What is the point P(1,4) transformed to by the matrix A?

[TsAmE]

Homework Statement

Find the 2 x 2 matrix A which represents a projection onto the x-axis, followed by a clockwise rotation through pi/4.

Homework Equations

None

The Attempt at a Solution

\begin{pmatrix} cos\frac{\pi}{4}&sin\frac{\pi}{4} \\ -sin\frac{\pi}{4}&cos\frac{\pi}{4}\end{pmatrix} \begin{pmatrix} 1&0 \\ 0&0\end{pmatrix}

\begin{pmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix} \begin{pmatrix} 1&0 \\ 0&0\end{pmatrix}

\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & 0 \\ -\frac{1}{\sqrt{2}} & 0 & 0\end{pmatrix}

but the correct step was \begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & \\ -\frac{1}{\sqrt{2}} & 0 & \end{pmatrix}

I don't understand why the 2nd column of the 2nd matrix hasnt been multiplied with the first matrix's row.

 

[Mark44]

Homework Statement

Find the 2 x 2 matrix A which represents a projection onto the x-axis, followed by a clockwise rotation through pi/4.

Homework Equations

None

The Attempt at a Solution

\begin{pmatrix} cos\frac{\pi}{4}&sin\frac{\pi}{4} \\ -sin\frac{\pi}{4}&cos\frac{\pi}{4}\end{pmatrix} \begin{pmatrix} 1&0 \\ 0&0\end{pmatrix}

\begin{pmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix} \begin{pmatrix} 1&0 \\ 0&0\end{pmatrix}

\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & 0 \\ -\frac{1}{\sqrt{2}} & 0 & 0\end{pmatrix}

Why do you have three columns in the matrix above? The product of two 2 x 2 matrices is a 2 x 2 matrix.

but the correct step was \begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & \\ -\frac{1}{\sqrt{2}} & 0 & \end{pmatrix}

I don't understand why the 2nd column of the 2nd matrix hasnt been multiplied with the first matrix's row.

 

[ZioX]

I am puzzled why you have the product of 2x2 matrices yielding a 2x3 matrix? Perhaps this is the source of your problem?

If not, then notice that the (dot) product of the 2nd column of the 2nd matrix and the 1st row of the 1st matrix will be the (1,2)-entry in the product of the two matrices. Which in your case is 0, as shown in the book.

So I am not really sure what you are asking.

 

[TsAmE]

Nevermind, I found what I did wrong . For the second part of the question it asked: To which point does the matrix A take the point P(1,4)?

Attempt:

\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & \\ -\frac{1}{\sqrt{2}} & 0 & \end{pmatrix} \begin{pmatrix}1 & 0 & \\ 0 &4&\end{pmatrix}

=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 0 & \\ -1 & 0 & \end{pmatrix} which is the same as:

<br /> \begin{pmatrix}\frac{1}{\sqrt{2}} &amp; 0 &amp; \\ -\frac{1}{\sqrt{2}} &amp; 0 &amp; \end{pmatrix}<br />

but I don't get why the answer is (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})

 

[Mark44]

Nevermind, I found what I did wrong . For the second part of the question it asked: To which point does the matrix A take the point P(1,4)?

Attempt:

\begin{pmatrix}\frac{1}{\sqrt{2}} &amp; 0 &amp; \\ -\frac{1}{\sqrt{2}} &amp; 0 &amp; \end{pmatrix} \begin{pmatrix}1 &amp; 0 &amp; \\ 0 &amp;4&amp;\end{pmatrix}

This is wrong. The matrix on the left above is your transformation matrix that projects a vector onto the x-axis and then rotates the projected vector 45 degrees clockwise. You should NOT be multiplying a matrix with that matrix. Instead, multiply the vector(1, 4). (This vector extends from (0, 0) to the point (1, 4).

As a sanity check, what do you get when you project <1, 4> onto the x-axis? What vector do you get when you rotate the projected vector by 45 degrees clockwise?

You should be drawing sketches of these vectors if you aren't doing so already.

=\frac{1}{\sqrt{2}}\begin{pmatrix}1 &amp; 0 &amp; \\ -1 &amp; 0 &amp; \end{pmatrix} which is the same as:

<br /> \begin{pmatrix}\frac{1}{\sqrt{2}} &amp; 0 &amp; \\ -\frac{1}{\sqrt{2}} &amp; 0 &amp; \end{pmatrix}<br />

but I don't get why the answer is (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})