SOLUTION: Can you please help me do this problem. I really just don't know where to start.It is solutions to linear systems using matrices. 12x-4y-7z=8 -8x-6y+9z=7 34x+6y-2z=5 Can y

Algebra ->  College  -> Linear Algebra -> SOLUTION: Can you please help me do this problem. I really just don't know where to start.It is solutions to linear systems using matrices. 12x-4y-7z=8 -8x-6y+9z=7 34x+6y-2z=5 Can y      Log On

 Click here to see ALL problems on Linear Algebra Question 81532This question is from textbook college algebra with modeling and visualization : Can you please help me do this problem. I really just don't know where to start.It is solutions to linear systems using matrices. 12x-4y-7z=8 -8x-6y+9z=7 34x+6y-2z=5 Can you show me how to do it step by step please. Thank you for your time This question is from textbook college algebra with modeling and visualization Found 3 solutions by longjonsilver, Edwin McCravy, rapaljer:Answer by longjonsilver(2297)   (Show Source): You can put this solution on YOUR website! have a look at my Lesson on Cramer's Rule - that will explain one method using determinants Jon Answer by Edwin McCravy(13211)   (Show Source): You can put this solution on YOUR website! ``` Well, how about that! It looks as though LongJonSilver started to do it, found it was too messy and quit. He suggested Cramer's rule, but that's the determinant method, not the matrix method. OK, I'm game. I'll do it by matrices Solution by Edwin McCravy: Erase all the letters, plus signs, and equal signs and put a line where the equal signs were: [12 -4 -7 |8] [-8 -6 9 |7] [34 6 -2 |5] The plan is seven-fold, in this order: 1. Use the 1st row to get a 0 where the -8 is in the 2nd row 2. Use the 1st row to get a 0 where the 34 is in the 3rd row 3. Use the new 2nd row to get a 0 where the 6 is in the 3rd row Your matrix will then look like this: [ # # # | #] [ 0 # # | #] [ 0 0 # | #] There will be numbers where the #'s are. This form is called the reduced echelon form. 4. Rewrite the matrix as three equations. 5. Solve the 3rd equation for z 6. Substitute that value for z into the 2nd equation and solve for y 7. Substitute both the value of z and the value for y in the 1st equation and solve for x. Here we go: [12 -4 -7 | 8] [-8 -6 9 | 7] [34 6 -2 | 5] To get a zero where the -8 is, to the side, multiply the 1st row by 2 and the 2nd row by 3, and add them vertically 1st row x2 24 -8 -14 | 16 2nd row x3 -24 -18 27 | 21 ---------------- new 2nd row 0 -26 13 | 37 Replace only the 2nd row by that bottom line in the matrix [12 -4 -7 | 8] [ 0 -26 13 | 37] [34 6 -2 | 5] To get a zero where the 34 is, to the side, multiply the 1st row by -17 and the 3rd row by 6, and add them vertically 1st row x -17 -204 68 119 | -136 2nd row x 6 204 36 -12 | 30 ---------------------- new 3rd row 0 104 107 | -106 Replace only the 3rd row by that bottom line in the matrix [12 -4 -7 | 8] [ 0 -26 13 | 37] [ 0 104 107 | -106] To get a zero where the 104 is, to the side, multiply the 2nd row by 4 and the 3rd row by 1, (that is, leave it as it is), and add them vertically 2nt row x 4 0 -104 52 | 148 2nd row x 1 0 104 107 | -106 ---------------------- new 3rd row 0 0 159 | 42 Replace only the 3rd row by that bottom line in the matrix [12 -4 -7 | 8] [ 0 -26 13 | 37] [ 0 0 159 | 42] This is reduced echelon form. If your teacher wants you to get it all the way to row-reduced echelon form, you'll have to get 0's where the -4, the -7 and the 4 are, too. But it can be solved from the reduced echelon form as follows. Erase the brackets and the line, and put the letters, plus signs, and equal signs back in: 12x - 4y - 7z = 8 0x - 26y + 13z = 37 0x + 0y + 159z = 42 Erase the 0 terms 12x - 4y - 7z = 8 -26y + 13z = 37 159z = 42 Solve the 3rd equation for z 159z = 42 z = 42/159 z = 14/53 Sunstitute 14/53 for z in the 2nd equation -26y + 13z = 37 -26y + 13(14/53) = 37 -26y + 182/53 = 37 Clear of fractions by multiplying through by 53 -1378y + 182 = 1961 -1378y = 1779 y = -1779/1378 Now substitute y = -1779/1378 and z = 14/53 in the 1st equation: 12x - 4y - 7z = 8 12x - 4(-1779/1378) - 7(14/53) = 8 12x + 7116/1378 - 98/53 = 8 Reduce the fraction 7116/1378 to 3558/689 12x + 3558/689 - 98/53 = 8 Clear of fractions by multiplying through by 689 8268x + 3558 - 1274 = 5512 8268x + 2284 = 5512 8268x = 3228 x = 3228/8268 x = 269/689 So the solution is (x, y, z) = (269/689, -1779/1378, 14/53) and I'll bet anything you copied it wrong, for no respectable teacher would expect you to do a problem that would come out with such a god-awful answer! No wonder LongJonSilver gave it up. But what I got is correct for the problem you posted. Edwin``` Answer by rapaljer(4670)   (Show Source): You can put this solution on YOUR website! Are you allowed to use graphing calculators? In particular, if you are allowed to use a TI 85 or 86, there is a function on your calculator called [2nd] [SIMULT] that solves systems of equations without having to use either the elimination method or Cramer's Rule. If you have a TI 83+ or a TI 84, look in [APPS] for a program called [POLYSMLT]. If you don't have this program on these calculators, you can get it from Texas Instruments and have it put on your calculator. It's the easy way to solve a problem if it is allowed and if you have a calculator that will do it. R^2 at SCC