Problem 15 - Precalculus: Mathematical for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{aligned} x-2 y+z &=1 \\ y+2 z &=5 \\ x+y+3 z &=8 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = 1\), \(y = 1\), \(z = 2\).

Step 1: Write the Augmented Matrix

Represent the given system of equations as an augmented matrix. The system \(\begin{aligned} x-2y+z &=1 \ y+2z &=5 \ x+y+3z &=8 \end{aligned}\) is written as the matrix: \[\begin{bmatrix} 1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \1 & 1 & 3 & | & 8 \end{bmatrix}\] This matrix contains coefficients of \(x\), \(y\), and \(z\) along with the constants from the equations.

Step 2: Transform to Row-Equivalent Matrix

Use row operations to transform the matrix into an upper triangular form and then into the reduced row-echelon form. Start by eliminating \(x\) from the third row with \(R_3 - R_1\):\[\begin{bmatrix} 1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \0 & 3 & 2 & | & 7 \end{bmatrix}\].

Step 3: Continue Gaussian Elimination

Next, eliminate \(y\) from the third row by substituting \(R_3 - 3R_2\):\[\begin{bmatrix} 1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \0 & 0 & -4 & | & -8 \end{bmatrix}\]. Ensure the new matrix has zeros below leading coefficients.

Step 4: Back Substitution

The matrix is now in upper triangular form. Start back substitution to find the values of variables:1. From the third row: \(-4z = -8\rightarrow z = 2\).2. Substitute \(z = 2\) into the second row: \(y + 2(2) = 5 \rightarrow y = 1\).3. Substitute \(y = 1\) and \(z = 2\) into the first row: \(x - 2(1) + 2 = 1 \rightarrow x = 1\).

Step 5: Verify the Solution

Verify by substituting \(x = 1\), \(y = 1\), \(z = 2\) back into the original equations to ensure all are satisfied: 1. \((1) - 2(1) + 2 = 1\) is true. 2. \(1 + 2(2) = 5\) is true. 3. \((1) + 1 + 3(2) = 8\) is true. Therefore, the solution satisfies all equations.

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. These variables can be represented as unknowns that you need to solve for. For example, in a system with two equations and two variables, it might look something like this:

• Equation 1: \( a_1x + b_1y = c_1 \)
• Equation 2: \( a_2x + b_2y = c_2 \)

The solution to a system of linear equations can be a single point, infinitely many points, or no solution at all, depending on how the lines intersect when graphed. In particular, a unique solution exists when one exact set of values for the variables satisfies all equations simultaneously, as in the case of the system solved here.

Row Reduction

Row reduction, or Gaussian elimination, is a method to solve systems of linear equations by transforming the corresponding augmented matrix into a simpler form. This simpler form is either the row-echelon form or the reduced row-echelon form. The row operations used in this process include:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row to another

These operations simplify the matrix and help in finding the values of the variables by making the system easier to work with. The goal in row reduction is to reach an upper triangular form, where all rows below the leading diagonal are zeros, making it simple to "back substitute" and find solutions.

Back Substitution

Back substitution is the process used after row reduction to determine the values of the unknown variables in a system of equations. Once you've transformed an augmented matrix into an upper triangular form, where all elements below the main diagonal are zeros, each equation corresponds to a row of the matrix, and the equations can be easily solved starting from the bottom:

• Begin with the last equation, which will have only one unknown with a non-zero coefficient. Solve it directly for that variable.
• Substitute this variable back into the equation directly above it to solve for the second variable.
• Continue this process until all variables are found.

This step-by-step method ensures that, given a unique solution exists, every variable can be solved for uniquely.

Augmented Matrix

An augmented matrix is a compact way of representing a system of linear equations. It consists of coefficients of the variables and the constants on the right-hand side of the equations, all combined into a single matrix. For instance, for a system:

• Equation 1: \( ax + by + cz = d \)
• Equation 2: \( ex + fy + gz = h \)

The augmented matrix would be represented as:\[\begin{bmatrix}a & b & c & | & d \e & f & g & | & h\end{bmatrix}\]Here, each row corresponds to an equation, and each column (except the last one) corresponds to a variable's coefficient. The vertical bar separates the variable coefficients from the constants, providing a clear and structured format to perform row operations in row reduction easier and more systematically.