Jump to content
Sign in to follow this  

Question on Matrices

Recommended Posts

Hey can anyone help me with this question? Solve the system of linear equations using the row reduction method:

x+2y-2z=5

3x+2y+z=0

-x+y-3z=6

Please try to show me the solving as well if you can.

thanks :D

Share this post


Link to post
Share on other sites

Hey can anyone help me with this question? Solve the system of linear equations using the row reduction method:

x+2y-2z=5

3x+2y+z=0

-x+y-3z=6

Please try to show me the solving as well if you can.

thanks :D

Very easy:

(1 2 -2) (x) (5)

(3 2 1) x (y ) = (0)

(-1 1 -3) (z) (6)

In order to get the values of x y z, you multiply both sides by the inverse of 3 x 3 matrix on the left side. I'm sure you'll be able to find out on internet how to do so and therefore it would look like:

(x) (7 -4 -6) (5)

(y ) = (-8 5 7) x (0)

(z) (-5 3 4) (6)

(x) (7x5)+(-4x0)+(-6x6) (-1)

(y ) = (-8x5)+(5x0)+(7x6) = (2)

(z) (-5x5)+(3x0)+(4x6) (-1)

Therefore, x=-1, y=2, z=-1

EDIT: haha i'm sorry for the messed up order, but i hope you get the idea...

Edited by shad0wboss

Share this post


Link to post
Share on other sites

Hey can anyone help me with this question? Solve the system of linear equations using the row reduction method:

x+2y-2z=5

3x+2y+z=0

-x+y-3z=6

Please try to show me the solving as well if you can.

thanks :D

Very easy:

(1 2 -2) (x) (5)

(3 2 1) x (y ) = (0)

(-1 1 -3) (z) (6)

In order to get the values of x y z, you multiply both sides by the inverse of 3 x 3 matrix on the left side. I'm sure you'll be able to find out on internet how to do so and therefore it would look like:

(x) (7 -4 -6) (5)

(y ) = (-8 5 7) x (0)

(z) (-5 3 4) (6)

(x) (7x5)+(-4x0)+(-6x6) (-1)

(y ) = (-8x5)+(5x0)+(7x6) = (2)

(z) (-5x5)+(3x0)+(4x6) (-1)

Therefore, x=-1, y=2, z=-1

EDIT: haha i'm sorry for the messed up order, but i hope you get the idea...

hmm thank you!

but isnt this different from the row reduction method we are supposed to use in IB? Because we are supposed to use 3 reduction methods (its called reduced row echelon form) and through this we must reach a matrix that looks like an identity matrix.. any idea on how to use that method? I know the three methods we are allowed to use which are switching the rows of the matrix, multiplying a row with a number and adding rows of the matrix to each other. I am having a hard time using this particular technique, so if youve got any ideas please do share :P

Share this post


Link to post
Share on other sites

Damn, I don't think I learned rref until university O.o And finding the inverse of a 3x3 matrices is so time consuming it's awful. You have to compute the determinant, and use the inverse of that multiplied by the adjoint of the matrice. (I don't believe you have to know this in IB2 SL Math) but you do it without knowing because you are expected to know how to compute a 2x2 matrix (determinant, inverse and all that). Unless of course the syllabus changed :confused: I sat my math exams in 2011.

Row reducing is very similar to how you solve a system of equations. I'll assume you know how it works:

1 2 -2 5

3 2 1 0

-1 1 -3 6

That is a matrix you have to set up. I will refer to the first row as Eq. 1, the second to Eq. 2, .... Notice the last column is the result of the 3 equations. The key is to obtain a "leading 1" and use that to easily manipulate the matrix to rref.

1 2 -2 5 Triple Eq. 1 - Eq. 2

0 4 -7 15 Eq 1+Eq 3

0 3 -5 11

1 2 -2 5 Eq.2 - Eq 3

0 1 -2 4

0 3 -5 11

1 2 -2 5 Eq 3 - Triple Eq 2

0 1 -2 4

0 0 1 -1

1 0 2 -3 Eq1- Double Eq 2

0 1 -2 4

0 0 1 -1

1 0 0 -1 Eq.1- Double Eq 3

0 1 0 2 Eq. 2 + Double Eq 3

0 0 1 -1

I hope the steps I showed are right, but if they aren't you should be able to see what I'm doing. So the solution column (transposed) is (-1 2 -1) and those values seem to be correct if you plug them back into the original equations. On the left of the solution column is a 3x3 identity matrix.

That took me way too long...

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.