Download How do you write a system of equations as a matrix

Question 1: How do you write a system of equations as a matrix equation?
In section 3.2, you learned how to multiply two matrices. This process involved
multiplying the entries in the row of one matrix by the columns of another matrix and
then adding those products. For instance, suppose we multiply
1 2   5 
3 4   6 

 
Since we are multiplying a 2 x 2 matrix by a 2 x 1 matrix, the product is a 2 x 1 matrix.
Working out the product, we get
1 2  5  1  5  2  6 
3 4   6   3  5  4  6 

  

When the product is written this way, we can easily identify the product of the numbers
in the rows of the first matrix with the numbers in the second matrix as well as the sum.
Instead of multiplying the first matrix by a constant column matrix, let’s try multiplying by
a matrix containing two variables x and y,
x
 y
 
When we carry out the multiplication now, we get
1 2   x  1x  2 y 
 3 4   y   3 x  4 y 

  

The entries in the product look a lot like the part of a system of equations. In fact, if we
set this equal to a 2 x 1 matrix with constants,
1 2   x  17 
3 4   y   39 

   
we can write the matrix equation as an equivalent system of equations.
These matrices are equal when the corresponding entries are equal or when
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1x  2 y  17
3 x  4 y  39
In other words,
1x  2 y  17
1 2   x  17 
3 4   y   39  is the same as 3 x  4 y  39

   
When a system of equations is written in terms of matrices, we call it a matrix equation.
In this matrix equation, the three matrices are typically called A, X and B.
1 2   x  17 
3 4   y   39 

    A
X  B
The matrix A is called the coefficient matrix and it entries are the coefficients on the
variables when they are written in the same order in each equation. The matrix X is
called the variable matrix and contains the two variables in the problem. The matrix B is
called the constant matrix and contains the constants from the right hand side of the
matrix equation.
To match the different matrices with their entries, each equation must have the
variables on the left side with the variable terms listed in the same order. The constants
must be on the right side. Once each equation has this format, we can read the entries
in each matrix.
Example 1
Write a Linear System as a Matrix Equation
Write the system of linear equations
x  y  5
3x  4 y  1
as the matrix equation AX  B .
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Solution The system is in the proper format to determine the coefficient
matrix and the constant matrix. The variable matrix for this system is
 x
X  
 y
Using the coefficients on the variables, define the coefficient matrix and
constant matrix as
1 1 
A

3 4 
 5 
B 
1
With these definitions, the product AX is equal to
1 1   x   x  y 
AX  
   

3 4   y  3 x  4 y 
When this matrix is set equal to the constant matrix B, we get
 x  y   5 
3 x  4 y    1 

  
These matrices are equal when x  y  5 and 3 x  4 y  1 . This means
the matrix equation AX  B is equivalent to the original system of
equations.
We can use the same strategy to write larger systems of equations as matrix equations.
This leads to larger matrices. We incorporate different variable names into the variable
matrix X as shown in the next example.
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Example 2
Write a Linear System as a Matrix Equation
Write the system of linear equations
x1  x2  x3  2
2 x1  2 x2  3x3  1
2 x3  7  x1
as a matrix equation AX  B .
Solution Before we can define the matrices, we need to put each
equation in the proper format. In the first and second equations, all
variable terms are on one side of the equation and the constant terms
are on the other side of the equations. In the third equation, there are
variable terms on both sides of the equation. We can put the third
equation in the proper format by subtracting x1 from both sides of the
equation,
2 x3  x1  7  x1  x1
 x1  2 x3  7
This leads to the system
x1  x2  x3  2
2 x1  2 x2  3 x3  1
 x1
 2 x3  7
The variable matrix is
 x1 
X   x2 
 x3 
The coefficients on the variables give the coefficient matrix,
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 1 1 1
A   2 2 3
 1 0 2 
Note that any signs are included in the coefficients and variables that
are missing correspond to a coefficient of zero.
The constant matrix,
2
B   1
 7 
matches the constants on the right hand of the system.
We can check to see that the matrix equation AX  B is equivalent to
the system by carrying out the product AX :
 1 1 1  x1   x1  x2  x3 
AX   2 2 3  x2    2 x1  2 x2  3 x3 
 1 0 2   x3    x1  2 x3 
If we set this matrix equal to B,
 x1  x2  x3   2 
 2 x  2 x  3 x    1
2
3
 1
 
  x1  2 x3   7 
the corresponding entries must be equal. This is equivalent to the
original system after it had been modified to put it into the proper
format.
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Once a system of linear equations is written as a matrix equation AX  B , we can use
the inverse of the matrix A to find the solution to the system. In the next question, we’ll
learn how to do this and use it to solve an application.
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