So it's one of those 'hence' questions that we cannot use the 'otherwise' method to solve?

In most 'hence or otherwise' questions, the 'hence' method is usually the obvious and straightforward one. It is easier to use the determinant to solve in this case, but not so easy to grasp the concept of *why* or *how* it works.

*wonders aloud how matrices are actually taught in secondary schools... students nowadays are expected to know so much more, sigh*

]]>With the migration of the majority of the Matrices topic from the old A-Maths 4018 Syllabus to the current E-Maths 4016 Syllabus, the only trace of matrices that's left in the current 4038 AMaths syllabus is the part that deals with expressing a pair of linear equations in matrix form and solving the equations by inverse matrix method, which basically covers concepts like inverse matrices, determinants, singular/non-singular matrices etc. - stuff that have been left out of EMaths.

As such, it is likely that one will be told and compelled to 杀 two simultaneous equations 用 inverse matrix method within the question, in order for the examiner to cover this portion of the syllabus in your AMaths paper.

Anyway it's a pretty fast, one-step process to form the quadratic equation for `m` using the determinant right?

**shape** of **location** of **size** of ...

But if so desired, Miss Loi's mouth will be open in the shape of , at the same scale.

And to be honest I've always thought of matrices as a method used to solve large numbers of simultaneous linear equations, via RREF. For 2 equations isn't this a bit 杀鸡用牛刀？

]]>As the roar of the car drew nearer, the Maths Tutor closed her eyes in fear, and another vision appeared to her ... a vision of a series of similar questions that has been appearing in some of her students' tests and exams, asking them to determine if there's an *intersection*, a *solution*, *common solution**, **unique solution* etc. to a pair of linear equations/lines.

Regardless of the phrase used, the scenarios of all kinds of 'solutions' between a pair of simultaneous equations `l`_{1} and `l`_{2} can be summarised in three cases as follows:

In our question, we are looking for the value of `m` such that the two *non-overlapping* lines *don't* meet i.e. in **mathslover**'s words

they must be parallel, and have different y-intercepts

which is the 'No Solutions' (Case 2(a)) described in the diagram above.

To obtain this value of `m`, one may either begin with equating the gradients of both lines as in mathlover's working or via the determinant of the **coefficient matrix** in an associated matrix equation (see below).

No matter which method is used, you'll almost certainly end up with a quadratic equation with *two* values of `m` (one for each scenario (a) & (b) under Case 2) since being parallel only means that there's no *unique* solution (or intersection point) but both equations can still belong to the same line with ∞ solutions (Case 2(b)).

∴ you'll need to substitute the values of `m` obtained into both equations (as shown in the second part of mathlover's working in order to ascertain the final value of that yields two parallel lines with *no common solution* as required by the question.

Note that **clarion**'s substitution approach won't work because we're finding `m` (instead of `x` and `y`) and this will lead to a single equation with two unknowns (`m` and `x`).

As an added note, this kind of question usually forms part of a problem that explicitly requires one to solve a pair of simultaneous linear equations using the inverse matrix method.

With this approach, we can convert our two equations:

`m``x` + (`m` − 1)`y` = 10

(`m` − 2)`x` + 3`m``y` = 20

to matrix form:

In order for the above to have **no unique solution**, the coefficient matrix must be singular i.e. its determinant = 0.

So determinant of

Expanding the above we have our quadratic equation for `m`:

2`m`² + 3`m` − 2 = 0

(`m` − 1/2)(`m` + 2) = 0

`m` = 1/2 or −2

Sub `m`=1/2 into the equations:

Eqn 1: (1/2)`x` − 1/2`y` = 10 ⇒ `y` = `x` − 20

Eqn 2: (−3/2)`x` + 3/2`y` = 20 ⇒ `y` = `x` + 40/3

⇒ two parallel lines with different `y`-intercepts

Sub `m`=−2 into the equations:

Eqn 1: −2`x` − 3`y` = 10

Eqn 2: −4`x` − 6`y` = 20 ⇒ −2`x` − 3`y` = 10

⇒ Eqn 2 is the same as Eqn 1!

∴ `m` = 1/2 if the two lines don't meet (no common solution).

Now to decide what to do with the ridiculous 轻功 and angsana tree that has suddenly sprung up ...

]]>In fact this may well be the Achilles' Heel of the formidable foreign *cyborgs* studying amongst us!

**clarion** + **mathslover**,

This is taking 無厘頭 to the extreme!!!

... on the scale of 1 unit = 1 metre ... she stared with her mouth open in the shape of

x² +y² = 1

So the radius of Miss Loi's wide open mouth is 1 m???

*FACEPALM*

]]>Glancing down, he quickly pictured the road to be positioned along , on the scale of 1 unit = 1 metre.

Miss Loi's car (when she stared with her mouth open in the shape of ) was situated on (-10,10) and the Angsana tree was situated at the origin. The Angsana tree was 2m tall.

If Miss Loi accelerated at a constant rate of 2m/s²,find the time *t* after Miss Loi started her car, that the henchman has to leap off the angsana tree, so that he will land on her car.

Let the acceleration of free fall be 10m/s²...

And then Miss Loi's eyebrows started moving into the shape of y = |x| as she returned to discover two overaged students hijack her question until 体无完肤...

]]>...He leapt up gently onto the branch of a tall Angsana tree by the road with a level of 轻功 that far surpassed any of the characters in the TVB 古装剧 that Miss Loi faithfully followed.

In a flash the henchman had disappeared, leaving Miss Loi's mouth agape in the shape of . What was that? She thought. It was no where near the Seventh Month...

P.S: I realise I've forgotten a lot of my O level Maths, but I haven't forgot my story-spinning skills!

]]>The 2nd step after cross multiplying, when you moved over you forgot to change sign. Also, when you use your equation must be quadratic. Perhaps you have overlooked the fact there is no term?

_________________________________________

Equation 1:

Equation 2:

Both equations are linear. For two linear equations to not cross paths, they must be parallel, and have different y-intercepts.

gradient of equation 1 = gradient of equation 2

When m=-2,

y intercept of equation 1 =

y intercept of equation 2 =

Hence m cannot be -2.

When

y intercept of equation 1 =

y intercept of equation 2 =

Hence .

The speeding roar of the car shocked the henchman to a standstill, he stared for 0.5 seconds at the oncoming car, and did the unthinkable...

]]>y=

From eqn 2,

y=

hence, =

cross multiplying,

If both paths don't meet,

simplifying,... (lazy to type out here)

m less than -1.99? O: i think its wrong leh...