Problem 21 - Precalculus: Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. \(\left\\{\begin{array}{r}x+y+z=4 \\ x+3 y+3 z=10 \\ 2 x+y-z=3\end{array}\right.\)

Step-by-Step Solution

Answer

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

Step 1: Write the System in Matrix Form

Firstly, express the system of equations as an augmented matrix. The equations are given as: \[ \begin{cases} x+y+z=4 \ x+3y+3z=10 \ 2x+y-z=3 \end{cases}\] This can be written in an augmented matrix form as: \[ \begin{bmatrix} 1 & 1 & 1 & | & 4 \ 1 & 3 & 3 & | & 10 \ 2 & 1 & -1 & | & 3 \end{bmatrix} \]

Step 2: Apply Row Reduction

Use Gaussian elimination to simplify the matrix to row-echelon form. Subtract row 1 from row 2: \[ \begin{bmatrix} 1 & 1 & 1 & | & 4 \ 0 & 2 & 2 & | & 6 \ 2 & 1 & -1 & | & 3 \end{bmatrix} \] Subtract 2 times row 1 from row 3: \[ \begin{bmatrix} 1 & 1 & 1 & | & 4 \ 0 & 2 & 2 & | & 6 \ 0 & -1 & -3 & | & -5 \end{bmatrix} \]

Step 3: Continue Row Reduction

Add half of row 2 to row 3 to eliminate the first element in row 3: \[ \begin{bmatrix} 1 & 1 & 1 & | & 4 \ 0 & 2 & 2 & | & 6 \ 0 & 0 & -2 & | & -2 \end{bmatrix} \]

Step 4: Solve the Simplified System

Now, the simplified matrix represents the equivalent system of equations: \[ \begin{cases} x + y + z = 4 \ 2y + 2z = 6 \ -2z = -2 \end{cases} \] From the third equation, solve for \(z\):\[ z = 1 \]

Step 5: Substitute to Solve for y

Substitute \(z = 1\) into the second equation: \[ 2y + 2(1) = 6 \] This simplifies to:\[ 2y + 2 = 6 \] \[ 2y = 4 \] \[ y = 2 \]

Step 6: Substitute to Solve for x

Substitute \(z = 1\) and \(y = 2\) into the first equation: \[ x + 2 + 1 = 4 \] This simplifies to: \[ x + 3 = 4 \] \[ x = 1 \]

Step 7: Verify the Solution

Check the solution by plugging \(x = 1\), \(y = 2\), and \(z = 1\) back into the original equations: - The first equation: \(1 + 2 + 1 = 4\), which holds true.- The second equation: \(1 + 3(2) + 3(1) = 10\), which holds true.- The third equation: \(2(1) + 2 - 1 = 3\), which holds true.Thus, the solution \(x = 1\), \(y = 2\), \(z = 1\) satisfies all equations.

Key Concepts

Gaussian Elimination

Gaussian elimination is a systematic method for solving systems of linear equations. It transforms the system into a simpler form called row-echelon form, making the equations easier to solve. The core idea is to perform operations that eliminate variables step by step.

• Start with an augmented matrix form of the system of equations.
• Use row operations such as row swapping, row addition, and scalar multiplication.
• Aim to get zeroes below each pivot position (the leading coefficient in each row).

In our case, we started with the equations:\[\begin{align*}1. \quad & x + y + z = 4 \2. \quad & x + 3y + 3z = 10 \3. \quad & 2x + y - z = 3 \\end{align*}\]Gaussian elimination involves using these operations to simplify the system, thereby allowing us to easily backsolve for the variables.

Matrix Row Reduction

Matrix row reduction, also known as Gaussian elimination, involves transforming a matrix into a simpler form to solve the linear system. In practice, matrix row reduction is done by applying row operations to achieve a row-echelon form. This process involves:

• Using elementary row operations: swap rows, multiply a row by a non-zero scalar, and add a multiple of one row to another.
• Ensuring each leading number in a row is one, and all numbers below it are zero.
• Working column by column, starting from the leftmost side.

In the original problem, we reduced our initial matrix by eliminating elements in specific columns using these row operations: - First, remove the first element in the second row by replacing row two with row two minus row one. - Remove the element in the third row by replacing it with twice row one subtracted from row three. These steps simplify solving the system by reducing it to a form where back substitution can be easily applied.

Augmented Matrices

Augmented matrices are a compact way to write systems of linear equations. They combine the coefficient matrix and the constant terms' vector into a single matrix.

• The coefficients of the variables form the left side of the augmented matrix.
• The constants from the equations form the right side, separated by a bar.

For our exercise, an augmented matrix for the system:\[\begin{align*}x + y + z = 4 & \x + 3y + 3z = 10 & \2x + y - z = 3 & \\end{align*}\]is represented as:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \1 & 3 & 3 & | & 10 \2 & 1 & -1 & | & 3 \\end{bmatrix}\]The bar divides the matrix into the coefficients and constants, helping to visualize and operate on the system more effectively. Through row operations, we aim to manipulate this augmented matrix into a form that aids in solving the system.