Question

Problem 2 Use the matrix EOMs at the right. Letz 2 a. Approximate B (by matching the FIRST value on the diagonal of B) such that system 3 ตn be decoupled. Also, specify your values for a and B 3 -0.3 b. Approximate B (by matching the SECOND value on 3 6- (3) the diagonal of B) such that system 3 can be decoupled Also, specify your values for α and β. 1.8-0.41 0.4 1A 20-100 4) c. Approximate Bp (by matching the FIRST value on the 10 10 diagonal of B) such that system 4 can be decoupled. Also, specify your values for a and B. d. Approximate B (by matching the SECOND value oa the diagonal of B) such that system 4 can be decoupled. Also, specify your values for a and B.

Answer 1

5) The line spanned by the vector
2
is given by
y=2x

Reflection of a point about the line
ax + by + c = 0
is given by the formula:

1 b22

Substituting we get

5(4 3 , y) - (0,0)

Thus, $(x,y)\mapsto \frac{1}{5}(-3x+4y,4x+3y)$

Thus, the transformation is given by:

$T(x,y)= \frac{1}{5}(-3x+4y,4x+3y)$

Suppose the basis $\mathfrak{B}$ for $\mathbb{R}^2$ is $\left \{ b_1,b_2 \right \}=\left \{ \begin{pmatrix} x_1\\ y_1 \end{pmatrix} ,\begin{pmatrix} x_2\\ y_2 \end{pmatrix}\right \}$

Let $T(b_1)=\begin{pmatrix} a\\ b \end{pmatrix}_\mathfrak{B}$ and$T(b_2)=\begin{pmatrix} c\\ d \end{pmatrix}_\mathfrak{B}$

Then, $T(\overrightarrow{x})=\begin{pmatrix} a & c\\ b & d \end{pmatrix}\overrightarrow{x}$

We want the above matrix to be diagonal. Hence, we must have b=c=0

Therefore, we must have $T(b_1)=\begin{pmatrix} a\\ 0 \end{pmatrix}_\mathfrak{B}$

Hence, $T(b_1)=\frac{1}{5}\begin{pmatrix} -3x_1+4y_1\\ 4x_1+3y_1 \end{pmatrix}=a\begin{pmatrix} x_1\\ y_1 \end{pmatrix}$

Thus, we have

Solving this, we get:

$y_1=\frac{1}{4}(5a+3)x_1,\,x_1=\frac{1}{4}(5a-3)y_1$

Thus, we have $y_1=\frac{(5a+3)(5a-3)}{16}y_1$ which implies $(5a+3)(5a-3)=16$

Therefore, $25a^2=25\Rightarrow a=\pm 1$

Hence, $y_1=2x_1$ so we can take $(x_1,y_1)=(1,2)$

Similarly,

$T(b_2)=\frac{1}{5}\begin{pmatrix} -3x_2+4y_2\\ 4x_2+3y_2 \end{pmatrix}=d\begin{pmatrix} x_2\\ y_2 \end{pmatrix}$ gives us the equations

Thus, $y_2=\frac{(5d+3)}{4}x_2,\,x_2=\frac{(5d-3)}{4}y_2$

Solving this again gives $d=\pm 1$

We take the negative sign as this will gives us a basis. Taking positive sign will give a linearly dependent set $\mathfrak{B}$.

Thus, $y_2=\frac{x_2}{2}$ and so we can take $(x_2,y_2)=(2,1)$

Thus, a basis for which the transformation T has a diagonal matrix with respect to that basis $\mathfrak{B}$ is

$\mathfrak{B}=\left \{ \begin{pmatrix} 1\\ 2 \end{pmatrix},\begin{pmatrix} 2\\ 1 \end{pmatrix} \right \}$

6) Let the basis $\mathfrak{B}=\left \{a,b\right \}$

Then, $\begin{pmatrix} 1\\ 2 \end{pmatrix}_\mathfrak{B}=\begin{pmatrix} 2\\ 1 \end{pmatrix}\Rightarrow (1,2)=2a+b$

And $\begin{pmatrix} 1\\ 1 \end{pmatrix}_\mathfrak{B}=\begin{pmatrix} 0\\ 1 \end{pmatrix}\Rightarrow (1,1)=b$

Hence, $b=(1,1)$ and $2a+(1,1)=(1,2)$

Therefore, $2a=(0,-1)\Rightarrow a=\left ( 0,-\frac{1}{2} \right )$

Thus, the required basis of $\mathbb{R}^2$ is $\mathfrak{B}=\left \{ \left ( 0,-\frac{1}{2} \right ),(1,1) \right \}$