Question

Find the image of an arbitrary vector . Please see picture for the full question.

Answer 1

Do you know how to compute the inverse of a matrix?

Answer 2

yes, I do.

Answer 3

Then this problem is similar to the previous one. We can find the matrix for T written with respect to the standard basis and then apply it to the vector (x, y)

Answer 4

We will do so as follows.

Let \(A = \left(\begin{array}{cc}
3 & 1 \\
-8 & -3
\end{array}\right)\)
Let \(B = \left(\begin{array}{cc}
19 & 8 \\
51 & 19
\end{array}\right)\)

Then we get the following \(BA^{-1}.(x, y)\)

Answer 5

Let me know what you end up with.

Answer 6

so the inverse of A is \[\left[\begin{matrix}3 & 1 \\ -8 &-3\end{matrix}\right]\]

Answer 7

Yes it is its own inverse.

Answer 8

Now compute \(BA^{-1}\)

Answer 9

that would be \[\left[\begin{matrix}-7 & -5 \\ 1 & -6\end{matrix}\right]\]

Answer 10

Yes, now apply it to the vector \([x, y]\)

Answer 11

\[\left(\begin{matrix}-12x \\ -5y\end{matrix}\right)\]

Answer 12

Not quite, it is \(x\) times the first column, plus \(y\) times the second column.

Answer 13

You are multiplying
\(\left(\begin{array}{cc}
-7 & -5 \\
1 & -6
\end{array}\right)\left(\begin{array}{c}
x \\ y
\end{array}\right)\)

Answer 14

\[\left(\begin{matrix}-7x-5y \\ x-6y\end{matrix}\right)\]

Answer 15

Yes exactly.

Answer 16

Sorry about before, formatted it incorrectly, (x, y) is a column vector.

Answer 17

Anyway, that should be the correct solution.

Answer 18

It is. Thank you for the help tonight.

Answer 19

another way to solve this
assume that T(x,y) is a matrix