SOLUTION: Solve using matrices 2x-3y=4 5x+4y=33

Question 54438: Solve using matrices
2x-3y=4
5x+4y=33
Answer by tiffanyc(9) (Show Source): You can put this solution on YOUR website!
Using matrices is simply using elementary row operations to get 1x=? and 1y=? so the original equation in matrix form becomes the co-efficients of x and y and the constant term:
2 -3 4
5 4 33
Ideally, we want a matrix that is(see below) where ? is the answer for each variable.
1 0 ?
0 1 ?
start by dividing the top row by 2 to get a 1 in the first column of the first row
1 (-3/2) 2
5 4 33
Then get a zero in the first column of the second row by taking row 2 and subtracting 5 times row 1: (note: only row two changes in this step)
1 (-3/2) 2 = 1 (-3/2) 2
0 (8/2)-((-3)(5)/2) 33 = 0 (23/2) 23

Next get a one in the second column of the second row by multiplying the row by (2/23):
1 (-3/2) 2
0 1 2
Now to get a zero in the 2nd column of the first row, take row 1 and add (3/2) times the second row (because the term we're looking to get rid of is negative)
Note: in the first column (3/2) times 0 is still 0 so 1-0 will still be one.
Note: multiplying the second row is only for applying it to row 1, after the operation has been completed, the second row goes back to its original form.
1 (-3/2) 2 = 1 0 5
0(3/2) 1(3/2) 2(3/2) = 0 1 2
Now putting that back into linear algebra form
1x +0y = 5 x=5
0x +1y = 2 y=2
Check this by substituting the x and y values into the original equation:
2x -3y =4 2(5)-3(2) =4 10-6=4
5x +4y =33 5(5)+4(2) =33 25+8=33
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