Menyeselesaikan Solusi Matriks Segitiga Atas

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Apr 11, 2025

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Solving Upper Triangular Matrix Solutions: A Comprehensive Guide

Upper triangular matrices are a special type of square matrix where all the entries below the main diagonal are zero. Solving a system of linear equations represented by an upper triangular matrix is significantly simpler than solving a general system, thanks to the inherent structure of the matrix. This guide provides a comprehensive walkthrough of the process, explaining the method and offering helpful tips for accurate and efficient solutions.

Understanding Upper Triangular Matrices

Before diving into the solution process, let's solidify our understanding of upper triangular matrices. Consider a square matrix A of size n x n. A is considered upper triangular if:

• a<sub>ij</sub> = 0 for all i > j (entries below the main diagonal are zero).

For example:

[ 2  3  1 ]
[ 0  4 -2 ]
[ 0  0  1 ]

is an upper triangular matrix.

Back Substitution: The Core Method

The key to solving a system of linear equations represented by an upper triangular matrix is a technique called back substitution. This method iteratively solves for the unknowns, starting from the last variable and working backward.

Let's consider a system of equations represented by an upper triangular matrix A and a solution vector b:

Ax = b

Where:

• A is an n x n upper triangular matrix.
• x is an n x 1 column vector of unknowns (the solution we seek).
• b is an n x 1 column vector of constants.

The back substitution algorithm proceeds as follows:

1. Solve for x_n: The last equation in the system will only involve x<sub>n</sub>, allowing direct calculation: x<sub>n</sub> = b<sub>n</sub> / a<sub>nn</sub>. (Assuming a<sub>nn</sub> ≠ 0, otherwise the system may be singular or inconsistent).

2. Solve for x_n-1: Substitute the value of x<sub>n</sub> into the second-to-last equation. This equation will now only involve x<sub>n-1</sub>, allowing its calculation.

3. Iterative Process: Continue this iterative process, substituting the already solved values into the subsequent equations. Each equation will only involve one new unknown, making the solution straightforward.

4. Solve for x₁: The process ends with solving for x<sub>1</sub> using the first equation.

Example: Solving an Upper Triangular System

Let's solve the system:

2x + 3y + z = 11
0x + 4y - 2z = 10
0x + 0y + z = 3

1. From the third equation: z = 3

2. Substitute z = 3 into the second equation: 4y - 2(3) = 10 => 4y = 16 => y = 4

3. Substitute y = 4 and z = 3 into the first equation: 2x + 3(4) + 3 = 11 => 2x = 0 => x = 0

Therefore, the solution is x = 0, y = 4, z = 3.

Computational Efficiency and Numerical Stability

Back substitution is computationally very efficient. The number of arithmetic operations required is significantly less compared to solving general linear systems using methods like Gaussian elimination. Furthermore, if the diagonal elements (a<sub>ii</sub>) are sufficiently large compared to the other elements in their respective rows, back substitution tends to be numerically stable, minimizing the propagation of rounding errors.

Conclusion

Solving upper triangular matrices using back substitution is a fundamental technique in linear algebra with numerous applications in various fields, including computer graphics, engineering, and scientific computing. Its simplicity and efficiency make it a preferred method whenever dealing with this special type of matrix. Understanding back substitution is a crucial step in mastering linear algebra techniques. Remember to always check for the possibility of a singular or inconsistent system where a diagonal element is zero.