Solving A 3 Variable System Of Equations

Solving a 3 variable system ofequations involves finding the unique intersection point of three planes in three‑dimensional space. Day to day, this process extends the familiar techniques used for two‑variable linear systems, such as substitution and elimination, to a higher dimension while preserving the core ideas of consistency, independence, and uniqueness. So naturally, in this article we will explore the conceptual background, present a clear step‑by‑step methodology, illustrate the approach with a concrete example, and answer common questions that arise when tackling these systems. By the end, readers will have a solid grasp of how to solve a 3 variable system of equations confidently and accurately.

Introduction

A 3 variable system of equations typically takes the form

[\begin{cases} a_1x + b_1y + c_1z = d_1 \ a_2x + b_2y + c_2z = d_2 \ a_3x + b_3y + c_3z = d_3\end{cases} ]

where (x, y,) and (z) are the unknowns and the coefficients (a_i, b_i, c_i, d_i) are known numbers. The solution, if it exists, is a triple ((x, y, z)) that satisfies all three equations simultaneously. Determining this triple requires systematic manipulation of the equations to reduce the problem to simpler, more manageable forms. The most common strategies are substitution, elimination, and matrix operations (including the use of the inverse matrix). Each method relies on the same fundamental principle: performing operations that preserve the solution set while gradually eliminating variables until only one remains.

Methods of Solving

Substitution The substitution method isolates one variable in a single equation and then replaces that variable in the other two equations. This creates a new 2‑variable system that can be solved using the same substitution technique or by elimination.

1. Choose an equation with a coefficient of 1 (or a simple coefficient) for one variable. 2. Solve for that variable in terms of the other two.
2. Substitute the expression into the remaining equations.
3. Solve the resulting 2‑variable system using substitution again or elimination.
4. Back‑substitute to find the original variable.

Advantages: Conceptually straightforward; useful when one equation is already simple.
Limitations: Can become algebraically messy if the chosen equation has large or fractional coefficients.

Elimination (Gaussian Elimination)

Elimination systematically removes variables by adding or subtracting multiples of equations. The goal is to produce an upper‑triangular form, where each subsequent equation contains fewer variables than the previous one.

1. Align the equations so that like terms are vertically stacked.
2. Create a leading 1 (or a convenient pivot) in the first equation, if possible.
3. Eliminate the first variable from the second and third equations by adding suitable multiples. 4. Repeat the process for the second variable using the new second equation.
4. Back‑substitute to obtain the values of the remaining variables.

This approach is especially powerful when dealing with systems that have many equations or when using computational tools.

Matrix Inverse Method

When the coefficient matrix (A) is square and invertible, the system can be written as

[ A\mathbf{x} = \mathbf{b} ]

where (\mathbf{x} = \begin{bmatrix}x \ y \ z\end{bmatrix}) and (\mathbf{b} = \begin{bmatrix}d_1 \ d_2 \ d_3\end{bmatrix}). If (A^{-1}) exists, the solution is simply

[ \mathbf{x} = A^{-1}\mathbf{b} ]

The steps are:

1. Form the coefficient matrix (A) from the left‑hand side coefficients.
2. Compute the inverse (A^{-1}) (using row reduction or a formula for 3×3 matrices).
3. Multiply (A^{-1}) by the constant vector (\mathbf{b}) to obtain (\mathbf{x}).

Key point: The matrix must be non‑singular (determinant ≠ 0) for the inverse to exist; otherwise, the system may have infinitely many or no solutions.

Step‑by‑Step Example

Consider the following system:

[ \begin{cases} 2x + y - z = 3 \ -,x + 4y + 2z = 5 \ 3x - y + 4z = 7 \end{cases} ]

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We will solve it using the elimination method.

1. Write the augmented matrix:

[ \left[\begin{array}{ccc|c} 2 & 1 & -1 & 3 \ -1 & 4 & 2 & 5 \ 3 & -1 & 4 & 7 \end{array}\right] ]

2. Make the first pivot 1 by dividing the first row

Continuing the EliminationExample

3. Scale the first row to obtain a leading 1 (optional but convenient).
   [ R_1 \leftarrow \tfrac{1}{2}R_1 ;\Longrightarrow; \left[\begin{array}{ccc|c} 1 & \tfrac12 & -\tfrac12 & \tfrac32 \ -1 & 4 & 2 & 5 \ 3 & -1 & 4 & 7 \end{array}\right] ]

4. Eliminate the (x)-terms from the second and third rows by adding suitable multiples of (R_1).

   • For the second row: [ R_2 \leftarrow R_2 + R_1 ] [ \left[\begin{array}{ccc|c} 1 & \tfrac12 & -\tfrac12 & \tfrac32 \ 0 & 4+\tfrac12 & 2-\tfrac12 & 5+\tfrac32 \ 3 & -1 & 4 & 7 \end{array}\right] ;\Longrightarrow; \left[\begin{array}{ccc|c} 1 & \tfrac12 & -\tfrac12 & \tfrac32 \ 0 & \tfrac{9}{2} & \tfrac{3}{2} & \tfrac{13}{2} \ 3 & -1 & 4 & 7 \end{array}\right] ]

   • For the third row:
     [ R_3 \leftarrow R_3 - 3R_1 ] [ \left[\begin{array}{ccc|c} 1 & \tfrac12 & -\tfrac12 & \tfrac32 \ 0 & \tfrac{9}{2} & \tfrac{3}{2} & \tfrac{13}{2} \ 0 & -\tfrac{5}{2} & \tfrac{11}{2} & \tfrac{5}{2} \end{array}\right] ]

5. Clear the fractions in the second and third rows by multiplying each by 2 (this does not affect the solution set).

   [ \left[\begin{array}{ccc|c} 1 & \tfrac12 & -\tfrac12 & \tfrac32 \ 0 & 9 & 3 & 13 \ 0 & -5 & 11 & 5 \end{array}\right] ]

6. Create a pivot in the second row (already 9) and eliminate the (y)-term from the third row.

   • Compute the multiplier: (\displaystyle m = \frac{-5}{9}).
   • Update the third row: [ R_3 \leftarrow R_3 + mR_2 ] [ \left[\begin{array}{ccc|c} 1 & \tfrac12 & -\tfrac12 & \tfrac32 \ 0 & 9 & 3 & 13 \ 0 & 0 & 11 - \tfrac{5}{3} & 5 + \tfrac{65}{9} \end{array}\right] ] Simplifying the arithmetic:
     [ 11 - \tfrac{5}{3}= \tfrac{33-5}{3}= \tfrac{28}{3},\qquad 5 + \tfrac{65}{9}= \tfrac{45+65}{9}= \tfrac{110}{9}= \tfrac{110}{9}. ] Thus the augmented matrix becomes
     [ \left[\begin{array}{ccc|c} 1 & \tfrac12 & -\tfrac12 & \tfrac32 \ 0 & 9 & 3 & 13 \ 0 & 0 & \tfrac{28}{3} & \tfrac{110}{9} \end{array}\right]. ]

7. Scale the third row to obtain a leading 1 for (z).
   [ R_3 \leftarrow \frac{3}{28}R_3 ;\Longrightarrow; \left[\begin{array}{ccc|c} 1 & \tfrac12 & -\tfrac12 & \tfrac32 \ 0 & 9 & 3 & 13 \ 0 & 0 & 1 & \tfrac{110}{84}= \tfrac{55}{42} \end{array}\right]. ]

8. Back‑substitute to eliminate (z) from the second row and then (y) from the first row.

   • Eliminate (z) from the second row:
     [ R_2 \leftarrow R_2 - 3R_3 ] [ \left[\begin{array}{ccc|c} 1 & \tfrac12 & -\tfrac1

Continuing the process, we verify that all equations are consistent and arrive at a unique solution. The final system reveals a clear pattern emerging from the row operations, guiding us toward the exact values of the variables.

• From the third row, we have (z = \tfrac{55}{42}).
• Substituting back, we solve for (y) using the second row and then for (x).
• The resulting solution highlights the consistency of the transformations applied throughout.

This sequence of operations demonstrates the power of Gaussian elimination in simplifying complex linear systems. Each step refines the matrix, bringing clarity to the underlying relationships among the variables.

Boiling it down, by methodically applying row operations, we not only transformed the matrix but also uncovered the structure of the solution space. In real terms, this approach is essential for tackling similar problems in advanced linear algebra. Conclusively, mastering these techniques empowers us to figure out detailed mathematical challenges with confidence.

Conclusion: The systematic application of row operations leads to a well-defined solution, reinforcing our understanding of matrix transformations and their practical implications.