Use Cramer's rule,if possible, to solve the system of linear equations. Algebra?

Question

x+y+z-2w=-4

2y+z+3w=4

2x+y-z+2w=5

x-y+w=4

Answer 1

x+y+z-2w=-4                                                              2y+z+3w=4    

ax+by=c

ax+by=c'

let find determining of the system:

D=a   b

     a'   b' =ab' -a'b

Dx=c   b

       c'  b'=cb'-c'b

Dy=a   c

       a'   c' =ac'- a'c

x=Dx/x =cb'-c'b/ab'-a'b

y=ac'-a'c/ab'-a'b

S={x;y} ={cb'-c'b /ab'-a'b; ac'-a'c /ab'-a'b}

                                                                                                                  Use these formulas ,hope it 'll help

Answer 2

you need to compute five  4 x 4 determinants....time consuming...-30 / -30 , 30 / -30, 0 / -30, -60/-30...if I made no mistakes

Answer 3

Solve, using Cramer:

x + y + z – 2w = –4;

2y + z + 3w = 4;

2x + y – z + 2w = 5;

x – y + w = 4.

Ted has altogether the correct answer: it is

x = 1; y = –1; z = 0; w = 2.

He is also right, in that it is jolly tedious to perform.

I want to show you something of Cramer’s method. It is undeniably laborious; in actual practice, I do strongly recommend another method, such as row operations with augmented matrices. But, for reasons having to do with the development of the theory of algebra (of interest more to mathematicians, probably, than to engineers), Cramer is in every way worthy of study, and I hope that you will benefit accordingly.

Of the five tedious 4 X 4 matrices of which determinants are required, I will go into detail with only one of them, and the rest, I think you will be able to see for yourself.

You will notice that we follow three rules:

(1) If you swap two rows in a matrix, the determinant will be unaffected.

(2) If you multiply one row by a constant c, leaving the other rows the same, the determinant will also be multiplied by c. You can do this for several rows at a time.

(3) If you change a row by adding to it any multiple of another row, the determinant will be unchanged. (Or you can subtract and get the same.)

First, you find the determinant of the co-efficient matrix:

(The numbers shown in round brackets are row numbers.)

(01) |+01 +01 +01 –02|

(02) |+00 +02 +01 +03|

(03) |+02 +01 –01 +02|

(04) |+01 –01 +00 +01|

= D.

(05) |+01 +01 +01 –02| (=(01))

(06) |+02 +01 –01 +02| (=(03))

(07) |+01 –01 +00 +01| (=(04))

(08) |+00 +02 +01 +03| (=(02))

= D. (This is equivalent to two swaps.)

(09) |+02 +02 +02 –04| (=(05) X 2)

(10) |+02 +01 –01 +02| (=(06))

(11) |+02 –02 +00 +02| (=(07) X 2)

(12) |+00 +02 +01 +03| (=(08))

= 4D. (This multiplies the previous determinant by 4.)

(13) |+02 +02 +02 –04| (=(09))

(14) |+00 –01 –03 +06| (=(10) - (09))

(15) |+00 –03 +01 +00| (=(11) - (10))

(16) |+00 +02 +01 +03| (=(12) - (11))

= 4D. (Previous determinant is unchanged.)

(17) |+02 +02 +02 –04| (=(13))

(18) |+00 +06 +18 –36| (=(14) X -6)

(19) |+00 +06 –02 +00| (=(15) X -2)

(20) |+00 +06 +03 +09| (=(16) X 3)

= 144D. (Previous determinant, times 36)

(21) |+02 +02 +02 –04| (=(17))

(22) |+00 +06 +18 –36| (=(18))

(23) |+00 +00 –20 +36| (=(19) - (18))

(24) |+00 +00 +05 +09| (=(20) - (19))

= 144D. (Previous determinant is unchanged.)

(25) |+01 +01 +01 –02| (=(21) / 2)

(26) |+00 +01 +03 –06| (=(22) / 6)

(27) |+00 +00 +05 –09| (=(23) / -4)

(28) |+00 +00 +05 +09| (=(24))

= -3D. (Previous determinant, divided by 48)

(29) |+01 +01 +01 –02| (=(25))

(30) |+00 +01 +03 –06| (=(26))

(31) |+00 +00 +05 –09| (=(27))

(32) |+00 +00 +00 +18| (=(28) - (27))

= -3D = 1 X 1 X 5 X 18 = 90.

Thus, D = -30.

Now, we take the determinants of the other four matrices:

Dx is obtained by replacing the first column with the constant vector, namely, -4, 4, 5, and 4, which appeared to the right of the equal marks in the original system.

|–04 +01 +01 –02|

|+04 +02 +01 +03|

|+05 +01 –01 +02|

|+04 –01 +00 +01|

= Dx.

Dy is obtained by replacing the second column with the same constant vector.

|+01 –04 +01 –02|

|+00 +04 +01 +03|

|+02 +05 –01 +02|

|+01 +04 +00 +01|

= Dy.

Dz and Dw are obtained by replacing the third and fourth columns with the same constant vector.

|+01 +01 –04 –02|

|+00 +02 +04 +03|

|+02 +01 +05 +02|

|+01 –01 +04 +01|

= Dz.

|+01 +01 +01 –04|

|+00 +02 +01 +04|

|+02 +01 –01 +05|

|+01 –01 +00 +04|

= Dw.

Using the same methods by which we found D, above, the determinant of the co-efficient matrix, we find that

Dx = -30, Dy = 30, Dz = 0, and Dw = -60.

I won’t trouble you with all the tedious calculations.

Here’s where we apply Cramer:

x = Dx/D = -30/-30 = 1;

y = Dy/D = 30/-30 = -1;

z = Dx/D = 0/-30 = 0; and

w = Dw/D = -60/-30 = 2.

Checking, we have:

x + y + z - 2w = 1 - 1 + 0 - 4 = -4.

2y + z + 3w = -2 + 0 + 6 = 4.

2x + y – z + 2w = 2 – 1 + 0 + 4 = 5.

x – y + w = 1 + 1 + 2 = 4.

As you can see, these give the right results.

I hope that you will find this of help.