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a matrix is just another way of thinking about elimination, lets use an example. Think about the following system of equations.

1) 3x+4y=12
2) 3x+2y=6

Lets use elimination to solve this. Multiply equation (1) by -1, and then add it to equation 2.

-3x-4y=-12
+(3x+2y=6)
=0-2y=-6
or -2y=-6,

which means y=3

now instead of just plugging that back in, you use elimation again, lets write down our two equations now. With our second equation replaced with the simplifed y=3
1)3x+4y=12
2)y=3

ok so to use elimination again, we have to multply equation (2) by -4 and add it to the first equation.

3x+4y=12
+(-4y=-12)
3x+ 0 = 0
or 3x = 0

which means x=0

so with a matrix you do the exact same thing, we just ignore the variables for a second.

we start with our 2 equations

1) 3x+4y=12
2) 3x+2y=6

our coefficents(the number we're multiplying x by) of x go in the first column, the coefficents of y in the second column, and the answers in a special answer column, we call an augment of the matrix. We write them in such a way that the rows represent the exact equations written.

so for equation (1)
3x+4y=12 we write [3 4 | 12]

and for equation (1)
3x+2y=6 we write [3 2 | 6]

lets put our matrix together
[3 4 | 12]
[3 2 | 6 ]

now again, the steps are exactly the same,
the first line by -1, and then add the second line.

-3 -4 |-12
3 2 | 6
0 -2 | -6 <- this becomes our new line 2, since we now have one of the variables eliminated.

[3 4 | 12]
[0 -2| -6]

remember this line 2 means -2y=-6 so we want to divide both sides by -2 to get y by itself

[3 4 | 12]
[0 1 | 3 ]

Now we have to multply line (2) by -4 and add it to the first equation.

3 4 | 12
0 -4 |-12
3 0 | 0 <- this becomes our new line 2, since we now have our sencond variable eliminated.

[3 0 | 0]
[0 1 | 3 ]

now divide by 3 in the first row

[1 0 | 0]
[0 1 | 3 ]

this means 1x+0y=0 or x=0
and 0x+1y=3 or y=3

with more variables you just do the same process of elimination, get rid of the the x's first, then the y's and so on.

Hope this helps, let me know if you have any more questions.

The strategy is very similuar.

Let's pretend your matrix looks like the following

[1 1 2 | 0 ]
[1 2 1 | 0 ]
[2 1 1 | 0 ]

your goal is to get the matrix to look like this

[ 1 1 2 |0]
[0 a b |c]
[0 0 d |e]

You have to use elimination in the first and the second row with the goal of canceling your first variable (your x variable) so that in the second row you have that 0 in front. So lets multiply row 1 by -1 and add it to row 2

-1(1) -1 -1 -2 |0
+(2) 1 2 1 |0
= 0 1 -1 |0

now we replace this for row 2


[1 1 2 | 0 ]
[0 1 -1 | 0 ]
[2 1 1 | 0 ]

now we want to do the same thing with row 3, again the intent is to eliminate the 1st variable, so that we only have two variables left. So lets use elimination on row 1 and row 3. -2 times the 1st row plus the 3rd row.

-2(1) -2 -2 -4 | 0
+(3) 2 1 1 |0
= 0 -1 -3|0

this becomes our new row 3 now

[1 1 2 | 0 ]
[0 1 -1 | 0 ]
[0 -1 -3 | 0 ]

ok lets now look at row 2 and row 3 quickly and remember what they mean. Row 2 is just y-z=0 and row 3 is just -y-3z=0. These are two equations with the same 2 variables, and we know how to solve this with elimination. So lets just add them together to get rid of that y variable.

(2) 0 1 -1|0
+(3) 0 -1 -3|0
= 0 0 -4|0

now lets replace this for row 3

[1 1 2 | 0 ]
[0 1 -1 | 0 ]
[0 0 -4 | 0 ]

Now we've reached our goal the matrix does look like we wanted it to look.

[ 1 1 2 |0]
[0 a b |c]
[0 0 d |e]

and if you just want to solve the matrix, it is sufficient to stop here and use substitution to solve the rest of it .

The last line reads -4z=0 so solve for z now, z=0
the second line reads y-z=0 but now we know z=0, so we can use that.
y-0=0 so y=0
and finally the first line is x+y+2z=0 but we can solve for x now by plugging in y and z,
x+0+2(0)=0 so x=0

Your matrix homework is going to be slightly harder, because i rigged this matrix to be all equal to zero, so that I could explain it easier, but the idea is the same.

Let me know if you have any more questions.