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# One more matrix problem

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Let A be the matrix that rotates about the origin by an angle of $$\theta$$ counter-clockwise, where 0° ≤ $$\theta$$ < 360°, and let B be the matrix that dilates(centered at the origin) by a scale factor of k > 0. If $$BA = \begin{pmatrix} -1&-1\\ 1&-1 \end{pmatrix}$$, then find $$\theta$$ and k.

Some hints would be greatly appreciated :)

Jun 1, 2022

#1
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Hints:

1-Think about the general form of a "Rotation Matrix" (Something to do with cos(theta),sin(theta),-sin(theta),cos(theta), arranged in a 2x2 matrix).

2-A 2x2 matrix that dilates by scale factor of k must have two elements out of the four as "k"  and the other two elements are 0, as it is centered about the origin.

3-BA means multiply matrix A by matrix B, and order matters (B first then A). Use (1) and (2) to find BA.

4-You are given BA, compare each element you got from (3) with the corresponding given element.

5-Look for the desired system of equations.

6-Solve the system (Further hint: Think of tan(theta)=sin(theta)/cos(theta))

7-It seems there exists two answers for theta... Are both valid? Why? Why not?
8-Hint: k is positive!

Solution:

Ok, if A is a rotation matrix then it must be:

$$A=\begin{bmatrix} cos\theta && -sin\theta \\ sin\theta && cos\theta \end{bmatrix}$$

and If B is a matrix that dilates then it must be: $$\begin{bmatrix} k && 0 \\ 0 && k \end{bmatrix}$$

So given: $$BA=\begin{bmatrix} -1 && -1 \\ 1 && -1 \end{bmatrix}$$  $$=$$    $$\begin{bmatrix} k && 0 \\ 0 && k \end{bmatrix}$$ * $$\begin{bmatrix} cos\theta && -sin\theta \\ sin\theta && cos\theta \end{bmatrix}$$

$$\begin{bmatrix} kcos\theta && -ksin\theta \\ ksin\theta && kcos\theta \end{bmatrix}$$

and just comparing each element of the matrix yields a system with sufficient equations to find both k and theta.

$$-1=kcos\theta \text{ And, }1=ksin\theta$$

Divide the latter by the former:

$$\frac{ksin\theta}{kcos\theta}=-1 \iff tan\theta=-1$$, therefore, $$\theta=135,315$$ Now, which one of these is the answer depends on the previous system of equations.

Well, since k is positive, then $$\frac{-1}{cos\theta},\frac{1}{sin\theta}$$ must be positive. If $$\theta=135$$, this is satisfied. But, If $$\theta=315$$ 1/sin(theta) will be negative! Hence k is negative, but we are given k>0. Thus we reject $$\theta=315$$.

Therefore, $$\theta=135$$ only, yielding: $$k=\sqrt{2}$$

Jun 1, 2022
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Ohhhh. This makes sense! I don't really remember how to solve these problems because I missed one class on this topic and didn't fully understand it. Thanks alot!

MobiusLoops  Jun 1, 2022
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anytime!

Guest Jun 2, 2022