One more matrix problem

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: \(k=\sqrt{2}\)