Question

i need help:
Express the simultaneous equations 2x+ky=7 and 4x-9y=4 in matrix form. Given that the above equations have no solutions/ find the possible value of k.

i am also wondering what does it mean when it say "the equations have no solution" ?

Answer 1

Pleas help me :/

Answer 2

yes

Answer 3

You know matrix multiplication?
So....
\[\huge A=\left[\begin{matrix}2 & k \\ 4 & -9\end{matrix}\right]\]
And let's consider this matrix...
\[\huge \left[\begin{matrix}x \\ y\end{matrix}\right]\]

Answer 4

What do you get when you multiply these two?
\[\huge \left[\begin{matrix}2 & k \\ 4 & -9\end{matrix}\right]\left[\begin{matrix}x \\ y\end{matrix}\right]\]

Answer 5

i will get the same equation as above

Answer 6

That's right :)
So it all boils down to this...
\[\huge \left[\begin{matrix}2 & k \\ 4 & -9\end{matrix}\right]\left[\begin{matrix}x \\ y\end{matrix}\right]=\left[\begin{matrix}7 \\ 4\end{matrix}\right]\]
This seem about right?

Answer 7

yup

Answer 8

Okay, so we want this to have no solution, right?
For convenience, let's set
\[\huge z=\left[\begin{matrix}x \\ y\end{matrix}\right] \ \ \ \ ; \ \ \ c=\left[\begin{matrix}7 \\ 4\end{matrix}\right]\]
ok?

Answer 9

okayyy....

Answer 10

So we can rewrite this as...
\[\huge Az=c\]
See how much simpler that looks? :P

Answer 11

yeah..

Answer 12

So.. you now have your matrix form...
Az=c
just replace the matrices as necessary with their actual values.

Now we're interested in when this system has no solution... any idea how to do it? :)

Answer 13

that is exactly what i don't know :s

Answer 14

Okay... forget about those for a bit, and remember matrices... specifically, you know of inverse matrices?

Answer 15

yes i know

Answer 16

So, what if A has an inverse?
Then...
\[\huge Az=c\]\[\huge A^{-1}Az=A^{-1}c\]\[\huge z=A^{-1}c\]
Catch me so far?

Answer 17

yes totally!

Answer 18

So... you realise that IF A has an inverse, this WILL have a solution, right? :D

Answer 19

maybe .. :P what am i suppose to say here? :s

Answer 20

What you're supposed to say... sort of... is that if we don't want this system to have a solution, then A had better not have an inverse :)

So... yeah, what values of k do you need so that the matrix A has no inverse?

Answer 21

i am sorry i didn't get anything :/

Answer 22

Well, let's go back to the matrix A
\[\huge A=\left[\begin{matrix}2 & k \\ 4 & -9\end{matrix}\right]\]
You've heard of determinants?

Answer 23

yes

Answer 24

A neat trick at finding out whether a (square) matrix has an inverse...
It has an inverse IF AND ONLY IF its determinant is not zero.

So... what's the determinant of A?

Answer 25

-18-4k

Answer 26

LOL that's right!
Now, so that the matrix A has no inverse, this determinant must be equal to 0
-18 - 4k = 0
Now solve for k

Answer 27

ahh i get it now so by no solution we mean that A should not have an inverse and that only happens when determinant is equal to 0 right?



i am getting -4 1/2
is that right cause in my book the answer is 4 1/2 ?

Answer 28

I'm sure you're right. Perhaps you overlooked a negative-sign in your book?

Answer 29

no i didn't and that was one of the reasons i posted the question here i thought i was doing wrong -.-

Answer 30

\[\huge A^{-1}=\left[\begin{matrix}\frac14 & \frac18 \\ \frac19 & \frac{-1}{18}\end{matrix}\right]\]

Answer 31

That's what happens if k=4 1/2
A would have an inverse, and this system would have a solution.

Answer 32

You have my support @terenzreignz , all your logic is correct, but the inverse of A is wrong! -_-

Nevermind, it doesn't matter! ^_^

Answer 33

-4 1/2 will give equal slopes and not 4 1/2

Answer 34

Inverting matrices was never my strong point... but I stand by this inverse...
Unless you have a better idea, @shubhamsrg ? :D

Answer 35

yeah i really didn't get how you got this inverse?

Answer 36

1/4 -1/9
-1/8 -1/18


that should be the inverse

Answer 37

I don't think so... Try multiplying it with A

Answer 38

leme try..
hmm

Answer 39

@terenzreignz is right :)

Answer 40

I see where I went wrong, sorry buddy.

Answer 41

Outvoted, @shubhamsrg
@Farheen28 I used an online calculator (read: I cheated)
There are ways to calculate the inverse of matrices, one way in particular that works with 2x2's

Answer 42

hmmm.. :|

Answer 43

I mean, works really well with 2x2 matrices :D
That involves first getting the adjoint of the matrix... know what that is?

Answer 44

anyways thanks a ton @terenzreignz you explained everything brilliantly ! :D
i now know everything clearly!


adjoint is determinant if i am not wrong :/

Answer 45

adjoint?
I am confused with its actual definition, but for 2x2 matrices
\[\large \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\]
Its adjoint is
\[\large \left[\begin{matrix}d & -b \\ -c & a\end{matrix}\right]\]
and to get the inverse of the original 2x2 matrix, divide each element of the adjoint by the determinant.

You see this can't be done if the determinant is zero :D

Answer 46

ah you cleared one more thing!
GENIUS :D

Answer 47

Nope... just lucky :D
And this concludes another episode of "Terence Teaches Matrices" tune in next time, see you around XD

Answer 48

hhahahah i so will tune in again :D