# 2x2 matrices - reflection & inverse

• Oct 17th 2012, 04:32 AM
froodles01
2x2 matrices - reflection & inverse
Hi
Problem is;
f is reflection in the y-axis and
g is the linear transformation represented by the matrix

2 3
1 1

Find the matrices representing the following linear transformations
i) f
ii) gof
iii) g^-1

now iii) is fine & ii) would be OK if I knew i)

Now, rule for a refletion has the form;
qθ: R2 --> R2
x--> qθ x

where qθ = cos(2θ) sin(2θ)
. . . . . . . sin(2θ) -cos(2θ)

So please, how do I get f from his rule.
• Oct 17th 2012, 05:27 AM
Deveno
Re: 2x2 matrices - reflection & inverse
the reflection through the y-axis, preserves the y-axis, right? this means that the matrix for f sends (0,1) to (0,1) (since (0,1) lies on the y-axis). so all we need to know is what f does to (1,0), and we're done (almost).

but clearly, f sends (1,0) to it's "reflection", (1,0) in the "opposite direction", which is -(1,0) = (-1,0).

so let's say the matrix for f is:

$\begin{bmatrix}a&b\\c&d \end{bmatrix}$.

then:

$\begin{bmatrix}a&b\\c&d \end{bmatrix} \begin{bmatrix}1\\0 \end{bmatrix} = \begin{bmatrix}-1\\0 \end{bmatrix}$

and actually doing the matrix multiplication on the LHS gives us a = -1, c= 0.

similarly, we see that:

$\begin{bmatrix}a&b\\c&d \end{bmatrix} \begin{bmatrix}0\\1 \end{bmatrix} = \begin{bmatrix}0\\1 \end{bmatrix}$

so b = 0, d = 1.

thus the matrix for f is:

$\begin{bmatrix}-1&0\\0&1 \end{bmatrix}$

which clearly sends the point (x,y) to (-x,y).

how does this relate to your qθ rule? note that the angle the y-axis forms with the x-axis is θ = π/2, so:

cos(2θ) = cos(π) = -1
sin(2θ) = sin(π) = 0

so we get the same matrix as above.