Problem 66 - College Algebra and Calculus: An Applied Approach

Question

Use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} 2 x-y+3 z &=24 \\ 2 y-z &=14 \\ 7 x-5 y &=6 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is x = 6/7, y = 7, z = 10/7.

Step 1: Forming matrix

First, form a 3x4 matrix \( A \) of coefficients of the variables and constants from the equations. This matrix \( A \) will be: \[A = \\begin{bmatrix}2 & -1 & 3 & 24 \\0 & 2 & -1 & 14 \\7 & -5 & 0 & 6\end{bmatrix}\]

Step 2: Gauss-Jordan elimination

Now, perform the Gauss-Jordan elimination. First, swap row 1 and row 3. Then, subtract 7/2 times the first row from the third (now, the first) row and divide the second row by 2. The new matrix \( A \) becomes:\[A = \\begin{bmatrix}7 & -5 & 0 & 6 \\0 & 1 & -0.5 & 7 \\2 & -1 & 3 & 24\end{bmatrix}\] Now, subtract 2 times the first row from the third row. The new matrix becomes:\[A =\\begin{bmatrix}7 & -5 & 0 & 6 \\0 & 1 & -0.5 & 7 \\0 & 1 & 3 & 12\end{bmatrix}\] Subtract second row from the third row:\[A =\\begin{bmatrix}7 & -5 & 0 & 6 \\0 & 1 & -0.5 & 7 \\0 & 0 & 3.5 & 5\end{bmatrix}\]Finally, divide the third row by 3.5, and the first row by 7. Now, the matrix A matches the form of an identity matrix, which means we've found our solution:\[A =\\begin{bmatrix}1 & -5/7 & 0 & 6/7 \\0 & 1 & -0.5 & 7 \\0 & 0 & 1 & 5/3.5\end{bmatrix}\]

Step 3: Reading the solution from the matrix

The resulting matrix now gives the solution to the original set of equations. The first row says x = 6/7, the second row y = 7, and the third row z = 5/(3.5) = 10/7. So the solution to the system of equations is x = 6/7, y = 7, z = 10/7.

Key Concepts

Gaussian Elimination

Gaussian elimination is a method for solving systems of linear equations by systematically transforming the system's matrix into an upper triangular format. This process makes it easier to solve for the unknown variables.

During Gaussian elimination, we perform elementary row operations to achieve a triangular matrix. These operations include swapping rows, multiplying a row by a scalar, and adding a multiple of one row to another row. Usually, this process results in a staircase form where each successive row has more zeroes to the left.

Why Gaussian Elimination Matters

Understanding this method is crucial because it provides a clear procedure to follow, which can be applied to systems of any size. Moreover, learning Gaussian elimination helps build the foundation for more advanced linear algebra topics and computational methods.

Gauss-Jordan Elimination

Gauss-Jordan elimination is an extension of Gaussian elimination. Instead of stopping at an upper triangular matrix, it continues the process to obtain a matrix in reduced row echelon form (RREF). In RREF, not only is the matrix upper triangular, but every leading coefficient (the first non-zero number from the left in a non-zero row) is 1, and is the only non-zero entry in its column.

This method simplifies the back-substitution process, theoretically eliminating it altogether, as the final matrix form directly reveals the values of the unknowns.

Application in Solving Equations

Gauss-Jordan is particularly helpful when solving for any number of solutions—if there are any solutions at all. It presents the complete solution clearly in the matrix without needing further substitutions.

Matrix Algebra

Matrix algebra involves operations with matrices, such as addition, subtraction, multiplication, and finding inverses. Matrix multiplication, a key operation in solving systems of equations, doesn't follow the commutative property (meaning A x B does not generally equal B x A), and this is fundamental to understanding how matrices transform data.

In the context of linear equations, matrix algebra simplifies complex systems into a modular and structured framework.

Importance of Matrix Algebra

It is crucial in various applications including computer graphics, economics, statistics, and engineering. Mastery of matrix algebra enables the application of linear algebra to real-world problems.

Back-Substitution

Back-substitution is the final step in the Gaussian elimination process. After transforming the matrix into an upper triangular form, we can find the values of the variables starting from the bottom row and moving upwards--substituting the known values back into the rows above to solve for the remaining variables.

For instance, if the bottom row of the matrix represents the equation 'z = 3', you can use that value to solve for 'y' in the equation of the second row, then use both 'y' and 'z' to solve for 'x' in the first row.

Completing the Calculation

Although Gauss-Jordan eliminates the need for back-substitution, it’s a critical step in Gaussian elimination, especially with larger matrices where Gauss-Jordan elimination might be more computationally intensive.