Problem 63 - Precalculus

Question

Solve the system of linear equations using Gaussian elimination with back- substitution. $$\begin{array}{rr} 2 x_{1}-x_{2}+x_{3}= & 3 \\ x_{1}-x_{2}+x_{3}= & 2 \\ -2 x_{1}+2 x_{2}-2 x_{3}= & -4 \end{array}$$

Step-by-Step Solution

Answer

The solution is \(x_1 = 1\), \(x_2 = 0\), \(x_3 = 1\).

Step 1: Write the Augmented Matrix

Convert the given system of equations into an augmented matrix. The system:\[\begin{aligned}2x_1 - x_2 + x_3 &= 3\x_1 - x_2 + x_3 &= 2\-2x_1 + 2x_2 - 2x_3 &= -4\end{aligned}\]becomes:\[\begin{pmatrix}2 & -1 & 1 & | & 3 \1 & -1 & 1 & | & 2 \-2 & 2 & -2 & | & -4\end{pmatrix}\]

Step 2: Perform Row Operations to Achieve Upper Triangular Form

Start by making the coefficient of \(x_1\) in the second row zero. Subtract row 2 from row 1:\[R2 - R1 \rightarrow R2\]Resulting matrix:\[\begin{pmatrix}2 & -1 & 1 & | & 3 \-1 & 0 & 0 & | & -1 \-2 & 2 & -2 & | & -4\end{pmatrix}\]Now, make the coefficient in row 3 zero by adding row 1 to row 3:\[R3 + R1 \rightarrow R3\]Resulting matrix:\[\begin{pmatrix}2 & -1 & 1 & | & 3 \-1 & 0 & 0 & | & -1 \0 & 1 & -1 & | & -1\end{pmatrix}\]

Step 3: Simplify to Further Triangular Form and Adjust Signs

Multiply row 2 by -1 to make it consistent:\[-R2 \rightarrow R2\]Then, multiply row 1 by 0.5 to simplify:\[0.5 \, R1 \rightarrow R1\]The matrix becomes:\[\begin{pmatrix}1 & -0.5 & 0.5 & | & 1.5 \1 & 0 & 0 & | & 1 \0 & 1 & -1 & | & -1\end{pmatrix}\]

Step 4: Back-Substitution to Solve the System

Now, perform back substitution. From row 3,\[x_2 - x_3 = -1\]From row 2,\[x_1 = 1\]Substitute \(x_1 = 1\) into row 1 equation:\[1 - 0.5x_2 + 0.5x_3 = 1.5\] Simplify to solve for \(x_2\) and \(x_3\). We already have,\[x_2 - x_3 = -1\]From row 1:\[0.5x_2 - 0.5x_3 = 0.5\]\[x_2 = 0\], \[x_3 = 1\].

Step 5: Verify the Solution

Substitute \(x_1 = 1\), \(x_2 = 0\), and \(x_3 = 1\) back into the original equations to verify:\[\begin{aligned}2(1) - 1(0) + 1(1) &= 3 \, \text{true}\1(1) - 0 + 1(1) &= 2 \, \text{true}\-2(1) + 2(0) - 2(1) &= -4 \, \text{also true}\end{aligned}\]All equations are satisfied.

Key Concepts

Augmented Matrix

When solving systems of linear equations, an augmented matrix provides a simple way to work with the equations without writing them out every time. To create an augmented matrix, take each coefficient and constant from your equations and arrange them into a rectangular array. Each row corresponds to an equation, and each column corresponds to a variable or constant term.

For example, consider the following system of equations:

• Equation 1: \(2x_1 - x_2 + x_3 = 3\)
• Equation 2: \(x_1 - x_2 + x_3 = 2\)
• Equation 3: \(-2x_1 + 2x_2 - 2x_3 = -4\)

These equations are transformed into an augmented matrix like this:\[\begin{pmatrix}2 & -1 & 1 & | & 3 \1 & -1 & 1 & | & 2 \-2 & 2 & -2 & | & -4\end{pmatrix}\]

Where the vertical line separates the coefficient part of the matrix from the constants. This setup allows you to use row operations to solve the system efficiently.

Back-Substitution

Back-substitution is a straightforward method used in conjunction with Gaussian elimination to solve a system of linear equations. Once you've transformed your matrix into an upper triangular form, you begin solving for the variables starting from the last equation upwards. This process involves isolating one variable at a time and substituting previously solved values into other equations.

For example, if your reduced system results in:

• Row 3: \(x_2 - x_3 = -1\)
• Row 2: \(x_1 = 1\)

The back-substitution process would start with row 3. From it, express one variable in terms of the others, such as \(x_2 = x_3 - 1\). Then move to row 2, which gives \(x_1 = 1\) directly.

Continuing the process, substitute \(x_1\) into row 1's equation. You eventually solve each variable by substituting known values back into the remaining equations. This neat step-by-step approach exploits the form of the triangular matrix to simplify solving significantly.

Row Operations

Row operations allow us to manipulate rows in a matrix to simplify it and reach a solution more conveniently. These operations help achieve the triangular form essential in the Gaussian elimination process. There are three types of row operations that you can perform:

• Row swapping: Exchange two rows. It helps in repositioning rows to simplify calculations.
• Row multiplication: Multiply a row by a non-zero constant to achieve needed coefficients.
• Row addition: Add or subtract a multiple of one row from another to create zeros in strategic positions.

These operations are crucial to transform the original matrix into an upper triangular form step by step.

For instance, to get zeros below the first leading coefficient in our example, perform these operations:

• Subtract row 1 from row 2 (\(R_2 - R_1 \rightarrow R_2\))
• Add row 1 to row 3 (\(R_3 + R_1 \rightarrow R_3\))

Using these operations, you simplify the system without changing its solution, setting the stage for back-substitution.

Triangular Form

Triangular form refers to a matrix that has been arranged such that all elements below the main diagonal (or leading diagonal) are zero. This arrangement is particularly useful for solving systems of equations because it allows for easier application of back-substitution. In a perfect triangular form, any matrix of size \(n \times n\) will have non-zero entries only in the diagonal and possibly above it.

Consider the system from our example matrix after row operations:\[\begin{pmatrix}1 & -0.5 & 0.5 & | & 1.5 \1 & 0 & 0 & | & 1 \0 & 1 & -1 & | & -1\end{pmatrix}\]

This matrix is close to triangular form. The zeros below the pivot in row 1 are evident, and this form facilitates solving through back-substitution, starting from the bottom row upward, maintaining the linear structure and ensuring accurate solutions.