matrices

[crystal star]

yeah so

does the matrix B

and the equations in iii)

anything look similar

@fierce lake

hint: ||look at the coefficients of x,y,z||

[fierce lake]

yes

i see the coeffient match

but what do i do

[crystal star]

mmm yep so

well so you can actually rewrite the sim equations as

[fierce lake]

rewrite it as a matrix?

[crystal star]

yep

ill try draw it

[fierce lake]

okay

[crystal star]

do you see how

if you multiplied this out

[fierce lake]

yep

[crystal star]

you would get the first, second and third equations

so what we can do is

set it equal to

the RHS in vector form

[fierce lake]

i seee

[crystal star]

and now well we want

(x y z) on its own

[fierce lake]

ahh so

B-1 x (1,4,-1)?

[crystal star]

perfect yep

[fierce lake]

B-1 would go on the left of (14-1) right

[crystal star]

it would yeah

cuz you need to left multiply by B^-1 to get rid of B on the lhs

[fierce lake]

okay cool

yep

[crystal star]

easier to visualise

[fierce lake]

yep

[crystal star]

awesome yep

then you just use part b

[fierce lake]

so xyz = 072?

[crystal star]

well

you should be able to plug it back into your equations

and they should work

[fierce lake]

wait no i think im wrong

[crystal star]

maybe double check your part ii)

wolfram alpha says its:

[fierce lake]

yeah wait somethings wrong

oh wait

i used b not b-1 lol

[crystal star]

woopsie lol

[fierce lake]

how would i get rid of the fraction?

[crystal star]

well you know what a is now

and then you can just

[fierce lake]

substitute?

[crystal star]

multiply everything in your final result by the fraction

yep

[fierce lake]

do i use a = -2/3 right

[crystal star]

nope

[fierce lake]

wait why

[crystal star]

so owhat would a need to be

to match your simultaneous equations

'a' corresponds to

the coefficient of x in the first equation

right?

[fierce lake]

yeah so -

[crystal star]

so -1

[fierce lake]

oh

i wouldnt use the stuff from the previous parts?

[crystal star]

you would because

in previous parts youve used a as a varaible

its just now

youre given a value for a

at least in part 2 youve not used a value for a

[fierce lake]

but in the first part we worked out a =-2/3

[crystal star]

in the first part you showed that

B is singular when a = -2/3

[fierce lake]

oh right

[crystal star]

ii) said if B is non-singular

[fierce lake]

im still doing something wrong

[crystal star]

if B is singular then it doesnt have an inverse

[fierce lake]

yep

got that

[crystal star]

so ii) you should have an inverse with the letter a in it

[fierce lake]

yep

[crystal star]

wolfram alpha says it should be this

[fierce lake]

ugh silly mistake i wrote -3 not 3

[crystal star]

rip

[fierce lake]

got the right answer now

[fierce lake]

tysm!