Problem 71 - 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. \( \left\\{ \begin{array}{l} x - 3z = -2 \\ 3x + y - 2z = 5 \\ 2x + 2y + z = 4 \end{array} \right. \)

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 4\), \(y = -3\), and \(z = 2\).

Step 1: Formulating the Matrix

Write the equations in matrix form, which is also known as the augmented matrix. It looks like this, where the column on the right side of the line is the numbers on the other side of the equations, and the other columns correspond to the coefficients of \(x\), \(y\), and \(z\) respectively:\[ \left[ \begin{array}{ccc|c}1 & 0 & -3 & -2 \3 & 1 & -2 & 5 \2 & 2 & 1 & 4\end{array} \right] \]

Step 2: First Row Transformation

Subtract 3 times the first row from the second and subtract 2 times the first row from the third:\[ \left[ \begin{array}{ccc|c}1 & 0 & -3 & -2 \0 & 1 & 7 & 11 \0 & 2 & 7 & 8\end{array} \right] \]

Step 3: Second Row Transformation

Subtract 2 times the second row from the third:\[ \left[ \begin{array}{ccc|c}1 & 0 & -3 & -2 \0 & 1 & 7 & 11 \0 & 0 & -7 & -14\end{array} \right] \]

Step 4: Third Row Transformation and Back substitution

Divide the third row by -7 to leave a 1 in the z place:\[ \left[ \begin{array}{ccc|c}1 & 0 & -3 & -2 \0 & 1 & 7 & 11 \0 & 0 & 1 & 2\end{array} \right] \]Back substitution is the process of finding the actual values of the variables \(x\), \(y\), and \(z\). Start from the bottom equation to find the value of \(z\) (which is \(z = 2\)). Then, substitute \(z\) into the second equation to find \(y\) (\(y = 11 − 7*2 = -3\)). After finding \(y\) and \(z\), substitute them into the first equation to find \(x\) (\(x = −2 + 3*2 = 4\)).

Key Concepts

Gaussian elimination

Gaussian elimination is a method of solving systems of linear equations. It's especially useful because it is systematic and can be turned into an algorithm, making it ideal for computation. The primary goal is to transform the system of equations into an upper triangular matrix, from which the solution can be determined easily.

• Firstly, you convert the given equations into an augmented matrix. This matrix includes the coefficients of variables and the constants from each equation.
• The next step is using row operations to transform this matrix. Row operations can include swapping rows, multiplying a row by a non-zero number, or adding a multiple of one row to another.

The target in Gaussian elimination is to make all the entries below the main diagonal zero, transforming the matrix into what is known as row echelon form. Once achieved, it simplifies the solving process dramatically.

Gauss-Jordan elimination

Gauss-Jordan elimination is an enhancement of Gaussian elimination. While the goal of Gaussian elimination is to convert a matrix into row echelon form, Gauss-Jordan elimination continues the process to achieve reduced row echelon form, where:

• The leading entry in each non-zero row is 1, which is called a "pivot".
• The pivot is the only non-zero entry in its column.

When a matrix is in this form, the system of equations it represents is almost solved. Each row corresponds to a simpler equation that can be read off directly to find the values of the variables. This makes Gauss-Jordan elimination an effective method, as it leads directly to the solution without requiring back-substitution.

Matrix Method

The matrix method involves representing a system of linear equations as a matrix equation. For instance, in the system given:\[\begin{bmatrix}1 & 0 & -3 \3 & 1 & -2 \2 & 2 & 1 \\end{bmatrix}\begin{bmatrix}x \y \z \\end{bmatrix}=\begin{bmatrix}-2 \5 \4 \\end{bmatrix}\]This matrix equation represents the same system of equations in a compact form:

• The first matrix is the coefficient matrix.
• The second is the column matrix of the variables.
• The third is the constant matrix.

The task is then to manipulate these matrices, using methods like Gaussian or Gauss-Jordan elimination, to find the solution to the system — that is, the values of the variables.

Back-substitution

Back-substitution is a step used after Gaussian elimination and involves solving a system of equations that has been transformed into an upper triangular form. Once the augmented matrix is simplified through row operations, it often results in the following form:

• Equations with one variable can be solved outright, starting from the bottom row.
• Then, the known values are substituted into the equations above them.

For example, with the upper triangular matrix obtained:\[x - 3z = -2 \y + 7z = 11 \z = 2\]You start from the bottom. Since you know \(z = 2\), substitute \(z\) in the second equation to find \(y\), then use both values to find \(x\). Back-substitution proceeds in reverse order, working backwards from the simpler equations to the more complex ones, arriving at the solution step by step.