uni algebra (matrics) (1 Viewer)

[...]

lol..sorry, another one

Consider the system of linear equations

x₁ - x₂ + x₃ = 1,
2x₁ +2x₂ + x₃ = 5+1,
x₁ + 3x₂ - x₃ = 4+2,
x₁ + 7x₂ - 2x₃ = 10-1

a) reduce its associated augmented matrix to row-echolen form
b) fine a value of
c) solve the system


its too tricky for me brain!

edit: that λ is suppose to be that gamma symbol..lol

 

[KeypadSDM]

2₁

You sure about that?

 

[...]

blah..curse to u modern computing!

 

[Ragerunner]

YOUR SCARING ME.

IM DOING MATRICES SOON!

 

[martin]

x1 - x2 + x3 = 1,
2x1 +2x2 + x3 = 5+1,
x1 + 3x2 - x3 = 4+2,
x1 + 7x2 - 2x3 = 10-1

augmented matrix is (using L for lambda)
(You can imagine a dotted line between 3rd and 4th columns if you want to seperate the coefficents on LHS from constants on RHS)

1, -1, L, 1
2, 2, 1, 5L+1
1, 3, -1, 4L+2
1, 7, -2L, 10L-1

now we can add or subtract multiples of rows so doing Gaussian elimination
R2 -> R2 - 2R1
R3 -> R3 - R1
R4 -> R4 - R1

1, -1, L, 1
0, 4, 1-2L, 5L-1
0, 4, -1-L, 4L+1
0, 8, -3L, 10L-2

then
R3 -> R3 - R2
R4 -> R4 - 2R2

1, -1, L, 1
0, 4, 1-2L, 5L-1
0, 0, -2+L, -L+2
0, 0, -2+L, 0

I think this is row echelon form (reduced row echelon form is when you go all the way back up, right?)


thanks,
Martin

 

[martin]

For part 2 write out system in full
x1 - x2 + Lx3 = 1
4x2+(1-2L)x3 = 5L-1
(-2+L)x3 = -L+2
(-2+L)x3 = 0

so -2+L=0
therefore L=2

so

x1-x2+2x3=1
4x2-3x3=9
0=0
0=0

now this isn't enough info to solve fully but if we just think of x3 as a real parameter ie let x3=t

x2=1/4(9+3t)
x1 = 1+x2-2x3 = 1 + 1/4(9+3t) - 2t
= 13/4 - 5/4 t


so my solution (after about a million mistakes) is
x1 = 13/4 - 5/4 t
x2 = 1/4 (9+3t)
x3 = t
L = 2

we'll check say t=1 so
x1 = 2, x2 = 3, x3 = 1, L=2

this satisfies all the equations so it seems right.

thanks,
Martin

 

[...]

Originally posted by martin
now we can add or subtract multiples of rows so doing Gaussian elimination
R2 -> R2 - 2R1
R3 -> R3 - R1
R4 -> R4 - R1

1, -1, L, 1
0, 4, 1-2L, 5L-1
0, 4, -1-L, 4L+1
0, 8, -3L, 10L-2

erk..how did u get -3L and 10L-2??

i got -2-L and 10L instead

 

[martin]

erk..how did u get -3L and 10L-2??

I had a mistake in the augmented matrix that I'd fixed later but forgotten to fix at the start (it should be right now)

-2L-L = -3L
10L-1-1 = 10L-2

 

[...]

Originally posted by martin
I had a mistake in the augmented matrix that I'd fixed later but forgotten to fix at the start (it should be right now)

-2L-L = -3L
10L-1-1 = 10L-2

ah, thx

ur a champ...

 

[+Po1ntDeXt3r+]

I did this for a friend in /UTS but i got Gamma = 4/3 for a unique soln..

OMG i hope i didnt fark it up!?!!!!

but for r4 i got
[0 8 -2- L 10L-2]

then using reduced row algorithms i get
r3 =[0 0 -2+L -L+2]
r4 =[0 0 -16+12L 0 ]

for a value of L where the matrix is solvable (.. it says soluble? .. 10ppm in water??? crap sooo geeky)

then the only real value that x3 can take is -1 (x3=-1)

so we can conclude that -16+12L = 0
then r4 is now a row of zeros..
L = 4/3
for a solvable equation set...

 

[turtle_2468]

Bachelor of Medicine (Math.)??

 

[+Po1ntDeXt3r+]

yer im aiming to get a math major..

 

[martin]

but for r4 i got
[0 8 -2- L 10L-2]

I originally did the same as you but I'm fairly sure now that
r1 = [1 -1 L 1]
r4 = [1 7 -2L 10L-1]

and r4 -> r4 - r1
= [0 8 -3L 10L-2]

 

[+Po1ntDeXt3r+]

Originally posted by martin
I originally did the same as you but I'm fairly sure now that
r1 = [1 -1 L 1]
r4 = [1 7 -2L 10L-1]

and r4 -> r4 - r1
= [0 8 -3L 10L-2]

yer ure right.. geez i did it too fast :S
r3 = [0 0 - 12 8 - 4L ]
r4 = [0 0 -16-4L 0 ]

for 3 unknowns then L= -4
since the bottom row has to be a row of zeros..

 

[+Po1ntDeXt3r+]

abdooooo!!!! read down.. there were mistakes made..
esp r3 and r4

about reduce row echelon form (rref)
1 0 0
0 1 0
0 0 1
0 0 0
would b the rref of a 4 x 3 matrix..

 

[...]

hmm..

i got another one i couldn't do

let B = 1 - 6
6 10 -

a) for which vaules(s) of is B invertible
b) If = -1, solve Bx = 0
c) If = -2, solve Bx = 0

hmm..i dunno what invertible means..lol...bloody past papers with no solutions

 

[wogboy]

Definition: Matrix A is invertible if and only if there exists an a matrix B such that: AB = I (I is the identity matrix)

Therefore det(AB) = det(I),
so det(A) * det(B) = det(I) (since the determinant of a product = the product of the determinants)
so det(A) * det(B) = 1 (determinant of identity matrix is 1)
det(B) = 1/det(A)

you can see that if det(A) = 0, the matrix B does not exist so B is not invertible. So that leads us to this theorem here:

Matrix A is invertible if & only if det(A) =/= 0

-------------------------------------

So for part a), find the determinant of your matrix B, and set this to equal zero. Note that this will give you all the values of such that A is NOT invertible.

For part b) and c), if = -1 or = -2 lies in the range that will make B invertible then x must be the zero vector (think about why this is true). Otherwise if B is not invertible, there will be infinitely many solutions including the zero vector. Use Gaussian elimination to solve in this case.

 

[+Po1ntDeXt3r+]

B=
a b
c d

by the definition of det(B) = ad-bc =
so for invertible det(B) = 0 as wogboy sed.

(1 - )(10-) - 36= 0

solve.. then thats ure answer

 

[...]

for some reason i can do inverse matrix..

like find ther inverse of

2,5,8,5
1,2,3,1
2,4,7,2
1,3,5,3

i think u have to make 1s digaonlly down right?

 

[Affinity]

what you do is
reduce the following to REDUCED row echelon form:

2 5 8 5 1 0 0 0
1 2 3 1 0 1 0 0
2 4 7 2 0 0 1 0
1 3 5 3 0 0 0 1

then the matrix on the right is the inverse.

there are otehr ways to do it .. but I think this is the best one in this case and most others