SOLUTION: Can you please help me with this problem? Solve the following systems of equations using matrix multiplication. 2x-5y+7z=4 3x+y-12z=-8 5x+2y-4z=3

Algebra ->  Matrices-and-determiminant -> SOLUTION: Can you please help me with this problem? Solve the following systems of equations using matrix multiplication. 2x-5y+7z=4 3x+y-12z=-8 5x+2y-4z=3      Log On


   



Question 389772: Can you please help me with this problem?
Solve the following systems of equations using matrix multiplication.
2x-5y+7z=4
3x+y-12z=-8
5x+2y-4z=3

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
2x-5y+7z=4
3x+y-12z=-8
5x+2y-4z=3


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%2C2%2C-5%2C7%2C3%2C1%2C-12%2C5%2C2%2C-4%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%2C4%2C-8%2C3%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 associative 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 find 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:
 
A%5E%28-1%29 to get the equation A%5E%28-1%29%28A%2AX%29=A%5E%28-1%29B:
 

 
When we perform the matrix multiplication we get:



The matrix on the left is the identity matrix

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%2C1%2C1%2C1%29%29

or x=1, y=1, z=1
 
Edwin