Matrices problem

Question:
The transpose of a matrix $M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is the matrix $M^{T}=\begin{pmatrix} a & c \\ b & d \end{pmatrix}$ and M is said to be orthogonal when $M^{T}M=I$, where I is the unit matrix. Given that the matrix
$N=\begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix}$
is orthogonal, find the value of k. Describe geometrically the transformation of x-y plane which is represented by N. Under a transformation S of the real plane into a itself, a point $P = (x,y)$ is mapped onto the point $S(P) = (ax+by, cx+dy)$. Show that when M is orthogonal, the distance between any two points P and Q is the same as the distance between their images S(P) and S(Q).
My attempt:

$N=\begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix}$
$N^{T}=\begin{pmatrix} \frac{2}{\sqrt{5}} &-\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & k \end{pmatrix}$

$N^{T}N=I$
$\begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix}\begin{pmatrix} \frac{2}{\sqrt{5}} &-\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & k \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$
$\begin{pmatrix} 5 &\frac{2}{5}- \frac{k}{\sqrt{5}} \\ \frac{2}{5}- \frac{k}{\sqrt{5}} & k^{2} - \frac{1}{\sqrt{5}} \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
$\frac{2}{5}- \frac{k}{\sqrt{5}} =0$
$k = \frac{2\sqrt{5}}{5}$

The transformation represented by N is a rotation by angle 26.6 (degrees) clockwise

I am having difficulty answering the last part of the question
Please I would like some help with the last part of the question

I noticed that $\begin{pmatrix}[br]a & b\\ c & d\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}[br]ax+by\\ [br]cx+dy[br]\end{pmatrix}$
but im dont know what to do from here

It's similar to your last point, but note that if c1 and c2 are the columns of P
P = x*c1 + y*c2
Similar for Q. Take their difference and calc the distance, noting the columns are orthogonal and unit length.

Working with the span of the columns is fairly common. Once you've got it sorted working with column vectors, see if you can write it directly with vectors and matrices and an inner product. Should just be 2, or 3 lines.

(Original post by mqb2766)

this is what i have done so far
i dont really understand what you wrote


$S(Q) = (ax+cy, bx+dy)$
$S(Q) = \begin{pmatrix}[br]a &c\\ b &d \end{pmatrix}\begin{pmatrix} x\\ y\end{pmatrix} = \begin{pmatrix}[br]ax+cy\\ [br]bx+dy[br]\end{pmatrix}$

the coordinates of $S(P) = (ax+by, cx+dy)$ and $S(Q) = (ax+cy, bx+dy)$
the matrix of S(P) and S(Q) : $\begin{pmatrix}ax+by & ax+cy\\ cx+dy & bx+dy \end{pmatrix}$
this is where i am stuck
can you explain a bit more I am finding it difficult.

thank you

As it stands, what I'm going to post isn't really valid, because it assumes (the standard result) that $(AB)^T = B^TA^T$ (*) for arbitrary matrices A B (such that the product AB makes sense) - from the question wording as posted, I don't think you should assume that result (although if it's in the formula booklet, you probably would get away with it).

However, the general concept is important enough that I think it's still worth making the post.

First, note that for a general vector x, we have $|{\bf x}|^2 = {\bf x} \cdot {\bf x} = {\bf x}^T {\bf x}$, where for the last part we are treating x as a 3x1 matrix.

Then $|S{\bf x}|^2 = (S{\bf x})^T(S{\bf x}) = {\bf x}^T S^T S {\bf x}$, where we have applied (*) to $(S{\bf x})^T$,

Since S is orthogonal $S^T S = I$, and so ${\bf x}^T S^T S {\bf x} = {\bf x}^T {\bf x} = |{\bf x}|^2$, and so $|S{\bf x}| = |{\bf x}|$ (**)

Finally, note that $S{\bf x} - S{\bf y} = S({\bf x} - {\bf y})$, so (**) also shows $|S{\bf x} - S{\bf y}| = |{\bf x} - {\bf y}|$ as desired.

Arguably, this is more a post-A-level approach than an A-level one, but I think the general idea of manipulating transposes is something worth being aware of at A-level, (and also the idea that ${\bf x} \cdot {\bf x} = {\bf x}^T {\bf x}$)).

You're not really thinking about it the right way. Also points P and Q will have different x and y values which You need to represent.

First write down the distance between P and Q in vector form.
Then think about how you do the same for the distance between the transformed points.

Post what you've done/where you get stuck?

here is what i have done so far

we know that $P= (x,y)$ and Q will have different x and y values but these values are unknown to us so
we will let $P= (x_{1}, y_{1})$ and $Q = (x_{2}, y_{2})$
the vector form of P and Q;
$\vec{OP} = \begin{pmatrix}x_{1}\\ y_{1}\end{pmatrix}$ and $\vec{OQ} = \begin{pmatrix}x_{2}\\ y_{2}\end{pmatrix}$

$\vec{PQ} =\vec{PO}+\vec{OQ} = \vec{OQ} -\vec{OP} = \begin{pmatrix}x_{2}\\ y_{2}\end{pmatrix} - \begin{pmatrix}x_{1}\\ y_{1}\end{pmatrix}$
$PQ = \left | \vec{PQ} \right | =\sqrt{(x_{2}- x_{1})^{2}+(y_{2}- y_{1})^{2}}$
thus the distance of $PQ = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}- y_{1})^{2}} ...(i)$
the question does not explicitly give the matrix for the transformation S, I assumed that M can be used especially after noticing that

if $S = \begin{pmatrix}a &b \\ c & d\end{pmatrix}$
then
if $S(P) = \begin{pmatrix}a &b \\ c & d\end{pmatrix} \begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}ax+by\\ cx+dy \end{pmatrix}$
but since $P = (x_{1}, y_{1})$ then $S(P) = (ax_{1}+by_{1}, cx_{1}+dy_{1})$

in vector form $\vec{S(P)} = \begin{pmatrix}ax_{1}+by_{1}\\ cx_{1}+dy_{1} \end{pmatrix}$
I am assuming that M can be used for the transofrmation S

$S(Q) = \begin{pmatrix}a &b \\ c & d\end{pmatrix} \begin{pmatrix}x_{2}\\ y_{2}\end{pmatrix} = \begin{pmatrix}ax_{2}+by_{2}\\ cx_{2}+dy_{2} \end{pmatrix}$
thus $S(Q) = (ax_{2}+by_{2},cx_{2}+dy_{2})$
in vector form $\vec{S(Q)} = \begin{pmatrix}ax_{2}+by_{2}\\ cx_{2}+dy_{2} \end{pmatrix}$

$S(P)S(Q)= \left | \overrightarrow{S(P)S(Q)} \right | = \begin{pmatrix}ax_{2}+by_{2}\\ cx_{2}+dy_{2} \end{pmatrix} - \begin{pmatrix}ax_{1}+by_{1}\\ cx_{1}+dy_{1} \end{pmatrix}$
$= \sqrt{((ax_{2}+by_{2})-(ax_{1}+by_{1}))^{2}+((cx_{2}+dy_{2})-(cx_1+dy_{1}))^{2}}$

$= \sqrt{(a(x_{2}-x_{1})+b(y_{2}-y_{1}))^{2}+(c(x_{2}-x_{1})+d(y_{2}-y_{1}))^{2}}$
$=\sqrt{(a^2+c^2)(x_{2}-x_{1})^{2}+2(ab+cd)(x_{2}-x_{1})(y_{2}-y_{1})+(b^2+d^2)(y_{2}-y_{1})^{2}} .....(ii)$

the distance between S(P) and S(Q) is $=\sqrt{(a^2+c^2)(x_{2}-x_{1})^{2}+2(ab+cd)(x_{2}-x_{1})(y_{2}-y_{1})+(b^2+d^2)(y_{2}-y_{1})^{2}} .....(ii)$
the question states that M is orthogonal thus $M^{T}M=I$
then $\begin{pmatrix}a & b\\ c &d \end{pmatrix}=\begin{pmatrix}1 &0 \\ 0& 1\end{pmatrix}$
$a=1 , b =0, c=0 , d= 1$
now I sub $a=1 , b =0, c=0 , d= 1$ into (ii)
I get $=\sqrt{(1^2+0^2)(x_{2}-x_{1})^{2}+2(1(0)+0(1))(x_{2}-x_{1})(y_{2}-y_{1})+(0^2+1^2)(y_{2}-y_{1})^{2}} .....(ii)$
$=\sqrt{(1^2)(x_{2}-x_{1})^{2}+2(0)(x_{2}-x_{1})(y_{2}-y_{1})+(1^2)(y_{2}-y_{1})^{2}} .....(ii) =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} .....(ii)$
since (i) = (ii)
thus when M is orthogonal the distance between any two points P and Q is the same the distance between their images S(P) and S(Q)

For the second half, if try and use the matrix vector notation.
So
M*(P-Q)
Is the transformed difference between points. To calculate the distance squared, take the inner product with itself
(M*(P-Q))'*(M*(P-Q))
Can you remove the transposed operator and evaluate the product?
It should be just a couple of lines?

the matrix vector notation for $M \times (P-Q)$
$[br]M = \begin{pmatrix}a &b \\ c& d\end{pmatrix}[br]P = \begin{pmatrix} x_{1}\\ y_{1} \end{pmatrix}[br]Q = \begin{pmatrix} x_{2}\\ y_{2}\end{pmatrix}[br]M \times (P-Q)=[br]\begin{pmatrix}a &b \\ c& d\end{pmatrix} \begin{pmatrix} x_{1} - x_{2}\\ y_{1} - y_{2} \end{pmatrix} =$
$= \begin{pmatrix} a(x_{1}-x_{2})+b(y_{1}-y_{2})\\ c(x_{1}-x_{2})+d(y_{1}-y_{2}) \end{pmatrix}$

distance $= \sqrt{(a(x_{1}-x_{2})+b(y_{1}-y_{2}))^{2} + c(x_{1}-x_{2})+d(y_{1}-y_{2}))^{2}} = \sqrt{a^{2}(x_{1}-x_{2})^{2}+2ab(x_{1}-x_{2})(y_{1}-y_{2})+b^{2}(y_{1}-y_{2})^{2} + c^{2}(x_{1}-x_{2})+ 2cd(x_{1}-x_{2})(y_{1}-y_{2})+d^{2}(y_{1}-y_{2}))^{2}}$

$distance ^{2} = a^{2}(x_{1}-x_{2})^{2}+2ab(x_{1}-x_{2})(y_{1}-y_{2})+b^{2}(y_{1}-y_{2})^{2} + c^{2}(x_{1}-x_{2})+ 2cd(x_{1}-x_{2})(y_{1}-y_{2})+d^{2}(y_{1}-y_{2})^{2} = (a^2+c^2)(x_{1}-x_{2})^2+ 2(ab+cd)(x_{1}-x_{2})(y_{1}-y_{2})+( b^2+d^2)(y_{1}-y_{2})$

replying to your suggestion of (M*(P-Q))*(M(P-Q)) i think it is
$(M \times (P-Q)) \times (M \times (P-Q)) = \begin{pmatrix} a^{2}(x_{1}-x_{2})^{2}+2ab(x_{1}-x_{2})(y_{1}-y_{2})+b^{2}(y_{1}-y_{2})^{2}\\ c^{2}(x_{1}-x_{2})+ 2cd(x_{1}-x_{2})(y_{1}-y_{2})+ d^{2}(y_{1}-y_{2})^{2} \end{pmatrix}$

I assume that it is the same as $distance^{2}$ but in matrix- vector
notation

but i noticed that
$M = \begin{pmatrix}a &b \\ c& d \end{pmatrix} M^{T} = \begin{pmatrix}a &c\\ b& d \end{pmatrix}$
$M^TM = \begin{pmatrix}a^2+c^2 &ab+cd \\ ab+cd& b^2+d^2 \end{pmatrix}[br]$
now if we multiply it by the matrix of (M(P-Q))

$= \begin{pmatrix} x_{1}-x_{2}\\ y_{1}-y_{2} \end{pmatrix}$

we get


$\begin{pmatrix}a^2+c^2 &ab+cd \\ ab+cd& b^2+d^2 \end{pmatrix} \begin{pmatrix} x_{1}-x_{2}\\ y_{1}-y_{2} \end{pmatrix} = \begin{pmatrix} a^2+c^2(x_{1}-x_{2})+(ab+cd)(y_{1}-y_{2})\\ (ab+cd)(x_{1}-x_{2})+(b^2+d^2)(y_{1}-y_{2}) \end{pmatrix}$


Since $M^TM(M(P-Q)) =\begin{pmatrix}a^2+c^2 &ab+cd \\ ab+cd& b^2+d^2 \end{pmatrix} \begin{pmatrix} x_{1}-x_{2}\\ y_{1}-y_{2} \end{pmatrix}$ is the matrix vector form of $distance ^2$ and $M^TM = I$
so; $I(M(P-Q)) = \begin{pmatrix} 1 & 0\\ 0&1 \end {pmatrix} \begin{pmatrix} x_{1}-x_{2}\\ y_{1}-y_{2} \end{pmatrix}$
[$= \begin{pmatrix} x_{1}-x_{2}\\ y_{1}-y_{2} \end{pmatrix}$

and the distance of $I(M(P-Q)) = \sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})}$
the distance between two images S(P) and S(Q) $= \sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})}$

and the distance of P and Q $= \sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})}$

therefore the distance between any two points P and Q is the same as the distance between their images S(P) and S(Q).
Thanks for the help PROSM is this the right way? I think i followed your instructions Please tell me if I am worng