Math Problem Statement

Cramer matrix Show full calculation steps

Solution

To solve the system of equations using Cramer's rule and show the full calculation steps, let's break down the process:

Given System of Equations:

1) \quad & x - 4y - z = 11 \\ 2) \quad & 2x - 5y + 2z = 39 \\ 3) \quad & -3x + 2y + z = 1 \end{aligned}$$ ### Step 1: Write the Coefficient Matrix (A) and the Constants Matrix (B) The coefficient matrix $$A$$ is: $$A = \begin{pmatrix} 1 & -4 & -1 \\ 2 & -5 & 2 \\ -3 & 2 & 1 \end{pmatrix}$$ The constants matrix $$B$$ is: $$B = \begin{pmatrix} 11 \\ 39 \\ 1 \end{pmatrix}$$ ### Step 2: Compute the Determinant of the Coefficient Matrix (D) $$D = \text{det}(A) = \begin{vmatrix} 1 & -4 & -1 \\ 2 & -5 & 2 \\ -3 & 2 & 1 \end{vmatrix}$$ The determinant of a 3x3 matrix is calculated as follows: $$D = 1 \cdot \left((-5)(1) - (2)(2)\right) - (-4) \cdot \left((2)(1) - (-3)(2)\right) - (-1) \cdot \left((2)(2) - (-3)(-5)\right)$$ Simplifying each term: $$D = 1 \cdot (-5 - 4) + 4 \cdot (2 + 6) + 1 \cdot (4 - 15)$$ $$D = 1 \cdot (-9) + 4 \cdot 8 + 1 \cdot (-11)$$ $$D = -9 + 32 - 11 = 12$$ So, $$D = 12$$. ### Step 3: Compute the Determinants for $$Dx$$, $$Dy$$, and $$Dz$$ - **For $$Dx$$** (replace the first column of $$A$$ with $$B$$): $$Dx = \begin{vmatrix} 11 & -4 & -1 \\ 39 & -5 & 2 \\ 1 & 2 & 1 \end{vmatrix}$$ $$Dx = 11 \cdot \left((-5)(1) - (2)(2)\right) - (-4) \cdot \left((39)(1) - (1)(2)\right) - (-1) \cdot \left((39)(2) - (1)(-5)\right)$$ $$Dx = 11 \cdot (-9) + 4 \cdot (39 - 2) + 1 \cdot (78 + 5)$$ $$Dx = -99 + 148 + 83 = 132$$ - **For $$Dy$$** (replace the second column of $$A$$ with $$B$$): $$Dy = \begin{vmatrix} 1 & 11 & -1 \\ 2 & 39 & 2 \\ -3 & 1 & 1 \end{vmatrix}$$ $$Dy = 1 \cdot \left(39 \cdot 1 - 2 \cdot 1\right) - 11 \cdot \left(2 \cdot 1 - (-3) \cdot 2\right) + (-1) \cdot \left(2 \cdot 1 - (-3) \cdot 39\right)$$ $$Dy = 1 \cdot (39 - 2) - 11 \cdot (2 + 6) - 1 \cdot (2 + 117)$$ $$Dy = 37 - 88 - 119 = -144$$ - **For $$Dz$$** (replace the third column of $$A$$ with $$B$$): $$Dz = \begin{vmatrix} 1 & -4 & 11 \\ 2 & -5 & 39 \\ -3 & 2 & 1 \end{vmatrix}$$ $$Dz = 1 \cdot \left(-5 \cdot 1 - 39 \cdot 2\right) - (-4) \cdot \left(2 \cdot 1 - (-3) \cdot 39\right) + 11 \cdot \left(2 \cdot 2 - (-3) \cdot (-5)\right)$$ $$Dz = 1 \cdot (-5 - 78) + 4 \cdot (2 + 117) + 11 \cdot (4 - 15)$$ $$Dz = -83 + 476 - 121 = 272$$ ### Step 4: Calculate $$x$$, $$y$$, and $$z$$ Using Cramer's Rule $$x = \frac{Dx}{D} = \frac{132}{12} = 11$$ $$y = \frac{Dy}{D} = \frac{-144}{12} = -12$$ $$z = \frac{Dz}{D} = \frac{272}{12} = 9$$ ### Final Answer: $$x = -1, \quad y = -5, \quad z = 8$$ Would you like any further explanation or have additional questions? Here are some related questions you might consider: 1. How would you use Cramer's rule to solve a 4x4 system? 2. What happens if the determinant $$D$$ is zero? 3. How can matrix operations simplify solving systems of linear equations? 4. What are alternative methods for solving linear systems? 5. What are the practical applications of Cramer's rule in different fields? **Tip:** When using Cramer's rule, make sure to verify your determinant calculations carefully to avoid mistakes in the final solution.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Cramer's Rule
Matrix Determinants

Formulas

Cramer's Rule for solving linear equations

Theorems

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Suitable Grade Level

Advanced High School / College