \\ -36\end{matrix}\right) \rightarrow \left(\begin{matrix} 3a +4b \\ 3c+4d\end{matrix}\right) = \left(\begin{matrix}-12 \\ -36\end{matrix}\right)\]Like amistre64 wrote, we now have a system of four equations to solve:
-1a -2b = 2
-1c -2d = 16

3a +4b = -12
3c +4d = -36

This can be solved by elimination or by using the property \[Ab = x \rightarrow b = A ^{-1}x\]if you have a matrix calculator.

\[A = \left(\begin{matrix}-1 & -2 & 0 & 0 \\ 0 & 0 & -1 & -2\\ 3 & 4 & 0 & 0\\ 0 & 0 & 3 & 4\end{matrix}\right)\]\[x = \left(\begin{matrix}2 \\ 16 \\ -12 \\ -36\end{matrix}\right) \]\[b = \left(\begin{matrix}a \\ b \\ c \\ d\end{matrix}\right)\]\[b = A ^{-1}x = \left(\begin{matrix}-1 & -2 & 0 & 0 \\ 0 & 0 & -1 & -2\\ 3 & 4 & 0 & 0\\ 0 & 0 & 3 & 4\end{matrix}\right)^{-1} \left(\begin{matrix}2 \\ 16 \\ -12 \\ -36\end{matrix}\right) = \left(\begin{matrix} -8 \\ 3 \\ -4 \\ -6\end{matrix}\right) = \left(\begin{matrix}a \\ b \\ c \\ d\end{matrix}\right)\]
\[\rightarrow T = \left[\begin{matrix}-8 & 3 \\ -4 & -6\end{matrix}\right]\]
Now let's check the answer
\[\left(\begin{matrix}-8 & 3 \\ -4 & -6\end{matrix}\right)\left(\begin{matrix}-1 \\ -2\end{matrix}\right) = \left(\begin{matrix}2 \\ 16\end{matrix}\right)\]\[\left(\begin{matrix}-8 & 3 \\ -4 & -6\end{matrix}\right)\left(\begin{matrix}3 \\ 4\end{matrix}\right) = \left(\begin{matrix}-12 \\ -36\end{matrix}\right)\] answer looks valid.

Answer 11

thank you

Answer 12

Yw :)