Which of the following options specifies the default matrix












0












$begingroup$


Which of the following options specifies the default matrix of the linear image $T$ corresponding to counterclockwise rotation $pi/6$ radians followed by mirroring in the $y$-axis?



Options:




  1. $dfrac12begin{bmatrix}-sqrt3&1\1&sqrt3end{bmatrix}$


  2. $dfrac12begin{bmatrix}-sqrt3&-1\-1&sqrt3end{bmatrix}$


  3. $dfrac12begin{bmatrix}1&sqrt3\sqrt3&-1end{bmatrix}$


  4. $dfrac12begin{bmatrix}1&-sqrt3\-sqrt3&-1end{bmatrix}$







matrices linear-transformations







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$endgroup$












  • $begingroup$
    What have you tried? Maybe you can multiply these matrices (in your mind) by a simple representation of a point, and see where it ends up.
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:48










  • $begingroup$
    Should I just multiply 1/2 to the different matrices and then add like in case 1: -sqrt(3)/2 +1/2 and 1/2+sqrt(3)/2 or am I thinking wrong?
    $endgroup$
    – anders
    Dec 20 '18 at 8:54












  • $begingroup$
    Yes, that's how matrix multiplication works. The terms look good because $cos{frac{pi}{6}} = frac{sqrt{3}}{2}$ etc.
    $endgroup$
    – Matti P.
    Dec 20 '18 at 9:53










  • $begingroup$
    could you show me how I should determine which of the options that are correct?
    $endgroup$
    – anders
    Dec 20 '18 at 10:08
















0












$begingroup$


Which of the following options specifies the default matrix of the linear image $T$ corresponding to counterclockwise rotation $pi/6$ radians followed by mirroring in the $y$-axis?



Options:




  1. $dfrac12begin{bmatrix}-sqrt3&1\1&sqrt3end{bmatrix}$


  2. $dfrac12begin{bmatrix}-sqrt3&-1\-1&sqrt3end{bmatrix}$


  3. $dfrac12begin{bmatrix}1&sqrt3\sqrt3&-1end{bmatrix}$


  4. $dfrac12begin{bmatrix}1&-sqrt3\-sqrt3&-1end{bmatrix}$







matrices linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried? Maybe you can multiply these matrices (in your mind) by a simple representation of a point, and see where it ends up.
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:48










  • $begingroup$
    Should I just multiply 1/2 to the different matrices and then add like in case 1: -sqrt(3)/2 +1/2 and 1/2+sqrt(3)/2 or am I thinking wrong?
    $endgroup$
    – anders
    Dec 20 '18 at 8:54












  • $begingroup$
    Yes, that's how matrix multiplication works. The terms look good because $cos{frac{pi}{6}} = frac{sqrt{3}}{2}$ etc.
    $endgroup$
    – Matti P.
    Dec 20 '18 at 9:53










  • $begingroup$
    could you show me how I should determine which of the options that are correct?
    $endgroup$
    – anders
    Dec 20 '18 at 10:08














0












0








0





$begingroup$


Which of the following options specifies the default matrix of the linear image $T$ corresponding to counterclockwise rotation $pi/6$ radians followed by mirroring in the $y$-axis?



Options:




  1. $dfrac12begin{bmatrix}-sqrt3&1\1&sqrt3end{bmatrix}$


  2. $dfrac12begin{bmatrix}-sqrt3&-1\-1&sqrt3end{bmatrix}$


  3. $dfrac12begin{bmatrix}1&sqrt3\sqrt3&-1end{bmatrix}$


  4. $dfrac12begin{bmatrix}1&-sqrt3\-sqrt3&-1end{bmatrix}$







matrices linear-transformations







share|cite|improve this question











$endgroup$




Which of the following options specifies the default matrix of the linear image $T$ corresponding to counterclockwise rotation $pi/6$ radians followed by mirroring in the $y$-axis?



Options:




  1. $dfrac12begin{bmatrix}-sqrt3&1\1&sqrt3end{bmatrix}$


  2. $dfrac12begin{bmatrix}-sqrt3&-1\-1&sqrt3end{bmatrix}$


  3. $dfrac12begin{bmatrix}1&sqrt3\sqrt3&-1end{bmatrix}$


  4. $dfrac12begin{bmatrix}1&-sqrt3\-sqrt3&-1end{bmatrix}$







matrices linear-transformations




matrices linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 8:55









user10354138

7,4322925




7,4322925










asked Dec 20 '18 at 8:46









andersanders

615




615












  • $begingroup$
    What have you tried? Maybe you can multiply these matrices (in your mind) by a simple representation of a point, and see where it ends up.
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:48










  • $begingroup$
    Should I just multiply 1/2 to the different matrices and then add like in case 1: -sqrt(3)/2 +1/2 and 1/2+sqrt(3)/2 or am I thinking wrong?
    $endgroup$
    – anders
    Dec 20 '18 at 8:54












  • $begingroup$
    Yes, that's how matrix multiplication works. The terms look good because $cos{frac{pi}{6}} = frac{sqrt{3}}{2}$ etc.
    $endgroup$
    – Matti P.
    Dec 20 '18 at 9:53










  • $begingroup$
    could you show me how I should determine which of the options that are correct?
    $endgroup$
    – anders
    Dec 20 '18 at 10:08


















  • $begingroup$
    What have you tried? Maybe you can multiply these matrices (in your mind) by a simple representation of a point, and see where it ends up.
    $endgroup$
    – Matti P.
    Dec 20 '18 at 8:48










  • $begingroup$
    Should I just multiply 1/2 to the different matrices and then add like in case 1: -sqrt(3)/2 +1/2 and 1/2+sqrt(3)/2 or am I thinking wrong?
    $endgroup$
    – anders
    Dec 20 '18 at 8:54












  • $begingroup$
    Yes, that's how matrix multiplication works. The terms look good because $cos{frac{pi}{6}} = frac{sqrt{3}}{2}$ etc.
    $endgroup$
    – Matti P.
    Dec 20 '18 at 9:53










  • $begingroup$
    could you show me how I should determine which of the options that are correct?
    $endgroup$
    – anders
    Dec 20 '18 at 10:08
















$begingroup$
What have you tried? Maybe you can multiply these matrices (in your mind) by a simple representation of a point, and see where it ends up.
$endgroup$
– Matti P.
Dec 20 '18 at 8:48




$begingroup$
What have you tried? Maybe you can multiply these matrices (in your mind) by a simple representation of a point, and see where it ends up.
$endgroup$
– Matti P.
Dec 20 '18 at 8:48












$begingroup$
Should I just multiply 1/2 to the different matrices and then add like in case 1: -sqrt(3)/2 +1/2 and 1/2+sqrt(3)/2 or am I thinking wrong?
$endgroup$
– anders
Dec 20 '18 at 8:54






$begingroup$
Should I just multiply 1/2 to the different matrices and then add like in case 1: -sqrt(3)/2 +1/2 and 1/2+sqrt(3)/2 or am I thinking wrong?
$endgroup$
– anders
Dec 20 '18 at 8:54














$begingroup$
Yes, that's how matrix multiplication works. The terms look good because $cos{frac{pi}{6}} = frac{sqrt{3}}{2}$ etc.
$endgroup$
– Matti P.
Dec 20 '18 at 9:53




$begingroup$
Yes, that's how matrix multiplication works. The terms look good because $cos{frac{pi}{6}} = frac{sqrt{3}}{2}$ etc.
$endgroup$
– Matti P.
Dec 20 '18 at 9:53












$begingroup$
could you show me how I should determine which of the options that are correct?
$endgroup$
– anders
Dec 20 '18 at 10:08




$begingroup$
could you show me how I should determine which of the options that are correct?
$endgroup$
– anders
Dec 20 '18 at 10:08










1 Answer
1






active

oldest

votes


















0












$begingroup$

Counter-clockwise rotation by angle $theta$ in the $xy$-plane is represented by the rotation matrix
$$
left[
begin{array}{cc}
cos{theta} & -sin{theta}\
sin{theta} & cos{theta}
end{array}
right]
$$
In other words, if you have a point or vector, whose (end) point is represented by a column vector, multiplying by this matrix from the left rotates the point or vector around the origin.



The problem definition also mentions mirroring about the line $y=0$. This corresponds to just flipping the sign of the $x$-coordinate, so the representative matrix of this operation is
$$
left[
begin{array}{cc}
-1 & 0\
0 & 1
end{array}
right]
$$
Putting this together, and plugging in $theta=frac{pi}{6}$ results in
$$
left[
begin{array}{cc}
-1 & 0\
0 & 1
end{array}
right]
left[
begin{array}{cc}
cos{theta} & -sin{theta}\
sin{theta} & cos{theta}
end{array}
right]
=
left[
begin{array}{cc}
-cos{theta} & sin{theta}\
sin{theta} & cos{theta}
end{array}
right]
=
left[
begin{array}{cc}
-frac{sqrt{3}}{2} & frac{1}{2}\
frac{1}{2} & frac{sqrt{3}}{2}
end{array}
right]
$$
Therefore, the correct answer is the first option on the list.






share|cite|improve this answer









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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Counter-clockwise rotation by angle $theta$ in the $xy$-plane is represented by the rotation matrix
    $$
    left[
    begin{array}{cc}
    cos{theta} & -sin{theta}\
    sin{theta} & cos{theta}
    end{array}
    right]
    $$
    In other words, if you have a point or vector, whose (end) point is represented by a column vector, multiplying by this matrix from the left rotates the point or vector around the origin.



    The problem definition also mentions mirroring about the line $y=0$. This corresponds to just flipping the sign of the $x$-coordinate, so the representative matrix of this operation is
    $$
    left[
    begin{array}{cc}
    -1 & 0\
    0 & 1
    end{array}
    right]
    $$
    Putting this together, and plugging in $theta=frac{pi}{6}$ results in
    $$
    left[
    begin{array}{cc}
    -1 & 0\
    0 & 1
    end{array}
    right]
    left[
    begin{array}{cc}
    cos{theta} & -sin{theta}\
    sin{theta} & cos{theta}
    end{array}
    right]
    =
    left[
    begin{array}{cc}
    -cos{theta} & sin{theta}\
    sin{theta} & cos{theta}
    end{array}
    right]
    =
    left[
    begin{array}{cc}
    -frac{sqrt{3}}{2} & frac{1}{2}\
    frac{1}{2} & frac{sqrt{3}}{2}
    end{array}
    right]
    $$
    Therefore, the correct answer is the first option on the list.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Counter-clockwise rotation by angle $theta$ in the $xy$-plane is represented by the rotation matrix
      $$
      left[
      begin{array}{cc}
      cos{theta} & -sin{theta}\
      sin{theta} & cos{theta}
      end{array}
      right]
      $$
      In other words, if you have a point or vector, whose (end) point is represented by a column vector, multiplying by this matrix from the left rotates the point or vector around the origin.



      The problem definition also mentions mirroring about the line $y=0$. This corresponds to just flipping the sign of the $x$-coordinate, so the representative matrix of this operation is
      $$
      left[
      begin{array}{cc}
      -1 & 0\
      0 & 1
      end{array}
      right]
      $$
      Putting this together, and plugging in $theta=frac{pi}{6}$ results in
      $$
      left[
      begin{array}{cc}
      -1 & 0\
      0 & 1
      end{array}
      right]
      left[
      begin{array}{cc}
      cos{theta} & -sin{theta}\
      sin{theta} & cos{theta}
      end{array}
      right]
      =
      left[
      begin{array}{cc}
      -cos{theta} & sin{theta}\
      sin{theta} & cos{theta}
      end{array}
      right]
      =
      left[
      begin{array}{cc}
      -frac{sqrt{3}}{2} & frac{1}{2}\
      frac{1}{2} & frac{sqrt{3}}{2}
      end{array}
      right]
      $$
      Therefore, the correct answer is the first option on the list.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Counter-clockwise rotation by angle $theta$ in the $xy$-plane is represented by the rotation matrix
        $$
        left[
        begin{array}{cc}
        cos{theta} & -sin{theta}\
        sin{theta} & cos{theta}
        end{array}
        right]
        $$
        In other words, if you have a point or vector, whose (end) point is represented by a column vector, multiplying by this matrix from the left rotates the point or vector around the origin.



        The problem definition also mentions mirroring about the line $y=0$. This corresponds to just flipping the sign of the $x$-coordinate, so the representative matrix of this operation is
        $$
        left[
        begin{array}{cc}
        -1 & 0\
        0 & 1
        end{array}
        right]
        $$
        Putting this together, and plugging in $theta=frac{pi}{6}$ results in
        $$
        left[
        begin{array}{cc}
        -1 & 0\
        0 & 1
        end{array}
        right]
        left[
        begin{array}{cc}
        cos{theta} & -sin{theta}\
        sin{theta} & cos{theta}
        end{array}
        right]
        =
        left[
        begin{array}{cc}
        -cos{theta} & sin{theta}\
        sin{theta} & cos{theta}
        end{array}
        right]
        =
        left[
        begin{array}{cc}
        -frac{sqrt{3}}{2} & frac{1}{2}\
        frac{1}{2} & frac{sqrt{3}}{2}
        end{array}
        right]
        $$
        Therefore, the correct answer is the first option on the list.






        share|cite|improve this answer









        $endgroup$



        Counter-clockwise rotation by angle $theta$ in the $xy$-plane is represented by the rotation matrix
        $$
        left[
        begin{array}{cc}
        cos{theta} & -sin{theta}\
        sin{theta} & cos{theta}
        end{array}
        right]
        $$
        In other words, if you have a point or vector, whose (end) point is represented by a column vector, multiplying by this matrix from the left rotates the point or vector around the origin.



        The problem definition also mentions mirroring about the line $y=0$. This corresponds to just flipping the sign of the $x$-coordinate, so the representative matrix of this operation is
        $$
        left[
        begin{array}{cc}
        -1 & 0\
        0 & 1
        end{array}
        right]
        $$
        Putting this together, and plugging in $theta=frac{pi}{6}$ results in
        $$
        left[
        begin{array}{cc}
        -1 & 0\
        0 & 1
        end{array}
        right]
        left[
        begin{array}{cc}
        cos{theta} & -sin{theta}\
        sin{theta} & cos{theta}
        end{array}
        right]
        =
        left[
        begin{array}{cc}
        -cos{theta} & sin{theta}\
        sin{theta} & cos{theta}
        end{array}
        right]
        =
        left[
        begin{array}{cc}
        -frac{sqrt{3}}{2} & frac{1}{2}\
        frac{1}{2} & frac{sqrt{3}}{2}
        end{array}
        right]
        $$
        Therefore, the correct answer is the first option on the list.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 6:31









        Matti P.Matti P.

        2,0381414




        2,0381414






























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