P and Q are two linear transformations in the X-Y plane defined by P: (x,...
P and Q are two linear transformations in the X-Y plane defined by
P: (x, y) → (-3x + 6y, 4x + y) and
Q: (x, y) → (2x-3y, -4x - 6y).
(a) Write down the matrices of P and Q. (b) What is the image of (-2,-3) under the transformation Q?
(c) Obtain a single transformation representing the transformation Q followed by P.
(d) Find the image of (1,4) when transformed by Q followed by P.
(e) Find the image P\(^1\) of the point (-√2,2√2) under an anticlockwise rotation of 225° about the origin.
Explanation
P: (x, y) → (2x+3y, 3x - y)
\(\begin{pmatrix} x\\ y \end{pmatrix}\) → \(\begin{pmatrix} -3x & 6y \\ 4x & + y \end{pmatrix}\)
= \(\begin{pmatrix} -3 & 6 \\ 4 & + 1 \end{pmatrix}\) \(\begin{pmatrix} x\\ y \end{pmatrix}\)
P = \(\begin{pmatrix} -3 & 6 \\ 4 & + 1 \end{pmatrix}\)
Q: (x, y) → (2x-3y, -4 - 6y)
\(\begin{pmatrix} x\\ y \end{pmatrix}\) → \(\begin{pmatrix} 2x & -3y \\ -4x & -6y \end{pmatrix}\)
= \(\begin{pmatrix} 2 & -3 \\ -4 & -6 \end{pmatrix}\) \(\begin{pmatrix} x\\ y \end{pmatrix}\)
Q = \(\begin{pmatrix} 2 & -3 \\ -4 & -6 \end{pmatrix}\)
(b) \(\begin{pmatrix} 2 & -3 \\ -4 & -6 \end{pmatrix}\) \(\begin{pmatrix} -2 \\ -3 \end{pmatrix}\)
= \(\begin{pmatrix} 2[-2] & -3[-3] \\ -4[-2] & -6[-3] \end{pmatrix}\)
= \(\begin{pmatrix} -4 & +9 \\ 8 & 18 \end{pmatrix}\)
= \(\begin{pmatrix} 5 \\ 26 \end{pmatrix}\)
: The image of (-2,-3) is (5,26)
(c) PQ = \(\begin{pmatrix} -3 & 6 \\ 4 & 1 \end{pmatrix}\) \(\begin{pmatrix} 2 & -3 \\ -4 & -6 \end{pmatrix}\)
\(\begin{pmatrix} -3[2] + 6[-4] & -3[-3] + 6[-6] \\ 4[2] + 1[-4] & 4[-3] + 1[-6] \end{pmatrix}\)
= \(\begin{pmatrix} -6 - 24 & 9 - 36 \\ 8 - 4 & -12 - 6 \end{pmatrix}\) = \(\begin{pmatrix} -30 & -27 \\ 4 & -18 \end{pmatrix}\)
= PQ: (x,y) → (-30x - 27y, 4x - 18y)
(d) \(\begin{pmatrix} -30 & -27 \\ 4 & -18 \end{pmatrix}\) \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\)
= \(\begin{pmatrix} -30[1] & + [-27][4] \\ 4[1] & + [-18][4] \end{pmatrix}\)
= \(\begin{pmatrix} -30 & -108 \\ 4 & -72 \end{pmatrix}\) = \(\begin{pmatrix} -138 \\ -68 \end{pmatrix}\)
: Image of the point (1,4) is (-138,-68)
(e) T = \(\begin{pmatrix} cos 225° & -sin 225 \\ sin 225° & cos 225° \end{pmatrix}\)
= \(\begin{pmatrix} -√2/2 & √2/2 \\ -√2/2 & -√2/2 \end{pmatrix}\) \(\begin{pmatrix} -√2 \\ 2√2 \end{pmatrix}\)
= \(\begin{pmatrix} 1 & +2 \\ 1 & -2 \end{pmatrix}\) - \(\begin{pmatrix} 3 \\ -1 \end{pmatrix}\)
: The image of P\(^1\) is (3,-1)