0
$\begingroup$

Can someone confirm whether I am correct or not for these two problems? (sorry if the format is hard to understand. Im not sure how to do everything here yet)

A) let A be the (3x2) matrix to where A=\begin{bmatrix}2&3\\1&1\\0&2\end{bmatrix} Recall that the function TA:R^2 -> R^3

Defined by TA(x)=A(dot)x is a linear transformation. Find the formula for T([x1,x2])

would I just take the matrix \begin{bmatrix}2&3\\1&1\\0&2\end{bmatrix} and multiply it accross\begin{bmatrix}x1\\x2\end{bmatrix} Thus getting this as the fomula for T([x1,x2])? \begin{bmatrix}2x1&3x2\\1x1&1x2\\0&2x2\end{bmatrix}


B) Another related question.

Let T:R^4->R^3 be a linear transformation defined by

T(x) = \begin{bmatrix}2x1-3x2\\5x1+6x3\\7x1-8x4\end{bmatrix}

where x= \begin{bmatrix}x1\\x2\\x3\\x4\end{bmatrix}

Find a matrix A such that T(x)=A(dot)x. Then verify the equality T(x)=A(dot)x

As an answer, would A = \begin{bmatrix}2&-3&0&0\\5&0&6&0\\7&0&0&-8\end{bmatrix}

If my A is correct, when i try to verify the equality T(x)=A(dot)x would i just get \begin{bmatrix}2x1-3x2\\5x1+6x3\\7x1-8x4\end{bmatrix} again? Thanks in advance! It has been a while since my professor covered this material.

$\endgroup$
  • $\begingroup$ B is fine. For A, you did the matrix multiplication incorrectly; you should end up with a $3 \times 1$ vector. $\endgroup$ – angryavian Apr 26 '17 at 22:56
  • $\begingroup$ Thankyou! would it be this? \begin{bmatrix}2x1+3x2\\x1+x3\\2x2\end{bmatrix} $\endgroup$ – Ash Pokemon Apr 26 '17 at 23:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.