Three Elementary Operations on Systems

\( \) \( \) \( \) \( \)
Two systems of equations are equivalent if they have the same solutions. Given a system of linear equations,
three different types of operations may be performed on it in order to obtain a system that has the same solutions. Using these operations, that are presented below, we may be able to solve systems of equations by elimination .

Elementary Operations on Systems of Equations

The three operations that may be performed on any system of linear equations and obtain an equivalent system are:

  1. Interchange any two equations
  2. Multiply all terms of any equation in the system by any number not equal to zero
  3. Add/Subtract (right sides together and left sides together) any two equations


Examples with Solutions

Example 1
Given that the following system of equations \( \left\{ \begin{array}{lcl} 2x - y - z& = & -2 \\ -x - 2y + z& = & -1 \\ -3x + 3y - z & = & 2 \end{array} \right. \) has the solution \( (0,1,1) \)

a) Interchange the top and the third equations of the given system and show that the solution of the system obtained is \( (0,1,1) \)
b) Multiply the top equation by \( 2 \) and the second equation by \( -1 \) of the given system and show that the solution of the system obtained is \( (0,1,1) \)
c) Add the second equation to the third equation of the given system and show that the solution of the system obtained is \( (0,1,1) \)
Solution to Example 1
a)
Interchange the top equation and the third equation: \( \left\{ \begin{array}{lcl} -3x + 3y - z & = & 2 \\ -x - 2y + z& = & -1 \\ 2x - y - z& = & -2 \end{array} \right. \)

Make substitutions of variables by their values \( (0,1,1) \) in the equations
\( \left\{ \begin{array}{lcl} -3(0) + 3(1) - (1) & = & 2 \\ -(0) - 2(1) + (1)& = & -1 \\ 2(0) - (1) - (1)& = & -2 \end{array} \right. \)

Simplify
\( \left\{ \begin{array}{lcl} 2 & = & 2 \\ - 1& = & -1 \\ - 2 & = & -2 \end{array} \right. \)

Hence \( (0,1,1) \) is the solution the system after the changes.

b)
Multiply the top equation by \( 2 \) and the second equation by \( -1 \): \( \left\{ \begin{array}{lcl} (2x - y - z& = & -2) \times 2 \\ (-x - 2y + z& = & -1) \times (-1) \\ -3x + 3y - z & = & 2 \end{array} \right. \)

Simplify
\( \left\{ \begin{array}{lcl} 4x - 2y - 2z& = & - 4 \\ x + 2y - z& = & 1 \\ -3x + 3y - z & = & 2 \end{array} \right. \)

Make substitutions of variables by their values \( (0,1,1) \) in the equations
\( \left\{ \begin{array}{lcl} 4(0) - 2(1) - 2(1)& = & - 4 \\ (0) + 2(1) - (1)& = & 1 \\ -3(0) + 3(1) - (1) & = & 2 \end{array} \right. \)

Simplify
\( \left\{ \begin{array}{lcl} -4& = & - 4 \\ 1& = & 1 \\ 2 & = & 2 \end{array} \right. \)

Hence \( (0,1,1) \) is the solution to the system after the changes made above.

c)
Add the second equation to the third to obtain a new system
\( \left\{ \begin{array}{lcl} 2x - y - z& = & -2 \\ -x - 2y + z& = & -1 \\ -4x + y & = & 1 \end{array} \right. \)
We already know that \( (0,1,1) \) is a solution to the first and second equations since they have not been modified.
Make substitutions of variables by their values \( (0,1,1) \) in the third equation
\( -4(0) + (1) = 1 \)
Simplify
\( 1 = 1 \)
Hence \( (0,1,1) \) is the solution to the system after changes.


Note that the three elementary operations may be combined as shown in examples 2 and 3 below.

Example 2
Given the following system of equations \( \left\{ \begin{array}{lcl} x - 4 y + z& = & - 4 \\ -2x + y + 3z& = & 6\\ 4x - 2y - z & = & -7 \end{array} \right. \) has the solution \( (-1,1,1) \)
Add \( 3 \) times the second equation to \( - 4 \) times the third equation and show that the solution of the system obtained is \( (-1,1,1) \)
Solution to Example 2
Add \( 3 \) times the second equation to \( - 4 \) times the third equation
\( (-2x + y + 3z = 6) \times 3 \\ + \\ (4x - 2y - z = -7) \times (-4) \)

Multiply
\( -6x + 3y + 9z = 18 \\ + \\ -16x + 8y + 4z = 28 \)

Add the two equations (left sides together and right sides together)
\( -6x + 3y + 9z + (-16x + 8y + 4z) = 18 + 28 \)

Group like terms and simplify
\( -22x + 11y + 13 z = 46 \)

After the changes, the system becomes
\( \left\{ \begin{array}{lcl} x - 4 y + z& = & - 4 \\ -2x + y + 3z& = & 6\\ -22x + 11y + 13 z & = & 46 \end{array} \right. \)

Substitute the variables by their values \( (-1,1,1) \) in the third equation
\( -22(-1) + 11(1) + 13 (1) = 46 \)

Simplify
\( 46 = 46 \)
Hence \( (-1,1,1) \) is a solution to the system after the changes made. We do not need to check that \( (-1,1,1) \) is a solution to the first and second equations because they have not been modified.


Example 3
Given the following system of equations \( \left\{ \begin{array}{lcl} x - y + z& = & 3 \\ 4x + y + 3z& = & 7\\ -12x - 4 y + 4 z & = & -20 \end{array} \right. \) has the solution \( (2,-1,0) \)

Make the following changes simultaneously
1) Add \( -3 \) times the top equation and \( 2 \) times the second equation to \( -\dfrac{1}{4} \) the third equation
2) Multiply the second equation by \( -2 \)
and show that the solution of the system obtained is \( (2,-1,0) \)
Solution to Example 3
1) First change
Add \( -3 \) times the top equation and \( 2 \) times the second equation to \( -\dfrac{1}{4} \) the third equation
\( (x - y + z = 3) \times (-3) \\ + \\ (4x + y + 3z = 7) \times (2) \\ +\\ (-12x - 4 y + 4 z = -20) \times (-\dfrac{1}{4}) \)

Multiply
\( -3x +3 y -3z = -9 \\ + \\ 8x + 2y + 6z = 14 \\ +\\ 3x + y - z = 5 \)

Add the left sides together and the right sides together of the above equations
\( -3x +3 y -3z + (8x + 2y + 6z) + (3x + y - z) = -9 + 14 + 5 \)

Group like terms and simplify
\( 8x + 6 y + 2 z = 10 \)

2) Second change
Multiply the second equation by \( -2 \)
\( (4x + y + 3z = 7) \times (-2) \)

Simplify
\( -8x - 2y - 6z = -14 \)

After the two changes requested, the system becomes
\( \left\{ \begin{array}{lcl} x - y + z& = & 3 \\ -8x - 2y - 6z& = & -14\\ 8x + 6 y + 2 z & = & 10 \end{array} \right. \)

Substitute the variables by their values \( (2,-1,0) \) in the system
\( \left\{ \begin{array}{lcl} 2 - (-1) + 0& = & 3 \\ -8(2) - 2(-1) - 6(0)& = & -14\\ 8(2) + 6(-1) + 2(0) & = & 10 \end{array} \right. \)

Simplify left sides in the above
\( \left\{ \begin{array}{lcl} 3& = & 3 \\ -14& = & -14\\ 10 & = & 10 \end{array} \right. \)

Hence \( (2,-1,0) \) is a solution to the system after the changes made. We do not need to check that \( (2,-1,0) \) is a solution to the first and second equations because they have not been modified.

Conclusion

If any of the elementary operations or any combinations of them is applied to a system of equations, we obtain an equivalent system of equations having the same solution as the original one. These three elementary operations are used to solve systems of equations by the method of elimination.



More References and links

  1. linear algebra
  2. Solve a system of linear equations by elimination
  3. inverse of a matrix
  4. elementary matrices
  5. Gaussian Elimination


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