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Question 20522: How do I perform the next required row operation on the following matrix and provide only the next table:
x y z
1 28 14 245
0 3 7 42
0 7 7 -38
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! trust you want to solve the equations for x,y and z and you are at this stage now....assuming that .....our objective is to finally get the matrix if possible into the following form ....(i am using ....to seperate the numbers with suitable gaps..your typing is giving raise to uneven gaps bringing a little lack of clarity)
1.....0.....0.....x
0.....1.....0.....y
0.....0.....1.....z
now we have
1......28.....14.....245
0.......3......7......42
0.......7......7.....-38
new row2=old row2/3.......to get 1 as required in row2.so we get...
1......28.....14.....245
0......3/3....7/3....42/3
0.......7......7.....-38
new row3=oldrow3-7*row2 to get 0 as required in row3
1......28.....14...........245
0.......1.....7/3...........14
0......7-7*1..7-7*7/3......-38-7*14
new row3 = old row3/(-28/3)..to get 1 as required in row3
1......28.....14...................245
0.......1.....7/3...................14
0.......0....(-28/3)/(-28/3).....(-136)/(-28/3)
this gives us finally in the following form
1......28.....14............245
0.......1.....7/3...........14
0.......0......1............102/7
now we go back in the same way to get 0 in row2 and row3
new row2=old row2-row3*7/3...and new row1=old row1-row3*14...so we get
1......28......14-1*14.......245-(102/7)*14
0.......1.......7/3-(7/3)*1...14-(102/7)*(7/3)
0.......0.........1.............102/7
the above on simplification gives us
1.......28.......0..........41
0........1.......0..........-20
0........0.......1..........102/7
now finally we take new row1=old row1-28*row2
1.......28-28*1......0.......41-(-28*20)
0........1.......0...........-20
0........0.......1...........102/7
so the final answer is
1......0.......0.......601
0......1.......0.......-20
0......0.......1.......102/7
which tells us that
1*x+0*y+0*z=x=601
0*x+1*y+0*z=y=-20
0*x+0*y+1*z=z=102/7
note that each and every transformation we did above can be interpreted as given in the last statement given above...this i hope will give you the insight of the process at every step.you can also substitute these values of x,y and z in each and every matrix above to see that they satify all the equations given by the different matrices..in general each mtrix can be taken as a set of simltanous equations in x,y and z...they can be written as follows..take column 1 is for x,column 2 is for y and column 3 is for z.so the first matrix you gave
1......28.....14.....245
0.......3......7......42
0.......7......7.....-38
tells us that
1*x+28*y+14*z=245....etc...
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