Problem 64 - Precalculus with Limits

Question

In Exercises 63-84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. \( \left\\{ \begin{array}{l} 2x + 6y = 16 \\ 2x + 3y = 7 \end{array} \right. \)

Step-by-Step Solution

Answer

The solution of the system of equations is \(x = 1\) and \(y = 3\).

Step 1: Build the Augmented Matrix

First, write the system of equations in matrix form. This augmented matrix represents our system of equations: \[\begin{bmatrix} 2 & 6 & 16 \ 2 & 3 & 7 \end{bmatrix}\]. Each row corresponds to an equation, and the columns correspond to the coefficients of the variables \(x\), \(y\) and the numbers on the right side of the equation. The first column corresponds to \(x\), the second to \(y\), and the third to the constants.

Step 2: Gaussian Elimination

To begin Gaussian elimination process, we aim to 'zero out' the value below the leading coefficient in the first column. As the first elements in both rows are the same, we can subtract the second row from the first. This leads us to: \[\begin{bmatrix} 0 & 3 & 9 \ 2 & 3 & 7 \end{bmatrix}\]. We can then swap the two rows so that the leading coefficient in the first row is not zero, which results in: \[\begin{bmatrix} 2 & 3 & 7 \ 0 & 3 & 9 \end{bmatrix}\].

Step 3: Back-Substitution

The second equation in the matrix form is equivalent to \(3y = 9\), which means \(y = 3\). Substituting \(y = 3\) back into the first equation \(2x + 3y = 7\), we find that \(x = 1\).

Key Concepts

Understanding Gaussian Elimination

Gaussian elimination is a method used to solve systems of linear equations. It's about transforming your system into a simpler one that is easier to solve.

To start, you write the system of equations as an augmented matrix. Each row represents an equation, and columns contain the coefficients and constants.

Next, the goal is to form an upper triangular matrix by creating zeros below the leading coefficients (the first non-zero number from the left in each row) in each column.

Here's how it works:

• Choose a pivot, the first element in the current row, and attempt to zero out all elements below it in the same column by adding suitable multiples of the pivot row to the other rows.
• Continue this process for each leading coefficient across all rows.

Once this triangular form is achieved, you can easily solve for the variables by back-substitution.

What is Back-Substitution?

Back-substitution is a technique used to find the values of variables after transforming the system into an upper triangular matrix.

Once the system is in this form, you start solving for unknowns from the last equation upwards, since each has fewer variables.

For example:

• If your final matrix gives a row like 3y = 9, you can directly solve for y: \( y = \frac{9}{3} = 3 \).
• Substitute this value back into previous equations to find other variables. If a preceding equation is 2x + 3y = 7, you'll know that \( 2x + 3 \times 3 = 7 \), so \( 2x = 7 - 9 \), which simplifies to \( x = \frac{-2}{2} = 1 \).

This step-by-step substitution ensures you find the correct solution for all variables.

Exploring Gauss-Jordan Elimination

Gauss-Jordan elimination is a further extension of Gaussian elimination. It's a method that transforms the matrix to a reduced row-echelon form, where each leading entry is 1 and the other entries in that column are zero.

This technique completely solves the system directly in one go, eliminating the need for back-substitution.

Here's what you do:

• Perform row operations to form zeros not just below, but also above the leading coefficients.
• Continue reducing until every leading coefficient becomes 1 and is the only non-zero number in its column.

This leads to an identity matrix on the left side of your augmented matrix, and directly shows the solution of the variables on the right.
It's a powerful method but can be more labor-intensive compared to Gaussian elimination, especially for larger systems.