Three Elementary Operations on Systems of Linear Equations
Two systems of linear equations are equivalent if they have exactly the same solution set. Starting from any system, three elementary operations can be applied to produce an equivalent system. These operations are the foundation of the method of elimination and of Gaussian elimination.
Table of Contents
The Three Elementary Operations on a System of Equations
The three elementary operations that can be performed on any system of linear equations to produce an equivalent system (same solution set) are:
- Interchange any two equations in the system.
- Multiply all terms of any one equation by a non-zero constant.
- Replace one equation by the sum of that equation and a multiple of another equation in the system.
Note: Combining more than two equations at once (e.g., replacing one equation with a linear combination of all three) is a sequence of elementary operations, not a single one. It preserves the solution set only when the resulting system remains equivalent — which requires care.
Examples with Detailed Solutions
Example 1: Applying Each Elementary Operation Individually
Consider the system:
\[ \begin{cases} 2x - y - z = -2 \quad (E_1)\\ -x - 2y + z = -1 \quad (E_2)\\ -3x + 3y - z = 2 \quad (E_3) \end{cases} \]which has the known solution \((0, 1, 1)\).
- a) Interchange \(E_1\) and \(E_3\), and verify that \((0, 1, 1)\) remains a solution.
- b) Multiply \(E_1\) by \(2\) and \(E_2\) by \(-1\), and verify the solution.
- c) Replace \(E_3\) by \(E_3 + E_2\), and verify the solution.
Solution
a) Interchange \(E_1\) and \(E_3\):
The equivalent system is:
\[ \begin{cases} -3x + 3y - z = 2 \\ -x - 2y + z = -1 \\ 2x - y - z = -2 \end{cases} \]Substitute \((0, 1, 1)\):
\[ \begin{cases} -3(0) + 3(1) - (1) = 2 \implies 2 = 2 \;\checkmark \\ -(0) - 2(1) + (1) = -1 \implies -1 = -1 \;\checkmark \\ 2(0) - (1) - (1) = -2 \implies -2 = -2 \;\checkmark \end{cases} \]The solution is preserved.
b) Multiply \(E_1\) by \(2\) and \(E_2\) by \(-1\):
\[ \begin{cases} 4x - 2y - 2z = -4 \\ x + 2y - z = 1 \\ -3x + 3y - z = 2 \end{cases} \]Substitute \((0, 1, 1)\):
\[ \begin{cases} 4(0) - 2(1) - 2(1) = -4 \implies -4 = -4 \;\checkmark \\ (0) + 2(1) - (1) = 1 \implies 1 = 1 \;\checkmark \\ -3(0) + 3(1) - (1) = 2 \implies 2 = 2 \;\checkmark \end{cases} \]The solution is preserved.
c) Replace \(E_3\) by \(E_3 + E_2\):
Compute the new equation:
\[ (-3x + 3y - z) + (-x - 2y + z) = 2 + (-1) \] \[ -4x + y = 1 \]The equivalent system is:
\[ \begin{cases} 2x - y - z = -2 \\ -x - 2y + z = -1 \\ -4x + y = 1 \end{cases} \]Substitute \((0, 1, 1)\) into the new third equation:
\[ -4(0) + (1) = 1 \implies 1 = 1 \;\checkmark \]The solution is preserved. \(E_3\) has been replaced by the new equation; the original \(E_3\) is no longer part of the system.
Example 2: A Sequence of Elementary Operations
Consider the system:
\[ \begin{cases} x - 4y + z = -4 \quad (E_1)\\ -2x + y + 3z = 6 \quad (E_2)\\ 4x - 2y - z = -7 \quad (E_3) \end{cases} \]which has the solution \((-1, 1, 1)\). Use a sequence of elementary operations to combine \(3E_2\) and \(-4E_3\) into a single new equation, and verify that the solution is preserved.
Solution
Step 1: Multiply \(E_2\) by \(3\) and \(E_3\) by \(-4\) (two applications of Operation 2):
\[ \begin{cases} -6x + 3y + 9z = 18 \\ -16x + 8y + 4z = 28 \end{cases} \]Step 2: Add the two resulting equations (Operation 3):
\[ (-6x + 3y + 9z) + (-16x + 8y + 4z) = 18 + 28 \] \[ -22x + 11y + 13z = 46 \]Step 3: Replace \(E_3\) with this new equation (the system becomes):
\[ \begin{cases} x - 4y + z = -4 \\ -2x + y + 3z = 6 \\ -22x + 11y + 13z = 46 \end{cases} \]Substitute \((-1, 1, 1)\) into the new equation:
\[ -22(-1) + 11(1) + 13(1) = 22 + 11 + 13 = 46 \implies 46 = 46 \;\checkmark \]The solution is preserved. Note that the operation was implemented as a sequence of elementary operations, not as a single "multi-equation" operation.
Example 3: Combining Operations on a Consistent System
Consider the system:
\[ \begin{cases} x - y + z = 3 \quad (E_1)\\ 4x + y + 3z = 7 \quad (E_2)\\ -12x - 4y + 4z = -20 \quad (E_3) \end{cases} \]which has the solution \((2, -1, 0)\). Perform the following sequence of operations and verify the solution:
- Replace \(E_3\) with \(-3E_1 + 2E_2 - \tfrac{1}{4}E_3\) (implemented as a sequence of elementary operations).
- Multiply \(E_2\) by \(-2\).
Solution
1) Replace \(E_3\) with \(-3E_1 + 2E_2 - \tfrac{1}{4}E_3\):
Scale each equation:
\[ \begin{aligned} -3E_1 &: -3x + 3y - 3z = -9 \\ 2E_2 &: 8x + 2y + 6z = 14 \\ -\tfrac{1}{4}E_3 &: 3x + y - z = 5 \end{aligned} \]Sum the three scaled equations:
\[ (-3x + 3y - 3z) + (8x + 2y + 6z) + (3x + y - z) = -9 + 14 + 5 \] \[ 8x + 6y + 2z = 10 \]Replace \(E_3\) with this new equation. The system becomes:
\[ \begin{cases} x - y + z = 3 \\ 4x + y + 3z = 7 \\ 8x + 6y + 2z = 10 \end{cases} \]2) Multiply \(E_2\) by \(-2\):
\[ -8x - 2y - 6z = -14 \]The updated system is:
\[ \begin{cases} x - y + z = 3 \\ -8x - 2y - 6z = -14 \\ 8x + 6y + 2z = 10 \end{cases} \]Substitute \((2, -1, 0)\):
\[ \begin{cases} 2 - (-1) + 0 = 3 \implies 3 = 3 \;\checkmark \\ -8(2) - 2(-1) - 6(0) = -14 \implies -14 = -14 \;\checkmark \\ 8(2) + 6(-1) + 2(0) = 10 \implies 10 = 10 \;\checkmark \end{cases} \]The solution \((2, -1, 0)\) is preserved. However, note that the combined operation in step 1 is a sequence of elementary operations; whether the final system is fully equivalent (i.e., has no additional solutions) depends on whether the coefficient matrix remains invertible. In this case, the solution set is preserved.
Conclusion
The three elementary operations — interchange, scalar multiplication, and replacement by a sum with a multiple of another equation — each produce an equivalent system with the same solution set. Any sequence of these operations also preserves the solution set. These operations are the building blocks used to solve systems of equations by elimination and to perform Gaussian elimination on augmented matrices.