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Use Gaussian elimination to solve the system

$$2x-3y=-5$$

$$3x+y=9$$

Find the value of x.

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Did you try anything? –  Charlie Nov 10 '12 at 20:05
    
is it like Simultaneous Equations? I dont know what Gaussian elimination is? –  Dara Nov 10 '12 at 20:07
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Do you know what Wikipedia is? –  joriki Nov 10 '12 at 20:09
    
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@Cameron: Press "help" under the "Add Comment" button to see a mini-guide to mini-Markdown. For a link, enclose the text to be linked in square brackets, immediately followed (no intervening space) by the URL in parentheses. –  joriki Nov 10 '12 at 20:19

2 Answers 2

Since we want to find the value of $x$, we need to eliminate $y$ from both the equations i.e. from the equations \begin{align} 2x-3y & = -5 & \text{Equation }1\\ 3x+y & = 9 & \text{Equation }2 \end{align} Though you can do it directly, without resorting to matrices, to enable you to understand how to work with matrices, I am going to do this in the matrix form.

The two equations can be written in a matrix form as shown below. \begin{align} \begin{bmatrix} 2 & -3\\ 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -5\\ 9 \end{bmatrix} \end{align} Typically, Gauss elimination ($LU$ factorization) involves making the entries below the main diagonal zero using row-operations i.e. \begin{align} \begin{bmatrix} 2 & -3\\ 3 & 1 \end{bmatrix} \to \begin{bmatrix} \times & \times\\ 0 & \times \end{bmatrix} \end{align} However, you could also do Gauss elimination by making the entries above the main diagonal zero using row-operations i.e. \begin{align} \begin{bmatrix} 2 & -3\\ 3 & 1 \end{bmatrix} \to \begin{bmatrix} \times & 0\\ \times & \times \end{bmatrix} \end{align} Since you are interested in $x$, it makes sense to proceed along the second line and convert \begin{align} \begin{bmatrix} 2 & -3\\ 3 & 1 \end{bmatrix} \end{align} to a lower triangular matrix.

This can be done as follows. If we denote the first equation as $R_1$ and the second equation as $R_2$, then the operation $R_1= R_1 + 3R_2$ converts \begin{align} \begin{bmatrix} 2 & -3\\ 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -5\\ 9 \end{bmatrix} \end{align} to \begin{align} \begin{bmatrix} 2+3 \times 3 & -3 + 3 \times 1\\ 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -5 + 3 \times 9\\ 9 \end{bmatrix} \text{ i.e. } \begin{bmatrix} 11 & 0\\ 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 22\\ 9 \end{bmatrix} \end{align} The first row/equation can now be used to read of the value for $x$. $$11x = 22 \implies x = 2$$ If you are interested in $y$ as well, plug in the value for $x$ and obtain $y$ from the second equation, i.e. $$3 \times 2 + y = 9 \implies y = 3$$

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Thanks a million Marvis, great help :) very clear explaination –  Dara Nov 10 '12 at 20:57
    
Sorry Marvis...I'm just getting the hang of typesetting matrices, and was wrapped up with trying to do so here, that I didn't see you had posted...sigh! –  amWhy Nov 10 '12 at 21:03

We have the system of equations given by:

$$2x-3y=-5$$

$$3x +y =9$$

Assuming you are being asked to represent the two equations in matrix form:

we use an augmented coefficent matrix, and using Gaussian elimination is just like solving simultaneous systems of equations. The values of each variable's coefficents appear in the first two columns, and the values to the RHS of the equations comprise the last column (i.e., they augment the matrix).

\begin{bmatrix} a_x & b_y & RHS1\\ c_x & d_y & RHS2 \end{bmatrix}

where row $(1)$ represents $a_xx + b_yy = RHS1$ and row $2$ represents $c_xx + d_yy = RHS2$.

So, given your equations, we are working with the following matrix: $$\begin{bmatrix} 2 & -3 &-5\\ 3 & 1&9 \end{bmatrix}\tag{1}$$ Then we can multiply row 2 by 3 in the above matrix and add that value to row 1: \begin{bmatrix} 11 & 0 &22\\ 3&1&9 \end{bmatrix}

If we divide row $1$ by 11, the above matrix simplifies to \begin{bmatrix}% 1 & 0 &2\\ 3 & 1& 9% \end{bmatrix}

We can multiply row 1 by $-3$ and add those values to row 2, and we end with:

$$\begin{bmatrix}% 1 & 0 & 2\\ 0 & 1& 3% \end{bmatrix}\tag{2}$$

This reduced matrix corresponds to the solutions

$$\text{row 1:}\;\;1\cdot x + 0\cdot y= 2 \iff x = 2$$ $$\text{row 2:}\;\;0\cdot x + 1\cdot y =3\iff y =3$$

These values are then solutions to your system of equations. If you were to operate on the original equations using the same operations used to row-reduce the augmented matrix, we'd get the same result. But using just the coefficient matrix (without confusing variables) - especially when you have a system of lots of equations and lots of unknowns, is usually more efficient.

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+1 Long tale like the ones Amzoti wrote. :) –  B. S. Aug 8 '13 at 12:54
    
Yes, indeed, @Babak. Our friend Amzoti writes excellent detailed answers and puts a lot of time into doing so! –  amWhy Aug 8 '13 at 12:57

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