Chapter 10: Problem 21
Find the complete solution of the linear system, or show that it is inconsistent. \(\left\\{\begin{array}{r}x+y+z=4 \\ x+3 y+3 z=10 \\ 2 x+y-z=3\end{array}\right.\)
Short Answer
Expert verified
The solution is \(x=1\), \(y=2\), \(z=1\).
Step by step solution
01
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: \[ \begin{bmatrix} 1 & 1 & 1 & | & 4 \ 1 & 3 & 3 & | & 10 \ 2 & 1 & -1 & | & 3 \end{bmatrix} \]
02
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: \[ \begin{bmatrix} 1 & 1 & 1 & | & 4 \ 0 & 2 & 2 & | & 6 \ 0 & -1 & -3 & | & -5 \end{bmatrix} \]
03
Continue Row Reduction
Add half of row 2 to row 3 to eliminate the first element in row 3: \[ \begin{bmatrix} 1 & 1 & 1 & | & 4 \ 0 & 2 & 2 & | & 6 \ 0 & 0 & -2 & | & -2 \end{bmatrix} \]
04
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 \(z\):\[ z = 1 \]
05
Substitute to Solve for y
Substitute \(z = 1\) into the second equation: \[ 2y + 2(1) = 6 \] This simplifies to:\[ 2y + 2 = 6 \] \[ 2y = 4 \] \[ y = 2 \]
06
Substitute to Solve for x
Substitute \(z = 1\) and \(y = 2\) into the first equation: \[ x + 2 + 1 = 4 \] This simplifies to: \[ x + 3 = 4 \] \[ x = 1 \]
07
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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).
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.
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.