Problem 32 - Precalculus: Mathematics for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{aligned} x+y+z &=4 \\ -x+2 y+3 z &=17 \\ 2 x-y &=-7 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (-2, 3, 3)\).

Step 1: Set up the augmented matrix

Write the system of equations as an augmented matrix. The matrix represents the coefficients of the variables and the constants: \[\begin{bmatrix}1 & 1 & 1 & | & 4 \-1 & 2 & 3 & | & 17 \2 & -1 & 0 & | & -7\end{bmatrix}\]

Step 2: Perform row operations to create zeros below the first pivot

Using the first row (pivot row), make the elements below the leading 1 in the first column zero by replacing \(R_2\) with \(R_2 + R_1\) and \(R_3\) with \(R_3 - 2R_1\):\[\begin{bmatrix}1 & 1 & 1 & | & 4 \0 & 3 & 4 & | & 21 \0 & -3 & -2 & | & -15\end{bmatrix}\]

Step 3: Create zeros below the second pivot

Using the second row as the pivot, add \(R_3 + R_2\) to create a zero below the leading entry in the second column.\[\begin{bmatrix}1 & 1 & 1 & | & 4 \0 & 3 & 4 & | & 21 \0 & 0 & 2 & | & 6\end{bmatrix}\]

Step 4: Solve for the last variable

With the matrix in row-echelon form, solve for the last variable in the third row. The equation is \(2z = 6\), so divide by 2 to solve for \(z\): \[ z = 3 \]

Step 5: Back substitution for the second variable

Substitute \(z = 3\) into the second row equation to find \(y\):\[ 3y + 4(3) = 21 \]Simplifying, we get:\[ 3y + 12 = 21 \]\[ 3y = 9 \]\[ y = 3 \]

Step 6: Back substitution for the first variable

Finally, substitute \(y = 3\) and \(z = 3\) into the first row equation to solve for \(x\):\[ x + 3 + 3 = 4 \]\[ x + 6 = 4 \]\[ x = -2 \]

Step 7: State the solution

The solution to the system of equations is \((x, y, z) = (-2, 3, 3)\). This solution satisfies all three original equations.

Key Concepts

Linear Equations

A linear equation is an equation that represents a straight line when graphed on a coordinate plane. The general format of a linear equation in three variables is \(ax + by + cz = d\). Here, \(a, b, c,\) and \(d\) are constants, and \(x, y,\) and \(z\) are variables. In the given problem, we have a system of three linear equations:

• \(x + y + z = 4\)
• \(-x + 2y + 3z = 17\)
• \(2x - y = -7\)

These equations represent three planes in three-dimensional space. The intersection of these planes will give us the solution to the system, which could be a unique point, a line, or a plane. In this exercise, we are tasked with finding the unique solution using Gaussian elimination.

Augmented Matrix

To solve a system of linear equations using Gaussian elimination, we first represent the system as an augmented matrix. An augmented matrix combines the coefficients of the variables with the constants from the right-hand side of the equation in one matrix format. For our system, the augmented matrix looks like this:\[\begin{bmatrix} 1 & 1 & 1 & | & 4 \-1 & 2 & 3 & | & 17 \2 & -1 & 0 & | & -7\end{bmatrix} \]The vertical line in the matrix is just a visual separator between the coefficients of the variables and the constant terms. This representation is a powerful tool because it allows us to use matrix row operations to systematically solve the equations.

Row Operations

Row operations are techniques used to manipulate the rows of an augmented matrix. These operations help simplify the system of linear equations to make it easier to solve. There are three types of row operations:

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

In the given solution, row operations are used to create zeros below the pivots (leading coefficients) to transform the matrix into row-echelon form. For instance:

• Adding \(R_1\) to \(R_2\) results in a new row \(R_2\) with a zero in the first column.
• Subtracting twice \(R_1\) from \(R_3\) modifies \(R_3\) to have a zero in the first column as well.

These operations gradually simplify the matrix, allowing us to solve the system efficiently through back substitution.

Unique Solution

A unique solution occurs when a system of linear equations has exactly one set of values for the variables that satisfies all the equations simultaneously. In the context of the Gaussian elimination method, achieving a unique solution means that the row operations have simplified the system to the point where we can clearly see the direct values for each variable. In our exercise, after applying row operations, the final matrix in row-echelon form enables us to find the values of \(z\), \(y\), and \(x\) through back substitution:

• From the third row, we solve for \(z = 3\).
• Substituting \(z\) into the second row, we find \(y = 3\).
• Finally, substituting both \(y\) and \(z\) into the first row provides \(x = -2\).

These values form the unique solution \((x, y, z) = (-2, 3, 3)\), which satisfies all original equations, confirming the system's consistency and independence.