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Question 64626This question is from textbook Holt Algebra 2 w/Trig
: I have tried for 2 days to solve this one! I have the answers, and they do solve the equations, so they are correct. However, I am unable to produce the answers using matrix transformations. Any step-by-step help would be appreciated.
-2r + 6s -3t + 3v = -5
4r - 3s + 2t -v = -3
-3r - 4s - 3t + 4v = -15
5r + 2s - 2t + 3v = -20
The solution is r,s,t,v = -2,1,3,-2 respectively.
This question is from textbook Holt Algebra 2 w/Trig
Found 2 solutions by venugopalramana, Fermat:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! see the following and try.if you are still in difficulty please come back
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Solve the system of linear equations by using the matrices: [9,3,4][x1]=7
[4,3,4][x2]=8
[1,1,1][x3]=3
9 3 4 7
4 3 4 8
1 1 1 3
R1=R1/9
1 0.333333333 0.444444444 0.777777778
4 3 4 8
1 1 1 3
R2=R2-4R1…..R3=R3-R1
1 0.333333333 0.444444444 0.777777778
0 1.666666667 2.222222222 4.888888889
0 0.666666667 0.555555556 2.222222222
R2=R2/(5/3)
1 0.333333333 0.444444444 0.777777778
0 1 1.333333333 2.933333333
0 0.666666667 0.555555556 2.222222222
R3=R3-R2*2/3
1 0.333333333 0.444444444 0.777777778
0 1 1.333333333 2.933333333
0 0 -0.333333333 0.266666667
R3=-R3*3
1 0.333333333 0.444444444 0.777777778
0 1 1.333333333 2.933333333
0 0 1 -0.8
R2=R2-R3*4/3….R1=R1-R3*4/9
1 0.333333333 0 1.133333333
0 1 0 4
0 0 1 -0.8
R1=R1-R2/3
1 0 0 -0.2
0 1 0 4
0 0 1 -0.8
HENCE X1=-0.2 X2=4 X3=-0.8
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Linear_Algebra/35842: Use the Gauss Jordan method to
solve the system of equations:
x+y+2z=7
3x-y+z=10
2x+y-3z=-6
1 solutions
Answer 21966 by venugopalramana(2734) About Me on
2006-05-02 06:35:47 (Show Source):
You can put this solution on YOUR website!
Use the Gauss Jordan method to solve the system of
equations:
x+y+2z=7.......... I
3x-y+z=10............II
2x+y-3z=-6................III
THE COEFFICIENT AND AUGMENTED MATRIX IS
X,Y,Z,CONSTANT
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1,1,2,7...............MEANS EQN.I
3,-1,1,10.............MEANS EQN.II
2,1,-3,-6.............MEANS EQN.III
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SEE THE FOLLOWING EXAMPLE SOLVED FOR A PROBLEM SIMILAR
TO YOURS WHICH IS
ESSENTIALLY SAME BUT FOR THE NUMBERS , FOLLOW STEP BY
STEP AND DO BY YOUR SELF,THERE BY YOU WILL LEARN HOW
TO DO AND BECOME AN EXPERT BY YOUR SELF!!!!IF YOU DO
NOT UNDERSTAND ASK WHERE IT IS NOT CLEAR...
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Linear_Algebra/20750: 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
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Answer 9969 by venugopalramana(390) 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...
Answer by Fermat(136)  (Show Source):
You can put this solution on YOUR website! I'm using Gauss-Jordan elimination method. You could follow these steps to produce your inverse matrix.
The augmented matrix is shown below, in sets of 4 equations, or rows.
The row operations, following a set of 4 eqns, apply to that set of eqns, with the result being in the following set of 4 eqns.
-2 6 -3 3 | -5 ---(1)
4 -3 2 -1 | -3 ---(2)
-3 -4 -3 4 |-15 ---(3)
5 2 -2 3 |-20 ---(4)
(3)x2, (4)x2
-2 6 -3 3 | -5 ---(5)
4 -3 2 -1 | -3 ---(6)
-6 -8 -6 8 |-30 ---(7)
10 4 -4 6 |-40 ---(8)
(6) + 2x(5), (7) - 3x(5), (8) + 5x(5)
-2 6 -3 3 | -5 ---(9)
0 9 -4 5 |-13 ---(10)
0-26 3 -1 |-15 ---(11)
0 34-19 21|-65 ---(12)
(12) + (11), (11) + 3x(10)
-2 6 -3 3 | -5 ---(13)
0 9 -4 5 |-13 ---(14)
0 1 -9 14 |-54 ---(15)
0 8-16 20 |-80 ---(16)
(14) - 9x(15), (16) - 8x(15)
-2 6 -3 3 | -5 ---(17)
0 9 77-121|473 ---(18)
0 1 -9 14 |-54 ---(19)
0 0 56-92 |352 ---(20)
(18) - (20), (20)÷4
-2 6 -3 3 | -5 ---(21)
0 0 21-29 |473 ---(22)
0 1 -9 14 |-54 ---(23)
0 0 14-23 | 88 ---(24)
(22) - (24)
-2 6 -3 3 | -5 ---(25)
0 0 7 -6 | 33 ---(26)
0 1 -9 14 |-54 ---(27)
0 0 14-23 | 88 ---(28)
(28) - 2x(26)
-2 6 -3 3 | -5 ---(29)
0 0 7 -6 | 33 ---(30)
0 1 -9 14 |-54 ---(31)
0 0 0-11 | 22 ---(32)
(32)÷ -11,
-2 6 -3 3 | -5 ---(33)
0 0 7 -6 | 33 ---(34)
0 1 -9 14 |-54 ---(35)
0 0 0 1 | 2 ---(36)
(33) - 3x(36), (34) + 6x(36), (35) - 14x(36)
-2 6 -3 0 | 1 ---(37)
0 0 7 0 | 21 ---(38)
0 1 -9 0 |-26 ---(39)
0 0 0 1 | 2 ---(40)
(38)÷7,
-2 6 -3 0 | 1 ---(41)
0 0 1 0 | 3 ---(42)
0 1 -9 0 |-26 ---(43)
0 0 0 1 | 2 ---(44)
(41) + 3x(42), (43) + 9x(42)
-2 6 0 0 | 10 ---(45)
0 0 1 0 | 3 ---(46)
0 1 0 0 | 1 ---(47)
0 0 0 1 | 2 ---(48)
(45) - 6x(47)
-2 0 0 0 | 4 ---(49)
0 0 1 0 | 3 ---(50)
0 1 0 0 | 1 ---(51)
0 0 0 1 | 2 ---(52)
(49)÷ -2
1 0 0 0 | -2 ---(53)
0 0 1 0 | 3 ---(54)
0 1 0 0 | 1 ---(55)
0 0 0 1 | 2 ---(56)
Rearrange,
1 0 0 0 | -2 ---(57)
0 1 0 0 | 1 ---(58)
0 0 1 0 | 3 ---(59)
0 0 0 1 | 2 ---(60)
The solution can now be read off as (-2, 1, 3, -2)
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The inverse matrix is,
|-29 -12 -9 3|
|33 22 -11 -11|
|181 256 19 -121|
|147 196 35 -77|
with a coefficient of (1/154).
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