1. 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

2. Re: Linear Equation

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
$\displaystyle \mathbf{A}$ is the matrix which consists of all the coefficients of your variables, $\displaystyle \mathbf{x}$ is a column vector containing your variables, and $\displaystyle \mathbf{b}$ 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?

3. Re: Linear Equation

Originally Posted by Prove It
$\displaystyle \mathbf{A}$ is the matrix which consists of all the coefficients of your variables, $\displaystyle \mathbf{x}$ is a column vector containing your variables, and $\displaystyle \mathbf{b}$ 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 ?

4. Re: Linear Equation

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...

$\displaystyle \mathbf{A} = \left[\begin{matrix}1 & -1 & \phantom{-}1 \\ 4 & -3 & -1 \\ 3 & \phantom{-}1 & \phantom{-}2\end{matrix}\right]$

$\displaystyle \mathbf{x} = \left[\begin{matrix} x \\ y \\ z \end{matrix}\right]$

$\displaystyle \mathbf{b} = \left[\begin{matrix} 3 \\ 6 \\ 23 \end{matrix}\right]$

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

5. Re: Linear Equation

Originally Posted by Prove It
I know what you have done, just to make it more readable...

$\displaystyle \mathbf{A} = \left[\begin{matrix}1 & -1 & \phantom{-}1 \\ 4 & -3 & -1 \\ 3 & \phantom{-}1 & \phantom{-}2\end{matrix}\right]$

$\displaystyle \mathbf{x} = \left[\begin{matrix} x \\ y \\ z \end{matrix}\right]$

$\displaystyle \mathbf{b} = \left[\begin{matrix} 3 \\ 6 \\ 23 \end{matrix}\right]$

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 ?

6. Re: Linear Equation

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 $x^2 + x + \sqrt{2}$

(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:

$\begin{bmatrix}1&-1&1&3\\0&1&-5&-6\\0&4&-1&14 \end{bmatrix}$

(this is the augmented matrix) can you continue?

7. Re: Linear Equation

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 $x^2 + x + \sqrt{2}$

(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:

$\begin{bmatrix}1&-1&1&3\\0&1&-5&-6\\0&4&-1&14 \end{bmatrix}$

(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

8. Re: Linear Equation

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:

$\begin{bmatrix}1&-1&1&3\\0&1&-5&-6\\0&4&-1&14 \end{bmatrix}$

(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 ?

9. Re: Linear Equation

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.

10. Re: Linear Equation

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.