Thread: Linear Equation and matrices.

It's been a ten years ago when I learning those equations, some how recently my boss just give me a question and asking for help, I d't have any ideal back to carry out those equation where all the knowledge already give back to teacher, i guess.

Therefore, hope some one here can give me a little help on it. Thank you.

Part 1
Writing those in linear equation,
a) x-y+z = 3
b) 4x-3y-z = 6
c) 3x +y+2z = 23

in the form of Ax=b. State the size of matrix A and its main diagonal elements

Thanks to Prove It Part 1 Solved

Part 2
Using the elementary row operation (ERO) on the augmented matrix A|b, solve the system of linear equation above.



<---Try to resolve but not correct, need help

Originally Posted by gilagila

It's been a ten years ago when I learning those equations, some how recently my boss just give me a question and asking for help, I d't have any ideal back to carry out those equation where all the knowledge already give back to teacher, i guess.

Therefore, hope some one here can give me a little help on it. Thank you.

Writing those in linear equation,
a) x-y+z = 3
b) 4x-3y-z = 6
c) 3x +y+2z = 23

in the form of Ax=b. State the size of matrix A and its main diagonal elements

is the matrix which consists of all the coefficients of your variables, is a column vector containing your variables, and is a column vector containing the values on the right-hand-side of the equation.

So do you think you can write down what values are in the matrices?

Originally Posted by Prove It

is the matrix which consists of all the coefficients of your variables, is a column vector containing your variables, and is a column vector containing the values on the right-hand-side of the equation.

So do you think you can write down what values are in the matrices?

Not sure is it correct or not, let me try some
a) x-y+z = 3
b) 4x-3y-z = 6
c) 3x +y+2z = 23

if A = 1 -1 1...........x = x.........b= 3
.........4 -3 -1 ..............y..............6
.........3 1 2...................z............23

so Size of matrix A = square matrix 3 x 3 and the main diagonal elements = 1, -3 , 2

am i right ?

Originally Posted by gilagila

Not sure is it correct or not, let me try some
a) x-y+z = 3
b) 4x-3y-z = 6
c) 3x +y+2z = 23

if A = 1 -1 1 x = x b= 3
4 -3 -1 y 6
3 1 2 z 23

so Size of matrix A = square matrix 3 x 3 and the main diagonal elements = 1, -3 , 2

am i right ?

I know what you have done, just to make it more readable...







And you are correct that the elements in the main diagonal are 1, -3 and 2

Originally Posted by Prove It

I know what you have done, just to make it more readable...







And you are correct that the elements in the main diagonal are 1, -3 and 2

new in this forum, just want to ask how you type those matrix ? using what function / icon in the edit area ?

it's called latex, there is a tutorial here. to use it, you use the "tex" tags. for example:

"[tex ]x^2 + x + \sqrt{2}[ /tex]" produces

(don't put the spaces after the tex or before the /tex)

when row-reducing, there are 3 operations we can perform:

1) switch two rows,
2) multiply one row by a non-zero number,
3) multiply one row by a non-zero number and add/subtract it from another row.

the idea is try to try get a leading 1 in each row, each leading 1 is rightward of the rows above, and we want all 0's below each leading 1 (you can also try to have 0's above each leading 1 as well, which brings the matrix into reduced row echelon form). we may wind up with 0 rows at the bottom.

in your matrix, since we already have a leading 1 in the first row, the first step is to eliminate the rest of the first column. subtracting 4 times the 1st row from the second, and then subtracting 3 times the first row form the third, we get:



(this is the augmented matrix) can you continue?

Originally Posted by Deveno

it's called latex, there is a tutorial here. to use it, you use the "tex" tags. for example:

"[tex ]x^2 + x + \sqrt{2}[ /tex]" produces

(don't put the spaces after the tex or before the /tex)

when row-reducing, there are 3 operations we can perform:

1) switch two rows,
2) multiply one row by a non-zero number,
3) multiply one row by a non-zero number and add/subtract it from another row.

the idea is try to try get a leading 1 in each row, each leading 1 is rightward of the rows above, and we want all 0's below each leading 1 (you can also try to have 0's above each leading 1 as well, which brings the matrix into reduced row echelon form). we may wind up with 0 rows at the bottom.

in your matrix, since we already have a leading 1 in the first row, the first step is to eliminate the rest of the first column. subtracting 4 times the 1st row from the second, and then subtracting 3 times the first row form the third, we get:



(this is the augmented matrix) can you continue?

thanks for the help. I just do it in MS Word and make it as a picture, somehow d't know wher eis wrong, cannot get the answer right. Pls advice. TQ

Originally Posted by Deveno

in your matrix, since we already have a leading 1 in the first row, the first step is to eliminate the rest of the first column. subtracting 4 times the 1st row from the second, and then subtracting 3 times the first row form the third, we get:



(this is the augmented matrix) can you continue?

when 4R1 - R2, in r2 it should be 4(1)-4, 4(-1)-(-3), 4(1)-(-1), 4(3)-6 = 0,-1,5,6, why your no.2 figure is +1 not -1 ? and your no.4 figure is -6 not +6 ?

there are different ways to row-reduce, you used slightly different operations than i did, but you can see that you (eventually) wound up with the same 2nd row as me, and the negative of the 3rd row.

the error you made in your row-reduction is in the very last step, instead of adding 8 to -3 (in the 4th column), you added 8 to 3, getting 11, instead of the 5 you should have gotten.

you may check that x = 5, y = 4, z = 2 is indeed a solution to your original equations.

Originally Posted by Deveno

there are different ways to row-reduce, you used slightly different operations than i did, but you can see that you (eventually) wound up with the same 2nd row as me, and the negative of the 3rd row.

the error you made in your row-reduction is in the very last step, instead of adding 8 to -3 (in the 4th column), you added 8 to 3, getting 11, instead of the 5 you should have gotten.

you may check that x = 5, y = 4, z = 2 is indeed a solution to your original equations.

oops..ya..finally i got the answer. Thanks for the tip on the mistake on last row deduction.

This problem has been solved. Thank you all.