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Give the standard matrix of the linear transformation that first sends (x, y, z) to (y, y, z), and

then rotates this vector 90 degrees counterclockwise about the origin in the x = y plane.

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    $\begingroup$ One and a half years being a member...one would expect you already learned how to properly (1) ask things in this site, and (2) write properly mathematics in this site. $\endgroup$ – DonAntonio Dec 9 '16 at 22:47
  • $\begingroup$ @DonAntonio I appreciate your help... This is all the context I have $\endgroup$ – Jake Mager Dec 9 '16 at 22:57
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Use a composition: T:$R^2$-->$R^2$. T(x,y,z)---->(y,y,z). What is the matrix for a rotation of 90 degrees? Not hard to find. Call it R:$R^2$-->$R^2$ defined by a matrix. (considers sin and cos as a 2x2 matrix). Using these facts, do you think you could come up with a composition involving the transformations T and R to find the desired linear transformation?

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  • $\begingroup$ I understand, but should I be using the identity matrix 3x3 to find this? $$ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ Times $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & cos(-90) & -sin(-90) \\ 0 & -sin(-90) & cos(-90) \\ \end{bmatrix} $$ $\endgroup$ – Jake Mager Dec 9 '16 at 22:54
  • $\begingroup$ Well, if we use this matrix, we get: T(e1,e2,e3) = A (your matrix ^)(e1,e2,e3) $\endgroup$ – wesssg Dec 9 '16 at 22:57
  • $\begingroup$ Your first matrix seems right, I tested it on the standard basis in $R^3$. What are we running into when we send the transformed vector from A to the next matrix B? Is there a problem? The work seems correct $\endgroup$ – wesssg Dec 9 '16 at 22:58
  • $\begingroup$ $$ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{bmatrix} $$ So what you are saying, this would be the final answer? I am not really following "T(e1,e2,e3) = A (your matrix ^)(e1,e2,e3)" $\endgroup$ – Jake Mager Dec 9 '16 at 23:04
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    $\begingroup$ Okay, after calculating this a few different times I think I finally got the right answer and I also gave the wrong rotation because sin(-90) should be positive in column 2 $$ \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{bmatrix} $$ $\endgroup$ – Jake Mager Dec 9 '16 at 23:21

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