Question 66335This question is from textbook An Incremental Development
: PLease help to solve
x + 2y + z = 7
3x - y + z = -12
4x + 3y - 2z = 9
This question is from textbook An Incremental Development
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! IF YOU WANT SOLUTION BY ELIMINATION
SEE THE FOLLOWING EXAMPLE ND TRY.IN CASE OF DIFFICULTY
PLEASE COME BACK
equations with 3 linear
variables
6x-4y+5z=31
5x+2y+2z=13
x+y+z=2
should you multiply the 2nd equation by 2?
show work
1 solutions
Answer 29601 by venugopalramana(2734) About Me on
2006-07-04 11:53:09 (Show Source):
You can put this solution on YOUR website!
OK..THE PROBLEM,WE FACE IN ANSWERING IS THAT WE ARE
NOT AWARE OF YOUR COURSE OF STUDY AND BACKGROUND.THE
SOLUTION I GAVE IS BY GAUSS JORDAN ELIMINATION METHOD
USED IN MATRICES AS WANTED BY SOME STUDENTS.IF YOU ARE
NOT TAUGHT THAT ,WE SHALL DO BY NORMAL ELIMINATION.
equations with 3 linear variables
6x-4y+5z=31.......................................I
5x+2y+2z=13...........................II
x+y+z=2.......................III
EQN.II - 2*EQN.III
5X+2Y+2Z-2X-2Y-2Z=13-2*2=9
3X=9
X=9/3=3
EQN.I+4*EQN.III
6X-4Y+5Z+4X+4Y+4Z=31+4*2=39
10X+9Z=39...BUT...X=3..SO
10*3+9Z=39
9Z=39-30=9
Z=9/9=1
PUTTING IN EQN.III
3+Y+1=2
Y=2-4=-2
HOPE YOU UNDERSTOOD.
-------------------------------------------------------------------------------
IF YOU WANT SOLUTION BY MATRICES THEN
SEE THE FOLLOWING EXAMPLE ND TRY.IN CASE OF DIFFICULTY
PLEASE COME BACK
Hi, I'm in homeschooling and I'm having trouble with
matrices. I was wondering how to solve the problem
where you have to find the x,y, and z values in the
matrix:
[7 -7 5 | 9]
[9 5 -7 | -17]
[6 1 -7 | -2]
I'd appreciate the help. Thank you!
Caitlyn Reese
1 solutions
Answer 9969 by venugopalramana(585) About Me on
2005-11-28 07:14:11 (Show Source):
the 4 column heads represent x,y,z and constant term
in the matrix of system of eqns.
then each row gives us one eqn.like say row 1 gives us
that 7x-7y+5z=9..etc�
hence if we can make the matrix to become
1 0 0 ?
0 1 0 ??
0 0 1 ???
then from the explantion given above it means
1x=?.1y=?? And 1z=???
so we try to transform the matrix in to that form..by
the following steps.
in fact using the above explanation,you can see that
what we do at each step is just
divide each eqn. with a constant/add/subtract etc
which does not change the basic
eqn.for ex. dividing row 1 by 7 means change the given
eqn.7x-7y+5z=9 to x-y+5z/7=9/7
legend:- or1 means old row 1..nr1 means new row 1�r1
means the existing row 1 please note that no changes
are made in rows other than those mentioned at each
step.
start with given matrix �
7...... -7..... 5...... 9
9...... 5...... -7..... -17
6...... 1...... -7..... -2
step 1�we want to make 1st.row 1st.column as
1�.so�.nr1=or1/7...
1...... -1..... (5/7).. (9/7)
9...... 5...... -7..... -17
6...... 1...... -7..... -2
step 2..we want to make 2nd/3rd.rows,col.1 as
0...so...nr2=or2-9*r1........nr3=or3-6*r1
1...... -1..... (5/7).. (9/7)
0... 14..... (-7-9*5/7).... (-17-9*9/7)
0...... 7...... (-7-6*5/7)..... (-2-6*9/7)
step 3�we want to make 2nd.row.2nd.col.as
1..so..nr2=or2/14
1...... -1..... 5/7.... 9/7
0...... 1...... (-94/7)/14..... (-200/7)/14
0...... 7...... (-7-6*5/7)..... (-2-6*9/7)
step 4..we want to make 3rd.row.2nd.col.as
0�so�.nr3=or3-7*r2
1...... -1..... (5/7).. (9/7)
0...... 1...... (-94/7)/14..... (-200/7)/14
0 0 (-79/7)-7*(-94/98) (68/7)-7*(-200/98)
step 5�.we want to make 3rd.row.3rd.col.as
1�so�.nr3=or3/(-32/7)
1...... -1..... (5/7).. (9/7)
0...... 1...... (-94/98)... (-200/98)
0...... 0..... 1...... -1
step 6�we want to make 1st/2nd.row 3rd.col.as
0..so..nr1=or1-5*r3/7...nr2=or2+94*r3/98
1...... -1..... 0..... 2
0...... 1...... 0...... -3
0...... 0...... 1...... -1
step7�.we want to make 1st.row 2nd.col.as
0..so�.nr1=or1+r2
1...... 0...... 0...... -1
0...... 1...... 0...... -3
0...... 0...... 1...... -1
so x=-1.....y=-3.....and z=-1...you can check back
YOU CAN SEE THE FOLLOWING ADDITIONAL MATERIAL FOR
REFERENCE
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
1 solutions
Answer 9892 by venugopalramana(370) About Me on
2005-11-25 08:01:32 (Show Source):
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...
|
|
|
