Hi guys,

Could i have a little help with the following problems?

Use matrix algebra to solve the following simultaneous equations:

Problem 1:

5x + y = 13

3x +2y = 5

Problem 2:

3x + 2y = -2

x + 4y = 6

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- Aug 7th 2006, 03:45 AMc00kyMatrix Algebra Problems...
Hi guys,

Could i have a little help with the following problems?

Use matrix algebra to solve the following simultaneous equations:

Problem 1:

5x + y = 13

3x +2y = 5

Problem 2:

3x + 2y = -2

x + 4y = 6 - Aug 7th 2006, 04:01 AMgalactus
I'll step through #1, you try #2 using the same technique.

Set it up like this:

$\displaystyle \begin{bmatrix}5&1&13\\3&2&5\end{bmatrix}$

1/5 times row 1:

$\displaystyle \begin{bmatrix}1&1/5&13/5\\3&2&5\end{bmatrix}$

-3 times row 1 add to row 2, replace row 2 with the result:

$\displaystyle \begin{bmatrix}1&1/5&13/5\\0&7/5&-14/5\end{bmatrix}$

5/7 times row 2:

$\displaystyle \begin{bmatrix}1&1/5&13/5\\0&1&-2\end{bmatrix}$

-1/5 times row 2 add to row 1, replace row 1 with the result:

$\displaystyle \begin{bmatrix}1&0&3\\0&1&-2\end{bmatrix}$

**x=3 and y=-2**

See what you have to do?. Whittle away until you get it in what is known as reduced row-echelon form. Your solutions are in the far right column and you have 0's and 1's in the other 2 columns. - Aug 7th 2006, 04:10 AMc00ky
Why did you multiply the rows by those certain numbers i.e. 1/5 & 5/7?

Thanks - Aug 7th 2006, 04:22 AMgalactus
Because that's what I used to get it into the form I wanted. It didn't have to be those particular numbers. Whatever works. Your method may be different.

You must have some clue of what you must do?. If you don't, you ought to see your teacher. Unless, let me guess, an on-line class.

You have to get the diagonal of 1's and 0's like I did on the 1st problem.

Give it a try with #2 and let's see how you done :) .

It's a matter of practice and observation. - Aug 7th 2006, 04:41 AMSoroban
Hello, c00ky!

Here is another method . . .

Given the matrix equation: .$\displaystyle A X \:=\:C$

. . find $\displaystyle A^{-1}$, the inverse of $\displaystyle A$, and multiply both sides.

We have: .$\displaystyle A^{-1}(A X) \;=\;A^{-1} C\quad\Rightarrow\quad(A^{-1}A)X\:=$ $\displaystyle \:A^{-1}C\quad\Rightarrow\quad IX\:=\:A^{-1}C$

. . Therefore, the solution is: .$\displaystyle X\:=\:A^{-1}C$

It helps if you know this handy formula:

. . The inverse of $\displaystyle A\:=\:\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is: .$\displaystyle A^{-1}\:=\:\frac{1}{ad-bc}\begin{pmatrix}d & -b\\-c & a\end{pmatrix}$

This is easily memorized:

. . (1)*Switch*on the main diagonal $\displaystyle (\searrow)$

. . (2)*Change signs*on the other diagonal $\displaystyle (\nearrow)$

. . (3) Divide by the determinant of $\displaystyle A.$

Quote:

Use matrix algebra to solve the following system:

$\displaystyle 1)\;\begin{array}{cc}5x + y \,= \,13 \\3x +2y \,= \,5\end{array}$

We have: .$\displaystyle \begin{pmatrix}5 & 1\\3 & 2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\;= \;\begin{pmatrix}13\\5\end{pmatrix}$

. . Then: .$\displaystyle A^{-1} \:=\:\frac{1}{7}\begin{pmatrix}2 & \text{-}1 \\ \text{-}3 & 5\end{pmatrix}$

Hence: .$\displaystyle \begin{pmatrix}x\\y\end{pmatrix} \:= \:\frac{1}{7}\begin{pmatrix}2&\text{-}1\\\text{-}3&5\end{pmatrix}\begin{pmatrix}13\\5 \end{pmatrix} $ $\displaystyle = \:\frac{1}{7}\begin{pmatrix}26-5\\\text{-}39 + 25\end{pmatrix} \;=\;\frac{1}{7}\begin{pmatrix}21\\ \text{-}14\end{pmatrix}

$

. . Therefore: /$\displaystyle \begin{pmatrix}x\\y\end{pmatrix}\;=\;\begin {pmatrix}3 \\ \text{-}2\end{pmatrix}$

- Aug 7th 2006, 04:46 AMc00ky
I shall give it a try, your almost right with your assumptions.. i was taught maths by a Media teacher! (I might aswell have been taught online by a giraffe)

- Aug 7th 2006, 04:46 AMgalactus
There ya' go. Soroban's method is more efficient with matrices of this size. They become more complicated as they get bigger, though.

- Aug 7th 2006, 10:18 AMc00ky
For the 2nd problem my answer is..

x 0.6

=

y 2.6

Could somebody check these please?

Thanks! - Aug 7th 2006, 10:22 AMThePerfectHackerQuote:

Originally Posted by**c00ky**

- Aug 7th 2006, 10:32 AMQuickQuote:

Originally Posted by**cookie**

note: rather than using latex, you can show us matrixes like this:

|2 3 5|

|3 4 5| - Aug 7th 2006, 10:39 AMc00ky
I mixed the 3 and 4 the wrong way around earlier on.

I now have x = -2 and y = 2 which seems to work!

Woohoo !