Question

Solve the following set of simultaneous equations using Gaussian elimination (including interchange where it is formally desirable), $$ \begin{aligned} x_{1}+3 x_{2}+4 x_{3}+2 x_{4} &=0 \\ 2 x_{1}+10 x_{2}-5 x_{3}+x_{4} &=6 \\ 4 x_{2}+3 x_{3}+3 x_{4} &=20 \\ -3 x_{1}+6 x_{2}+12 x_{3}-4 x_{4} &=16. \end{aligned} $$

Short answer

The solutions are \(x_1 = -64\), \(x_2 = 78\), \(x_3 = 22\), and \(x_4 = -56\).

Step-by-step solution

Write the augmented matrix

Translate the system of equations into an augmented matrix:\[\left[\begin{array}{cccc|c}1 & 3 & 4 & 2 & 0 \2 & 10 & -5 & 1 & 6 \0 & 4 & 3 & 3 & 20 \-3 & 6 & 12 & -4 & 16\end{array}\right]\]

Make the first element of the first row a leading 1

The first element in the first row is already 1. No changes needed.

Eliminate the first element of the second, third, and fourth rows

Subtract 2 times the first row from the second row: \[\left[\begin{array}{cccc|c}1 & 3 & 4 & 2 & 0 \0 & 4 & -13 & -3 & 6 \0 & 4 & 3 & 3 & 20 \-3 & 6 & 12 & -4 & 16\end{array}\right]\]Add 3 times the first row to the fourth row: \[\left[\begin{array}{cccc|c}1 & 3 & 4 & 2 & 0 \0 & 4 & -13 & -3 & 6 \0 & 4 & 3 & 3 & 20 \0 & 15 & 24 & 2 & 16\end{array}\right]\]

Make the element in the second row, second column a leading 1

Divide the second row by 4:\[\left[\begin{array}{cccc|c}1 & 3 & 4 & 2 & 0 \0 & 1 & -\frac{13}{4} & -\frac{3}{4} & \frac{3}{2} \0 & 4 & 3 & 3 & 20 \0 & 15 & 24 & 2 & 16\end{array}\right]\]

Eliminate the second element of the third and fourth rows

Subtract 4 times the second row from the third row: \[\left[\begin{array}{cccc|c}1 & 3 & 4 & 2 & 0 \0 & 1 & -\frac{13}{4} & -\frac{3}{4} & \frac{3}{2} \0 & 0 & 16 & 6 & 14 \0 & 15 & 24 & 2 & 16\end{array}\right]\]Subtract 15 times the second row from the fourth row: \[\left[\begin{array}{cccc|c}1 & 3 & 4 & 2 & 0 \0 & 1 & -\frac{13}{4} & -\frac{3}{4} & \frac{3}{2} \0 & 0 & 16 & 6 & 14 \0 & 0 & 63 & \frac{10}{4} & -8.5\end{array}\right]\]

Make the element in the third row, third column a leading 1

Divide the third row by 16:\[\left[\begin{array}{cccc|c}1 & 3 & 4 & 2 & 0 \0 & 1 & -\frac{13}{4} & -\frac{3}{4} & \frac{3}{2} \0 & 0 & 1 & \frac{3}{8} & \frac{7}{8} \0 & 0 & 63 & \frac{10}{4} & -8.5\end{array}\right]\]

Eliminate the third element of the fourth row

Subtract 63 times the third row from the fourth row: \[\left[\begin{array}{cccc|c}1 & 3 & 4 & 2 & 0 \0 & 1 & -\frac{13}{4} & -\frac{3}{4} & \frac{3}{2} \0 & 0 & 1 & \frac{3}{8} & \frac{7}{8} \0 & 0 & 0 & \frac{1}{4} & -14\end{array}\right]\]

Back substitution to find the variable values

From the last row, solve for \(x_4\):\[\frac{1}{4} x_4 = -14 \Rightarrow x_4 = -56\]Substitute \(x_4\) into the third row equation and solve for \(x_3\):\[x_3 + \frac{3}{8} (-56) = \frac{7}{8} \Rightarrow x_3 = 22\]Substitute \(x_3\) and \(x_4\) into the second row equation and solve for \(x_2\):\[x_2 - \frac{13}{4} (22) - \frac{3}{4} (-56) = \frac{3}{2} \Rightarrow x_2 = 78\]Substitute \(x_2\), \(x_3\), and \(x_4\) into the first row equation and solve for \(x_1\):\[x_1 + 3 (78) + 4 (22) + 2 (-56) = 0 \Rightarrow x_1 = -64\]

Key concepts

Simultaneous Equations

Simultaneous equations are a set of equations with multiple variables, which are solved together because the equations share these variables. In our given problem, we have four equations with variables \(x_1, x_2, x_3, \) and \(x_4\). The goal is to find the values of these variables that satisfy all equations simultaneously.

Here's an example of the equations we're working with:
\(x_1 + 3x_2 + 4x_3 + 2x_4 = 0\)
\(2x_1 + 10x_2 - 5x_3 + x_4 = 6\)
\(4x_2 + 3x_3 + 3x_4 = 20\)
\(-3x_1 + 6x_2 + 12x_3 - 4x_4 = 16\)

Solving simultaneous equations can be challenging, but methods like Gaussian elimination simplify this task. Gaussian elimination works by systematically transforming the equations to make the system easier to solve. By converting the system into an augmented matrix form and performing row operations, we transform the equations into a simpler form, which can then be used to determine the values of the variables.

Augmented Matrix

An augmented matrix is a condensed way of writing down the coefficients and constants of a system of linear equations. This matrix representation makes it easier to perform row operations systematically. The augmented matrix for our given problem looks like this:
\[ \begin{array}{cccc|c} 1 & 3 & 4 & 2 & 0 \ 2 & 10 & -5 & 1 & 6 \ 0 & 4 & 3 & 3 & 20 \ -3 & 6 & 12 & -4 & 16 \]

Each row in the augmented matrix represents an equation, while the columns correspond to the coefficients of the variables. The last column on the right side holds the constants from the equations. By transforming this matrix using row operations, we reduce the complexity of the system and move towards identifying the values of the variables systematically.

Back Substitution

Back substitution is a technique used after transforming the system of equations into an upper triangular form through Gaussian elimination. It involves solving the equations starting from the last equation and working upwards. Let's look at how this works:

• Step 1: Start with the last row: \( \frac{1}{4} x_4 = -14 \Rightarrow x_4 = -56 \)
• Step 2: Substitute \(x_4\) into the third row and solve for \(x_3: \)
  \( x_3 + \frac{3}{8}(-56) = \frac{7}{8} \Rightarrow x_3 = 22 \)
• Step 3: Substitute \(x_3 \) and \(x_4 \) into the second row and solve for \(x_2:\)
  \( x_2 - \frac{13}{4}(22) - \frac{3}{4}(-56) = \frac{3}{2} \Rightarrow x_2 = 78 \)
• Step 4: Finally, substitute \(x_2, x_3, \) and \(x_4\) into the first row and solve for \(x_1:\)
  \( x_1 + 3(78) + 4(22) + 2(-56) = 0 \Rightarrow x_1 = -64 \)

Back substitution completes the solution process, providing the values of all variables that satisfy the initial set of simultaneous equations.

Leading Coefficient

A leading coefficient is the first non-zero number from the left in each row of a matrix after performing row operations. It plays a crucial role in Gaussian elimination as it helps in reducing the matrix to a simpler form. For our matrix, the steps involve:

• Step 1: The leading coefficient in the first row is already \(1\). No changes needed.
• Step 2: Make the leading coefficient in the second row by dividing the entire second row by \(4\).
• Step 3: For the third row, divide it by \(16\) to make the leading coefficient \(1\).

These steps ensure that each row has a leading coefficient of \(1\), making it easier to perform further row operations and eventually reach the stage where back substitution can be applied. Ensuring leading coefficients are \(1\) simplifies the entire elimination process and paves the way for solving for each variable methodically.