Using determinants (1 Viewer)

leehuan · Apr 19, 2016

Already known:

$\det{\left( \begin{matrix} z & 1 & 2 \\ 1 & z & 3 \\ 1 & 1 & z+1 \end{matrix} \right) }=\left(z-1\right)\left({z}^{2}+2z-4\right)$

And hence solve:
zx+y=2
x+zy=3
x+y=z+1

Can't draw the link between the statements. All I get is this:

$\text{By using the exact same calculations:}\\\left( \begin{matrix} z & 1 \\ 1 & z \\ 1 & 1 \end{matrix}\left| \begin{matrix} 2 \\ 3 \\ z \end{matrix} \right) \longrightarrow \left( \begin{matrix} 1 & 1 \\ 0 & z-1 \\ 0 & 0 \end{matrix}\left| \begin{matrix} z+1 \\ 2-z \\ -{ z }^{ 2 }-2z+4 \end{matrix} \right)$

leehuan · Apr 19, 2016

Also for some reason I can't get the right things going on for this one

$$Let $A\left(x_1,y_1\right),B\left(x_2,y_2\right),C\left(x_3,y_3\right)$ be three points in the plane.$\\$Suppose that $A, B$ and $C$ are collinear. Show that$ \\ \det{\left( \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix} \right) }=0$

I got this determinant. Did I make a mistake cause I'm not too sure how the gradient formula ties with it as I end up with something containing in the form x_iy_i when doing gradients.

$$determinant$=\left(x_2y_3-x_3y_2\right)+\left(x_1y_3-x_3y_1\right)+\left(x_1y_2-x_2y_1\right)$

InteGrand · Apr 19, 2016

leehuan said:
Also for some reason I can't get the right things going on for this one

$$Let $A\left(x_1,y_1\right),B\left(x_2,y_2\right),C\left(x_3,y_3\right)$ be three points in the plane.$\\$Suppose that $A, B$ and $C$ are collinear. Show that$ \\ \det{\left( \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix} \right) }=0$

I got this determinant. Did I make a mistake cause I'm not too sure how the gradient formula ties with it as I end up with something containing in the form x_iy_i when doing gradients.

$$determinant$=\left(x_2y_3-x_3y_2\right)+\left(x_1y_3-x_3y_1\right)+\left(x_1y_2-x_2y_1\right)$

$\noindent Since they're collinear, remember that $\overrightarrow{AB}$ is a multiple of $\overrightarrow{AC}$, so this imposes some conditions on the coordinates.$

$\noindent We don't need to expand it either. We can just row-reduce a tiny bit. Doing $R_2 \to R_2 - R_2$ and $R_3 \to R_3 - R_1$, the determinant of that matrix is equal to $\det{\left( \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } -x_1& { y }_{ 2 } -y_1& 0 \\ { x }_{ 3 }-x_1 & { y }_{ 3 }-y_1 & 0 \end{matrix} \right) }$. Now, the second row is a multiple of the third row, because $\overrightarrow{AB}$ is a multiple of $\overrightarrow{AC}$, which tells us that $x_2 - x_1 = \lambda \left(x_3 - x_1\right)$ for some real number $\lambda$, and similarly for the $y$-values with the same $\lambda$. The 0's are also going to satisfy $0 = \lambda \cdot 0$ of course. So row 3 is a multiple of row 2, so the determinant is 0.$

InteGrand · Apr 19, 2016

Incidentally, that determinant is twice the (signed) area of the triangle formed by the three points A, B, C. Knowledge of this determinant formula for triangle areas makes some HSC conics Q's really easy to do (the ones like 'show the area is independent of theta' etc.).

leehuan · Apr 19, 2016

Oh wow that's quite some thing, I did consider vectorial methods and the scalar multiple but I completely missed the benefit of doing a row reduction here. (Also I'm assuming the third column having 1 entries and not 0 for R₂ and R₃ was an accident.)

Ok next part I have no idea where to start. Which trapezia?

"Now suppose that A, B, C are not collinear. By considering the areas of some trapezia (or otherwise), show that the area of the triangle with vertices A,B,C is given by |D| where...

2D = det. (same matrix)"

InteGrand · Apr 19, 2016

leehuan said:
(Also I'm assuming the third column having 1 entries and not 0 for R₂ and R₃ was an accident.)

Oh yes, that was a typo (I copy-pasted your determinant but forgot to change those last entries). I'll fix it up.

InteGrand · Apr 19, 2016

leehuan said:
Oh wow that's quite some thing, I did consider vectorial methods and the scalar multiple but I completely missed the benefit of doing a row reduction here. (Also I'm assuming the third column having 1 entries and not 0 for R₂ and R₃ was an accident.)

Ok next part I have no idea where to start. Which trapezia?

"Now suppose that A, B, C are not collinear. By considering the areas of some trapezia (or otherwise), show that the area of the triangle with vertices A,B,C is given by |D| where...

2D = det. (same matrix)"

OK This one they basically want us to prove what I said about the determinant being (twice the) the triangle area (signed). A proof of this may be found here: https://people.richland.edu/james/lecture/m116/matrices/area.html .

Here's a proof using trapezia: http://mathforum.org/library/drmath/view/55063.html .

KingOfActing · Apr 19, 2016

InteGrand said:
Incidentally, that determinant is twice the (signed) area of the triangle formed by the three points A, B, C. Knowledge of this determinant formula for triangle areas makes some HSC conics Q's really easy to do (the ones like 'show the area is independent of theta' etc.).

That totally reminds me, do you happen to know if the use of matrices is allowed in 4U HSC exams? I couldn't find anything about it online.

InteGrand · Apr 19, 2016

KingOfActing said:
That totally reminds me, do you happen to know if the use of matrices is allowed in 4U HSC exams? I couldn't find anything about it online.

Not really sure. I might have a search later about it. Someone used it last year I remember reading on the 2015 HSC 4U Exam Thoughts thread for a Q. about showing area independent of theta. I'm pretty sure they'll know what you're doing if you use determinants at least, (unlike some school teachers who may not know, like I read on a thread once someone used it in a school exam and got 0). But yeah, not sure if it's allowed.

Edit: Found the post about the person who lost marks in a school assessment for using determinants: http://community.boredofstudies.org...-hsc/285064/scabbing-marks-2.html#post5925818 . (Rereading that, I see that the person actually ended up getting the marks for it, by showing the Maths Head Teacher his Algebra textbook.)

I probably would only keep determinants as a last-resort method in the HSC unless you can get confirmation that it's allowed.

leehuan · Apr 20, 2016

InteGrand said:
I probably would only keep determinants as a last-resort method in the HSC unless you can get confirmation that it's allowed.

Probably tremendously unlikely.

Though I honestly wonder, if some people want to show how capable they are at maths and use beyond HSC stuff why not just have taken the action sooner and transferred to an IB school?

Paradoxica · Apr 20, 2016

leehuan said:
Probably tremendously unlikely.

Though I honestly wonder, if some people want to show how capable they are at maths and use beyond HSC stuff why not just have taken the action sooner and transferred to an IB school?

distance, probably.

RealiseNothing · Apr 20, 2016

KingOfActing said:
That totally reminds me, do you happen to know if the use of matrices is allowed in 4U HSC exams? I couldn't find anything about it online.

You have to stay within the syllabus.

dan964 · Apr 20, 2016

leehuan said:
Already known:

$\det{\left( \begin{matrix} z & 1 & 2 \\ 1 & z & 3 \\ 1 & 1 & z+1 \end{matrix} \right) }=\left(z-1\right)\left({z}^{2}+2z-4\right)$

And hence solve:
zx+y=2
x+zy=3
x+y=z+1

Can't draw the link between the statements. All I get is this:

$\text{By using the exact same calculations:}\\\left( \begin{matrix} z & 1 \\ 1 & z \\ 1 & 1 \end{matrix}\left| \begin{matrix} 2 \\ 3 \\ z \end{matrix} \right) \longrightarrow \left( \begin{matrix} 1 & 1 \\ 0 & z-1 \\ 0 & 0 \end{matrix}\left| \begin{matrix} z+1 \\ 2-z \\ -{ z }^{ 2 }-2z+4 \end{matrix} \right)$

Isn't it z+1 not z in the bottom row?

KingOfActing · Apr 20, 2016

RealiseNothing said:
You have to stay within the syllabus.

InteGrand said:
Not really sure. I might have a search later about it. Someone used it last year I remember reading on the 2015 HSC 4U Exam Thoughts thread for a Q. about showing area independent of theta. I'm pretty sure they'll know what you're doing if you use determinants at least, (unlike some school teachers who may not know, like I read on a thread once someone used it in a school exam and got 0). But yeah, not sure if it's allowed.

Edit: Found the post about the person who lost marks in a school assessment for using determinants: http://community.boredofstudies.org...-hsc/285064/scabbing-marks-2.html#post5925818 . (Rereading that, I see that the person actually ended up getting the marks for it, by showing the Maths Head Teacher his Algebra textbook.)

I probably would only keep determinants as a last-resort method in the HSC unless you can get confirmation that it's allowed.

I messaged my maths teacher last night, and she told me that matrices are allowed in 4U as far as she knows. I'll probably still call someone up at BOSTES to be sure, but it seems like a 'yes'.

Paradoxica · Apr 20, 2016

KingOfActing said:
I messaged my maths teacher last night, and she told me that matrices are allowed in 4U as far as she knows. I'll probably still call someone up at BOSTES to be sure, but it seems like a 'yes'.

Use the determinant to show X is true for scenario Y of conical section Z

leehuan · Apr 20, 2016

dan964 said:
Isn't it z+1 not z in the bottom row?

Yes my bad

RealiseNothing · Apr 20, 2016

KingOfActing said:
I messaged my maths teacher last night, and she told me that matrices are allowed in 4U as far as she knows. I'll probably still call someone up at BOSTES to be sure, but it seems like a 'yes'.

Definitely check with BOSTES because from all accounts I've heard it's a no.

But I did do my HSC 3 years ago so maybe it's changed, who knows.

leehuan · May 24, 2016

So far in the course all they say is if det = 0 then matrix is not invertible.

What does the magnitude of the determinant tell us otherwise?

dan964 · May 24, 2016

leehuan said:
So far in the course all they say is if det = 0 then matrix is not invertible.

What does the magnitude of the determinant tell us otherwise?

The determinant of it basically is the product of all the eigenvalues. (Eigenvalues are useful, because using them we can "diagonalise" matrices which are easily to compute for application purposes). I cannot remember much more about it.

The determinant being zero, means that the matrix is non-invertible, and that if we consider the matrix as representing a space of vectors (it is called the column space R(A)), it tells us that one of the vectors is a multiple of another and is called linearly dependent.

For multivariable calculus, the determinant of the derivative matrix (called the Jacobian) is useful in telling us the existence of a local inverse function for a function in Euclidean space (by what is called the inverse function theorem); and is also useful when integrating by substitution, integrals over regions, lines (contour integrals); involving multiple integrals and variables.

I'll see if I can find an example.

For example:

$\iint_S x^{2}+y \,dx\,dy$$
$where $ S $ is the parallelogram determined by the lines$
$x-y=0, x-y=1, x-2y=0, x-2y=1$
$$ setting $ u=x-y $ and $ v=x-2y$
$$ The new region $ T = \{(u,v): 0 \leq u \leq 1 \wedge 0 \leq v\leq 1\}$
which can be evaluated using the substitution.

The general formula is
$\iint_{g(\Omega)} f $$ = $\iint_g f \circ g \ || det(g') || $$
===
Also two similar matrices have the same determinant.
where A is similar to B if:
$B=X^{-1}AX$

Similar matrices can be used to represent the same linear mapping/operator L on a given vector space V.

InteGrand · May 24, 2016

leehuan said:
So far in the course all they say is if det = 0 then matrix is not invertible.

What does the magnitude of the determinant tell us otherwise?

One geometric interpretation is that it tells us how much a linear transformation scales areas by.

$\noindent For example, if $T$ is the linear map from $\mathbb{R}^{2} \to \mathbb{R}^{2}$, with $T\left(\bold{x}\right) = A\bold{x}$, then $\left |\det \left(A\right)\right |$ is the factor by which a parallelogram in the domain is enlarged when it is transformed by $T$.$

$\noindent If $T$ is from $\mathbb{R}^{n} \to \mathbb{R}^{n}$, with $n\times n$ matrix $A$, then $\left |\det \left(A\right)\right |$ is the factor by which a `hyper-parallelogram' in the domain is enlarged when it is transformed by $T$.$

$\noindent As special cases, you may recall that in the $2\times 2$ case, $\left |\det \left(A\right)\right |$ is the area of the parallelogram spanned by the columns (or rows) of $A$. In the the $3\times 3$ case, $\left |\det \left(A\right)\right |$ is the volume of the \textsl{parallelipiped} spanned by the columns (or rows) of $A$.$

Using determinants (1 Viewer)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

lukewarm mess

Well-Known Member

Well-Known Member

-insert title here-

what is that?It is Cowpea

what

lukewarm mess

-insert title here-

Well-Known Member

what is that?It is Cowpea

Well-Known Member

what

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)