Download Solving Systems of Equation in Three Variables

MATRICES
Using matrices to solve
Systems of Equations
Solving Systems with Matrices
We can use matrices to solve systems that
involve 2 x 2 (2 equations, 2 variables) and 3
x 3 (3 equations, 3 variables) systems. We
will look at two methods:
•Cramer’s Rule (uses determinants)
•Matrix Equations (uses inverse matrices)
Cramer’s Rule - 2 x 2


Cramer’s Rule relies on determinants
Consider the system below with
variables x and y:
a1 x  b1 y  C1
a2 x  b2 y  C2
Cramer’s Rule - 2 x 2

The formulae for the values of x and y are
shown below. The numbers inside the
determinants are the coefficients and
constants from the equations.
C1 b1
C2 b2
x
a1 b1
a2 b2
a1 C1
a2 C2
y
a1 b1
a2 b2
Cramer’s Rule - 3 x 3

Consider the 3 equation system below with
variables x, y and z:
a1 x  b1 y  c1z  C1
a2 x  b2 y  c2 z  C2
a3 x  b3 y  c3 z  C 3
Cramer’s Rule - 3 x 3

The formulae for the values of x, y and z are
shown below. Notice that all three have the
same denominator.
C1 b1 c1
C2 b2 c2
C3 b3 c3
x
a1 b1 c1
a2 b2 c2
a3 b3 c3
a1 C1 c1
a2 C2 c2
a 3 C 3 c3
y
a1 b1 c1
a2 b2 c2
a3 b3 c3
a1 b1 C1
a2 b2 C2
a3 b3 C3
z
a1 b1 c1
a2 b2 c2
a3 b3 c3
Cramer’s Rule


Not all systems have a definite solution. If the
determinant of the coefficient matrix is zero, a
solution cannot be found using Cramer’s Rule because
of division by zero.
When the solution cannot be determined, one of two
conditions exists:


The planes graphed by each equation are parallel and there
are no solutions.
The three planes share one line (like three pages of a book
share the same spine) or represent the same plane, in which
case there are infinite solutions.
Cramer’s Rule

Example:
Solve the system
9
5
2
x
3
1
1
2
2
1
2
2
1
1
2
4 23

1
1
23
2
4
3x - 2y + z = 9
x + 2y - 2z = -5
x + y - 4z = -2
3 9
1
1 5 2
1 2 4
69
y

 3
3 2 1
23
1 2 2
1 1 4
Cramer’s Rule
3x - 2y + z = 9
x + 2y - 2z = -5
x + y - 4z = -2
3 2 9
1 2 5
1 1 2
0
z

0
3 2 1
23
1 2 2
1 1 4
The solution is
(1, -3, 0)
Matrix Equations

Step 1: Write the system as a matrix
equation. A three equation
system is shown below.
a1 x  b1 y  c1z  C1
a2 x  b2 y  c2 z  C2
a3 x  b3 y  c3 z  C 3
 a1
a
 2
 a3
b1
b2
b3
c1   x   C1 
c2   y   C2 
c3   z   C3 
Matrix Equations

Step 2: Find the inverse of the
coefficient matrix.

This can be done by hand for a 2 x 2
matrix; most graphing calculators can
find the inverse of a larger matrix.
Matrix Equations

Step 3: Multiply both sides of the matrix
equation by the inverse.
The inverse of the coefficient matrix times
the coefficient matrix equals the identity matrix.
 x   a1 b1
 y   a b
   2 2
 z   a3 b3
1
c1 
 C1 



c2   C2 
 C3 
c3 
Note: The multiplication order on the right side is very
important. We cannot multiply a 3 x 1 times a 3 x 3 matrix!
Matrix Equations

Example: Solve the system
 3 2   x   9 
3x - 2y = 9
1 2   y    5 

   
x + 2y = -5
1
 3 2 
1  2 2
1 2   8  1 3




x 1  2 2  9 
 y   8  1 3   5 
 

 
Matrix Equations
x 1  2 2  9 
 y   8  1 3   5 
 

 
Multiply the matrices (a ‘2 x 2’ times a
‘2 x 1’) first, then distribute the scalar.
x 1  8 
 y   8  24 
 


x  1 
 y    3
   
Matrix Equations

Example #2: Solve the 3 x 3 system
 3 2 1   x   9 
1 2 2   y    5 

   
1 1 4   z   2 
3x - 2y + z = 9
x + 2y - 2z = -5
x + y - 4z = -2
Using a graphing calculator:
1
 236
 3 2 1 
1 2 2     2
 23


 231
1 1 4 
7
23
13
23
5
23
 232 

 237 
 238 
Matrix Equations
 x   236
 y    2
   23
 z   231
 x  1 
 y    3
   
 z   0 
7
23
13
23
5
23
 232 
7 
 23 
 238 
9
 5
 
 2