Math Problem Statement

solve this in matrix gaussian reduction: x+y+z=24, 2y+2z-x=0,y+9=x

Solution

To solve the system of equations using Gaussian elimination, let's first represent the system as a matrix. The given system is:

  1. x+y+z=24x + y + z = 24
  2. 2y+2zx=02y + 2z - x = 0
  3. y+9=xy + 9 = x

Step 1: Rewrite the system in standard form

We rewrite the equations to ensure all variables are on one side and constants on the other.

  1. x+y+z=24x + y + z = 24x+y+z=24x + y + z = 24
  2. x+2y+2z=0-x + 2y + 2z = 0x+2y+2z=0-x + 2y + 2z = 0
  3. xy=9x - y = 9xy+0z=9x - y + 0z = 9

Now, let's represent this system in augmented matrix form:

1 & 1 & 1 & | & 24 \\ -1 & 2 & 2 & | & 0 \\ 1 & -1 & 0 & | & 9 \end{pmatrix}$$ ### Step 2: Apply Gaussian elimination #### First, use row operations to get a leading 1 in the first row: The matrix is already in a good form, so we will focus on eliminating the first element in row 2 and row 3. - Add row 1 to row 2: $$R_2 = R_2 + R_1$$ - Subtract row 1 from row 3: $$R_3 = R_3 - R_1$$ This results in the following matrix: $$\begin{pmatrix} 1 & 1 & 1 & | & 24 \\ 0 & 3 & 3 & | & 24 \\ 0 & -2 & -1 & | & -15 \end{pmatrix}$$ #### Next, eliminate the leading element in the third row: - Multiply row 2 by $$\frac{1}{3}$$ to make the leading coefficient in the second row a 1: $$R_2 = \frac{1}{3}R_2$$ This gives: $$\begin{pmatrix} 1 & 1 & 1 & | & 24 \\ 0 & 1 & 1 & | & 8 \\ 0 & -2 & -1 & | & -15 \end{pmatrix}$$ - Now, add 2 times row 2 to row 3: $$R_3 = R_3 + 2R_2$$ This results in the matrix: $$\begin{pmatrix} 1 & 1 & 1 & | & 24 \\ 0 & 1 & 1 & | & 8 \\ 0 & 0 & 1 & | & 1 \end{pmatrix}$$ ### Step 3: Perform back-substitution From the matrix, we can now read off the equations: 1. $$z = 1$$ 2. $$y + z = 8$$ → $$y + 1 = 8$$ → $$y = 7$$ 3. $$x + y + z = 24$$ → $$x + 7 + 1 = 24$$ → $$x = 16$$ ### Solution: $$x = 16, \quad y = 7, \quad z = 1$$ Would you like more details on any of the steps or have any questions? Here are some related questions: 1. What is Gaussian elimination, and why is it useful? 2. How does back-substitution work in solving systems of equations? 3. Can Gaussian elimination be applied to non-square matrices? 4. What is the difference between row echelon form and reduced row echelon form? 5. How does Gaussian elimination relate to matrix inverses? **Tip:** Always check your solution by substituting values back into the original system to verify correctness.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Gaussian Elimination
Matrix Operations
System of Linear Equations

Formulas

Gaussian Elimination Process
Row Echelon Form
Back-substitution

Theorems

Gaussian Elimination Theorem

Suitable Grade Level

Grades 10-12