leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

Nailgun · Mar 20, 2016

Rathin said:
Can you tell me how do you do it?

Was bored

${ x }^{ 2 }-18x+{ y }^{ 2 }+20y=-60\\ ({ x }^{ 2 }-18x+81)+({ y }^{ 2 }+20y+100)=-60+181\\ (x-9)^{ 2 }+{ (y+10) }^{ 2 }={ 11 }^{ 2 }\\ \therefore \quad C(9,-10)\quad and\quad r=11$

Paradoxica · Mar 20, 2016

Nailgun said:
Was bored

${ x }^{ 2 }-18x+{ y }^{ 2 }+20y=-60\\ ({ x }^{ 2 }-18x+81)+({ y }^{ 2 }+20y+100)=-60+181\\ (x-9)^{ 2 }+{ (y+10) }^{ 2 }={ 11 }^{ 2 }\\ \therefore \quad C(9,-10)\quad and\quad r=11$

If you're that bored, go finish my question on the 2U marathon.

Nailgun · Mar 20, 2016

Paradoxica said:
If you're that bored, go finish my question on the 2U marathon.

I've tried it so many times lelel
I always end up with pronumerals in the ratio rather than just 4/3

Paradoxica · Mar 20, 2016

Nailgun said:
I've tried it so many times lelel
I always end up with pronumerals in the ratio rather than just 4/3

I did suggest the use of Simpson's Rule. Try that.

Nailgun · Mar 20, 2016

Paradoxica said:
I did suggest the use of Simpson's Rule. Try that.

That's what I was using lel

btw was the triangle area correct?

it almost cancels out perfectly to 4/3 so i think ive made a mistake somewhere

Paradoxica · Mar 20, 2016

Nailgun said:
That's what I was using lel

btw was the triangle area correct?

it almost cancels out perfectly to 4/3 so i think ive made a mistake somewhere

Pretty sure the area of the triangle has 4, not 8.

This can be observed by assuming the result desired is true.

leehuan · Mar 21, 2016

$\\$Find conditions on the coefficients ${a}_{11},{a}_{12},{a}_{21},{a}_{22},{b}_{1},{b}_{2}$ so that the system of equations$ \\ \begin{align*}{a}_{11}{x}_{1}+{a}_{12}{x}_{2}&={b}_{1}\\{a}_{21}{x}_{1}+{a}_{22}{x}_{2}&={b}_{2}\end{align*}\\ $has a unique solution, with the added condition ${a}_{11}\neq0$

$\\$Answer: For ${a}_{11}\neq 0$, if ${a}_{11}{a}_{22}-{a}_{12}{a}_{21}\neq 0$, then the solution is unique.$$

I do not compute. How did they derive this?

InteGrand · Mar 21, 2016

leehuan said:
$\\$Find conditions on the coefficients ${a}_{11},{a}_{12},{a}_{21},{a}_{22},{b}_{1},{b}_{2}$ so that the system of equations$ \\ \begin{align*}{a}_{11}{x}_{1}+{a}_{12}{x}_{2}&={b}_{1}\\{a}_{21}{x}_{1}+{a}_{22}{x}_{2}&={b}_{2}\end{align*}\\ $has a unique solution, with the added condition ${a}_{11}\neq0$

$\\$Answer: For ${a}_{11}\neq 0$, if ${a}_{11}{a}_{22}-{a}_{12}{a}{21}\neq 0$, then the solution is unique.$$

I do not compute. How?

$\noindent Call the first equation (1) and the second one (2). From (1), since $a_{11}\neq 0$, we have $x_1 = \frac{b_1 - a_{12}x_2}{a_{11}}$. Now, sub. this in (2):$

$\noindent $a_{21}\cdot \frac{b_1 - a_{12}x_2}{a_{11}} + a_{22}x_2 = b_2$.$

$\noindent Multiply by $a_{11}\neq 0$: $a_{21}b_1 - a_{21}a_{12}x_2 + a_{22}a_{11}x_2 = b_2 a_{11}\Longleftrightarrow \left(a_{11}a_{22}-a_{12}a_{21}\right) x_2 = b_2 a_{11}-a_{21}b_1$.$

$\noindent Writing $\Delta \equiv a_{11}a_{22}-a_{12}a_{21}$, we have $\Delta x_2 = b_2 a_{11}-a_{21}b_1$. This has a unique solution iff $\Delta \neq 0$ $\Big{(}$since then $x_2 = \frac{b_2 a_{11}-a_{21}b_1}{\Delta}$$\Big{)}$. If $\Delta = 0$, we do not have a unique solution, because we have the equation $0x_2 =b_2 a_{11}-a_{21}b_1$, which has no solution if $b_2 a_{11}-a_{21}b_1 \neq 0$, and infinitely many solutions if $b_2 a_{11}-a_{21}b_1 = 0$ (so either way not a unique solution). Since all our steps in solving were reversible, it follows that the given system has a unique solution iff $\Delta \neq 0$.$

$\noindent \textbf{Remark.} The quantity $\Delta = a_{11}a_{22}-a_{12}a_{21}$ is called the \textit{determinant} of the system (or matrix of coefficients). It is in fact true that the system will have a unique solution iff $\Delta \neq 0$ even if $a_{11}=0$.$

leehuan · Mar 22, 2016

$$Find and geometrically describe the solutions for the following systems of linear equations$ \\ $ c) $ \\ \begin{align*}4{x}_{1}+5{x}_{2}-2{x}_{3}&=16\\8{x}_{1}+10{x}_{2}-4{x}_{3}&=32\end{align*}$

The question is obviously trivial

$$My answer:$ \\ $By inspection, multiplying the first plane's equation by 2 gives the exact same equation as for the second plane. Hence, they are the same plane and the solutions are$ \\ {x}_{1},{x}_{2},{x}_{3}\in\mathbb R$

$Given answer:$ \left\{ \left( \begin{matrix} 4-\frac { 5 }{ 2 } \lambda +\frac { 1 }{ 2 } \mu \\ \lambda \\ \mu \end{matrix} \right) :\lambda ,\mu \in \mathbb R \right\}, $Equations represent the same plane.$

Obviously I can tell how they got their answer. But would there be any reason as to mine being incorrect here and do I have to give my answer the same way they did?

InteGrand · Mar 22, 2016

leehuan said:
$$Find and geometrically describe the solutions for the following systems of linear equations$ \\ $ c) $ \\ \begin{align*}4{x}_{1}+5{x}_{2}-2{x}_{3}&=16\\8{x}_{1}+10{x}_{2}-4{x}_{3}&=32\end{align*}$

The question is obviously trivial

$$My answer:$ \\ $By inspection, multiplying the first plane's equation by 2 gives the exact same equation as for the second plane. Hence, they are the same plane and the solutions are$ \\ {x}_{1},{x}_{2},{x}_{3}\in\mathbb R$

$Given answer:$ \left\{ \left( \begin{matrix} 4-\frac { 5 }{ 2 } \lambda +\frac { 1 }{ 2 } \mu \\ \lambda \\ \mu \end{matrix} \right) :\lambda ,\mu \in \mathbb R \right\}, $Equations represent the same plane.$

Obviously I can tell how they got their answer. But would there be any reason as to mine being incorrect here and do I have to give my answer the same way they did?

$\noindent The solutions aren't just $x_1,x_2,x_3\in \mathbb{R}$. That would be saying that ANY triple $\left(x_1,x_2,x_3\right)$ would satisfy those equations, but that's not the case (instead, we need to choose triples from the plane described by those equations. This is what the given answer did.).$

leehuan · Mar 22, 2016

Oh of course. Thanks.

(This is why nobody should do maths whilst having a headache)

leehuan · Mar 23, 2016

The entire question:
$$For which values of $\lambda$ do the equations$ \\ \begin{align*}x+2y+\lambda z&=1\\-x+\lambda y-z&=0\\\lambda x-4y+\lambda z&=-1\end{align*} \\ $have a) no solutions, b) infinitely many solutions, c) a unique solution?$

Given answers:
$$a) $\lambda=\pm2$, b) $\lambda=1$, c) all other values of $\lambda$

Now, my working thus far has everything besides lambda=-2 at this point.

$$Rewrite as the augumented matrix $\left[A\left|\textbf{b}\right]$ and converting into row echelon form:$ \\ \left[ \begin{matrix} 1 & 2 & \lambda \\ 0 & \lambda +2 & \lambda +1 \\ 0 & 0 & \left( 1-\lambda \right) \left( \lambda -2 \right) \end{matrix}\left| \begin{matrix} 1 \\ 1 \\ 1-\lambda \end{matrix} \right \right]$

$In ${R}_{3}$, just by inspection, the case of $\lambda=2$ yields a trivial contradiction (meaning no solution) and $\lambda=1$ changes all the terms in the bottom row to 0. When all the terms are 0, it will just so happen that not every variable ${x}_{j}$ becomes a leading variable anymore, thus there are an infinite number of solutions for this case.$

$$Is there, then, a shortcut that will deduce that when $\lambda=-2$ we also have no solution? So far I'm looking at ${a}_{22}$ but I don't know what to do with it because of ${a}_{23}$ also consisting of $\lambda$

seanieg89 · Mar 23, 2016

When lambda=-2, the bottom two rows will give you contradictory expressions for z.

Not sure if you have learned about determinants yet, but they give you a v. quick way of determining when a matrix is invertible (ie when Ax=y has a unique solution). This reduces such problems to checking the isolated values of lambda that make A singular to check whether the resulting systems have no solutions or infinitely many.

Drongoski · Mar 23, 2016

leehuan said:
$$Find and geometrically describe the solutions for the following systems of linear equations$ \\ $ c) $ \\ \begin{align*}4{x}_{1}+5{x}_{2}-2{x}_{3}&=16\\8{x}_{1}+10{x}_{2}-4{x}_{3}&=32\end{align*}$

The question is obviously trivial

$$My answer:$ \\ $By inspection, multiplying the first plane's equation by 2 gives the exact same equation as for the second plane. Hence, they are the same plane and the solutions are$ \\ {x}_{1},{x}_{2},{x}_{3}\in\mathbb R$

$Given answer:$ \left\{ \left( \begin{matrix} 4-\frac { 5 }{ 2 } \lambda +\frac { 1 }{ 2 } \mu \\ \lambda \\ \mu \end{matrix} \right) :\lambda ,\mu \in \mathbb R \right\}, $Equations represent the same plane.$

Obviously I can tell how they got their answer. But would there be any reason as to mine being incorrect here and do I have to give my answer the same way they did?

To add to the discussion:

The 2 equations, i.e. the 2 planes are actually the 1 plane as the 2 planes are identical. So the solution set consists of all the points that lie on the single plane. The given answer is the usual way of specifying this set.

leehuan · Mar 23, 2016

seanieg89 said:
When lambda=-2, the bottom two rows will give you contradictory expressions for z.

Not sure if you have learned about determinants yet, but they give you a v. quick way of determining when a matrix is invertible (ie when Ax=y has a unique solution). This reduces such problems to checking the isolated values of lambda that make A singular to check whether the resulting systems have no solutions or infinitely many.

Yea problem, we haven't. So I wouldn't have been able to spot that lambda=-2 gives two contradictions just by looking at the matrix

InteGrand · Mar 23, 2016

leehuan said:
Yea problem, we haven't. So I wouldn't have been able to spot that lambda=-2 gives two contradictions just by looking at the matrix

Don't need to know about determinants for this. Sub. lambda = -2, then equation 2 (row 2) says that -z = 1, whilst row 3 says that -12z = 3, a contradiction.

leehuan · Mar 23, 2016

InteGrand said:
Don't need to know about determinants for this. Sub. lambda = -2, then equation 2 (row 2) says that -z = 1, whilst row 3 says that -12z = 3, a contradiction.

Yea but I wouldn't have seen that just by staring at my matrix

InteGrand · Mar 23, 2016

leehuan said:
Yea but I wouldn't have seen that just by staring at my matrix

What do you mean? Plug. in lambda = 2 to the matrix you posted. Then interpret the rows as equations as usual.

Green Yoda · Mar 23, 2016

quick question:
for 1/|x-1| > 1/|x+1|
I got the answer as x>0 and x≠1 or x≠-1
but the answer is just x>0 and x≠1
why can x=-1?

InteGrand · Mar 23, 2016

Rathin said:
quick question:
for 1/|x-1| > 1/|x+1|
I got the answer as x>0 and x≠1 or x≠-1
but the answer is just x>0 and x≠1
why can x=-1?

Because then the denominator of the RHS would be 0, and we can't divide by 0.

Edit: oh you said why *can*. Well, it can't be -1 in fact, as you had.

leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

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