Question
This question will diagonalize a matrix and apply this to the transformation of a curve.
Let the matrix $M=\left(\begin{array}{cc}\frac{5}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{5}{2}\end{array}\right)$.
Let $\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)=\mathbf{R}^{-1}$.
Let $\mathbf{R}\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}X \\ Y\end{array}\right)$.
Let $\left(\begin{array}{cc}\frac{1}{\sqrt{3}} & 0 \\ 0 & \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{l}X \\ Y\end{array}\right)=\left(\begin{array}{l}u \\ v\end{array}\right)$.
Hence state the geometrical shape represented by
a. Find the eigenvalues for M. For each eigenvalue find the set of associated eigenvectors.
b. Show that the matrix equation $(x \quad y) \mathbf{M}\left(\begin{array}{l}x \\ y\end{array}\right)=(6)$ is equivalent to the Cartesian equation $\frac{5}{2} x^2+x y+\frac{5}{2} y^2=6$.
c.i.
Show that $\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{array}\right)$ and $\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)$ are unit eigenvectors and that they correspond to different eigenvalues.
c.ii.
Hence, show that $M\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{ll}2 & 0 \\ 0 & 3\end{array}\right)$.
d.i. Find matrix R.
d.ii
Show that $\mathbf{M}=\mathbf{R}^{-1}\left(\begin{array}{ll}2 & 0 \\ 0 & 3\end{array}\right) \mathbf{R}$.
e.i. Verify that $\left(\begin{array}{ll}X & Y\end{array}\right)=\left(\begin{array}{ll}x & y\end{array}\right) \mathbf{R}^{-1}$.
e.ii.Hence, find the Cartesian equation satisfied by $X$ and $Y$.
f. Find the Cartesian equation satisfied by u and v and state the geometric shape that this curve represents.
g. State geometrically what transformation the matrix R represents.
h.i. the curve in $X$ and $Y$ in part (e) (ii), giving a reason.
h.ii.the curve in $x$ and $y$ in part (b).
i. Write down the equations of two lines of symmetry for the curve in x and y in part (b).
▶️Answer/Explanation
a. $\left|\begin{array}{cc}\frac{5}{2}-\lambda & \frac{1}{2} \\ \frac{1}{2} & \frac{5}{2}-\lambda\end{array}\right|=0 \Rightarrow\left(\frac{5}{2}-\lambda\right)^2-\left(\frac{1}{2}\right)^2=0 \Rightarrow \frac{5}{2}-\lambda= \pm \frac{1}{2} \Rightarrow \lambda=2$ or $3 \quad$ M1M1A1A1
$\lambda=2\left(\begin{array}{cc}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{array}\right)\left(\begin{array}{l}p \\ q\end{array}\right)=\left(\begin{array}{l}0 \\ 0\end{array}\right) \Rightarrow q=-p \quad$ eigenvalues are of the form $t\left(\begin{array}{c}1 \\ -1\end{array}\right) \quad$ M1A1 $\lambda=3\left(\begin{array}{cc}-\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{array}\right)\left(\begin{array}{l}p \\ q\end{array}\right)=\left(\begin{array}{l}0 \\ 0\end{array}\right) \Rightarrow q=p \quad$ eigenvalues are of the form $t\left(\begin{array}{l}1 \\ 1\end{array}\right) \quad$ M1A1 [8 marks]
b
$$
\begin{aligned}
& \left(\begin{array}{ll}
x & y
\end{array}\right)\left(\begin{array}{cc}
\frac{5}{2} & \frac{1}{2} \\
\frac{1}{2} & \frac{5}{2}
\end{array}\right)\left(\begin{array}{l}
x \\
y
\end{array}\right)=(6) \Rightarrow\left(\begin{array}{ll}
\frac{5}{2} x+\frac{1}{2} y & \frac{1}{2} x+\frac{5}{2} y
\end{array}\right)\left(\begin{array}{l}
x \\
y
\end{array}\right)=(6) \quad \text { M1A1 } \\
& \Rightarrow\left(\frac{5}{2} x^2+\frac{1}{2} x y+\frac{1}{2} x y+\frac{5}{2} y^2\right)=(6) \Rightarrow \frac{5}{2} x^2+x y+\frac{5}{2} y^2=6 . \quad \text { AG }
\end{aligned}
$$
[2 marks]
c.i. $\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1 \\ -1\end{array}\right)$ corresponding to $\lambda=2, \quad\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ 1\end{array}\right)$ corresponding to $\lambda=3 \quad \boldsymbol{R 1 R 1}$ [2 marks]
c.ii.
$M\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{array}\right)=2\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{array}\right)$ and $M\left(\begin{array}{l}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)=3\left(\begin{array}{l}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right) \Rightarrow M\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{ll}2 & 0 \\ 0 & 3\end{array}\right) \quad \boldsymbol{A 1 A G}$
[1 mark]
d.i.
Determinant is 1. $\mathbf{R}=\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right) \quad$ M1A1
[2 marks]
d.ii. $\mathbf{M R}^{-1}=\mathbf{R}^{-1}\left(\begin{array}{ll}2 & 0 \\ 0 & 3\end{array}\right)$ so post multiplying by $\mathbf{R}$ gives $\mathbf{M}=\mathbf{R}^{-1}\left(\begin{array}{ll}2 & 0 \\ 0 & 3\end{array}\right) \mathbf{R} \quad$ M1AG
[1 mark]
e.i.
$$
\left(\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{array}\right)\left(\begin{array}{l}
x \\
y
\end{array}\right)=\left(\begin{array}{l}
X \\
Y
\end{array}\right) \Rightarrow\left(\begin{array}{l}
\frac{1}{\sqrt{2}} x-\frac{1}{\sqrt{2}} y \\
\frac{1}{\sqrt{2}} x+\frac{1}{\sqrt{2}} y
\end{array}\right)=\left(\begin{array}{l}
X \\
Y
\end{array}\right) \Rightarrow\left(\begin{array}{ll}
X & Y
\end{array}\right)=\left(\begin{array}{ll}
\frac{1}{\sqrt{2}} x-\frac{1}{\sqrt{2}} y & \frac{1}{\sqrt{2}} x+\frac{1}{\sqrt{2}} y
\end{array}\right) \quad \text { M1A1 }
$$
and $\left(\begin{array}{ll}x & y\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)=\left(\begin{array}{ll}\frac{1}{\sqrt{2}} x-\frac{1}{\sqrt{2}} y & \frac{1}{\sqrt{2}} x+\frac{1}{\sqrt{2}} y\end{array}\right)$ completing the proof A1AG
[3 marks]
e.ii.
$$
\left(\begin{array}{ll}
x & y
\end{array}\right) \mathbf{M}\left(\begin{array}{l}
x \\
y
\end{array}\right)=(6) \Rightarrow\left(\begin{array}{ll}
x & y
\end{array}\right) \mathbf{R}^{-1}\left(\begin{array}{ll}
2 & 0 \\
0 & 3
\end{array}\right) \mathbf{R}\left(\begin{array}{l}
x \\
y
\end{array}\right)=(6) \Rightarrow\left(\begin{array}{ll}
X & Y
\end{array}\right)\left(\begin{array}{ll}
2 & 0 \\
0 & 3
\end{array}\right)\left(\begin{array}{l}
X \\
Y
\end{array}\right)=(6)
$$
$$
\Rightarrow\left(2 X^2+3 Y^2\right)=(6) \Rightarrow \frac{X^2}{3}+\frac{Y^2}{2}=1 \quad \text { M1A1 }
$$
[2 marks]
(f) $\frac{X}{\sqrt{3}}=u, \frac{Y}{\sqrt{2}}=v \Rightarrow u^2+v^2=1$, a circle (centre at the origin radius of 1 )
A1A1
[2 marks]
(g)A rotation about the origin through an angle of $45^{\circ}$ anticlockwise. A1A1
[2 marks]
h(i). an ellipse, since the matrix represents a vertical and a horizontal stretch R1A1
[2 marks]
h(ii).an ellipse $\boldsymbol{A 1}$
[1 mark]
(i) $y=x, y=-x \quad$ A1A1
[2 marks]