SOLUTION: hello given the following system of three equations 3x-2y+z=-12 -x+y-2z=10 4x+3y+2z= -1 find the values of x, y, and z by i)cramers rule ii)matrix inversion

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Question 243147: hello
given the following system of three equations
3x-2y+z=-12
-x+y-2z=10
4x+3y+2z= -1

find the values of x, y, and z by i)cramers rule ii)matrix inversion

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
hello
given the following system of three equations
3x-2y+z=-12
-x+y-2z=10
4x+3y+2z= -1
find the values of x, y, and z by i)cramers rule ii)matrix inversion

Cramer's rule:

I will assume you already know how to get the value of a 3x3
determinant.  If you don't know how, post again asking how
it's done.

There are 4 columns,

1. The column of x-coefficients matrix%283%2C1%2C3%2C-1%2C4%29

2. The column of y-coefficients matrix%283%2C1%2C-2%2C1%2C3%29

3. The column of z-coefficients matrix%283%2C1%2C1%2C-2%2C2%29 

4. The column of constants:     matrix%283%2C1%2C-12%2C10%2C-1%29

There are four determinants:

1. The determinant D consists of just the three columns
of x, y, and z coefficients. in that order, but does not
contain the column of constants.

D=abs%28matrix%283%2C3%2C3%2C-2%2C1%2C-1%2C1%2C-2%2C4%2C3%2C2%29%29. 

It has value D=29. 

2. The determinant D%5Bx%5D is like the determinant D
except that the column of x-coefficients is replaced by the
column of constants.  D%5Bx%5D does not contain the column 
of x-coefficients.

D%5Bx%5D=abs%28matrix%283%2C3%2C-12%2C-2%2C1%2C10%2C1%2C-2%2C-1%2C3%2C2%29%29.

It has value D%5Bx%5D=-29.

3. The determinant D%5By%5D is like the determinant D
except that the column of y-coefficients is replaced by the
column of constants.  D%5By%5D does not contain the column 
of y-coefficients.

D%5By%5D=abs%28matrix%283%2C3%2C3%2C-12%2C1%2C-1%2C10%2C-2%2C4%2C-1%2C2%29%29.

It has value D%5By%5D=87.

4. The determinant D%5Bz%5D is like the determinant D
except that the column of z-coefficients is replaced by the
column of constants.  D%5Bz%5D does not contain the column 
of z-coefficients.

D%5Bz%5D=abs%28matrix%283%2C3%2C3%2C-2%2C12%2C-1%2C1%2C10%2C4%2C3%2C-1%29%29.

It has value D%5Bx%5D=-87.

Now the formulas for x, y and z are

x=D%5Bx%5D%2FD=-29%2F29=-1
y=D%5By%5D%2FD=87%2F29=3
x=D%5Bz%5D%2FD=-87%2F3=-3

---------------------------

By matrix inversion.

I will assume you already know how to find the inverse
of a matrix, and how to multiply two matrices. If you don't, 
post again asking how.

First we form three matrices, A, X, and B.

1. Matrix A is the 3x3 coefficient matrix A, which consists 
of just the three columns of x, y, and z coefficients. in 
that order, but does not contain the column of constants.

A=%28matrix%283%2C3%2C3%2C-2%2C1%2C-1%2C1%2C-2%2C4%2C3%2C2%29%29. 

2. Matrix X is the 3x1 matrix of variables X=%28matrix%283%2C1%2Cx%2Cy%2Cz%29%29

3. Matrix B is the 3x1 matrix, whose only column is the
column of constants: %28matrix%283%2C1%2C-12%2C10%2C-1%29%29

Next we form the matrix equation:

       A%2AX+=+B

or



To solve the equation

       A%2AX+=+B

we left-multiply both sides by A%5E%28-1%29, the inverse of A.

 A%5E%28-1%29%2A%28A%2AX%29+=+A%5E%28-1%29%2AB

Then since the associatitive principle holds for matrix multiplication,
(even though the commutative principle DOES NOT!!!), we can move
the parentheses on the left around the first two matrix factors:

%28A%5E%28-1%29%2AA%29%2AX+=+A%5E%28-1%29%2AB

Now since %28A%5E%28-1%29%2AA%29=I, where I is the identity matrix, the
above becomes:

I%2AX+=+A%5E%28-1%29%2AB

and by the identity property:

X=A%5E%28-1%29%2AB

Performing these operations with the actual matrices we have
the equation AX=B



Next we form the inverse of A, which is written A-1.



Then we indicate the left multiplication of both sides by
A%5E%28-1%29 to get the equation A%5E%28-1%29%28A%2AX%29=A%5E%28-1%29B:



Next we use the associative principle to move the parentheses so that
they are around the first two factors to get the equation %28A%5E%28-1%29%2AA%29%2AX=A%5E%28-1%29%2AB:



Now we perform the actual multiplications and we get the equation IX=A%5E%28-1%29%2AB:



Then when we multiply the identity matrix I by the column matrix of
variables, we just get the matrix of variables, or the 
equation X=A%5E%28-1%29B

%28matrix%283%2C1%2Cx%2Cy%2Cz%29%29=%28matrix%283%2C1%2C-1%2C3%2C-3%29%29

Edwin