Math Problem Statement
Solve the System. Give answer as (x,y,z).
{6x-1y+1z=27 12x+1y-5z=33 -24x+1y+4z=-84
Solution
To solve the system of equations, we have:
6x - y + z = 27 \\ 12x + y - 5z = 33 \\ -24x + y + 4z = -84 \end{cases}$$ ### Step 1: Write the system in augmented matrix form The augmented matrix for this system is: $$\begin{bmatrix} 6 & -1 & 1 & | & 27 \\ 12 & 1 & -5 & | & 33 \\ -24 & 1 & 4 & | & -84 \\ \end{bmatrix}$$ ### Step 2: Use row operations to simplify the matrix 1. Divide the first row by 6 to make the coefficient of $$x$$ in the first row equal to 1. $$R_1 = \frac{1}{6} R_1 \Rightarrow \begin{bmatrix} 1 & -\frac{1}{6} & \frac{1}{6} & | & 4.5 \end{bmatrix}$$ So the new augmented matrix becomes: $$\begin{bmatrix} 1 & -\frac{1}{6} & \frac{1}{6} & | & 4.5 \\ 12 & 1 & -5 & | & 33 \\ -24 & 1 & 4 & | & -84 \\ \end{bmatrix}$$ 2. Make the entries below the leading 1 in the first column zero by adjusting rows 2 and 3. - For row 2: $$R_2 = R_2 - 12 \times R_1$$ - For row 3: $$R_3 = R_3 + 24 \times R_1$$ After these operations, we get: $$\begin{bmatrix} 1 & -\frac{1}{6} & \frac{1}{6} & | & 4.5 \\ 0 & 3 & -7 & | & -21 \\ 0 & 5 & 8 & | & 24 \\ \end{bmatrix}$$ ### Step 3: Make the leading coefficient of the second row a 1 by dividing $$R_2$$ by 3. $$R_2 = \frac{1}{3} R_2 \Rightarrow \begin{bmatrix} 0 & 1 & -\frac{7}{3} & | & -7 \end{bmatrix}$$ The matrix now is: $$\begin{bmatrix} 1 & -\frac{1}{6} & \frac{1}{6} & | & 4.5 \\ 0 & 1 & -\frac{7}{3} & | & -7 \\ 0 & 5 & 8 & | & 24 \\ \end{bmatrix}$$ 3. Make the entry below the leading 1 in the second column zero by adjusting row 3: - For row 3: $$R_3 = R_3 - 5 \times R_2$$ After this operation, we get: $$\begin{bmatrix} 1 & -\frac{1}{6} & \frac{1}{6} & | & 4.5 \\ 0 & 1 & -\frac{7}{3} & | & -7 \\ 0 & 0 & \frac{71}{3} & | & 59 \\ \end{bmatrix}$$ ### Step 4: Make the leading coefficient of the third row a 1 by dividing $$R_3$$ by $$\frac{71}{3}$$. $$R_3 = \frac{3}{71} R_3 \Rightarrow \begin{bmatrix} 0 & 0 & 1 & | & \frac{177}{71} \end{bmatrix}$$ The matrix now is: $$\begin{bmatrix} 1 & -\frac{1}{6} & \frac{1}{6} & | & 4.5 \\ 0 & 1 & -\frac{7}{3} & | & -7 \\ 0 & 0 & 1 & | & \frac{177}{71} \\ \end{bmatrix}$$ ### Step 5: Back-substitute to solve for $$x$$, $$y$$, and $$z$$ 1. Substitute $$z = \frac{177}{71}$$ into the second equation to solve for $$y$$. 2. Substitute both $$y$$ and $$z$$ values into the first equation to solve for $$x$$. After simplifying, we find: $$(x, y, z) = \left(3, -7, \frac{177}{71}\right)$$ Let me know if you need further details or have any questions! Here are some related questions for further practice: 1. How do row operations help in solving systems of linear equations? 2. Can every system of linear equations be solved using the augmented matrix method? 3. How can Gaussian elimination be applied to larger systems of equations? 4. What are the possible outcomes of a system of equations (consistent, inconsistent, dependent)? 5. How would you verify the solution to this system of equations? **Tip:** In systems with fractions, handling each fraction step-by-step or converting to decimal can simplify calculations.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Systems of Linear Equations
Matrix Operations
Formulas
Augmented Matrix
Row Operations for Gaussian Elimination
Theorems
Gaussian Elimination
Back-substitution in Linear Systems
Suitable Grade Level
Grades 10-12