Cramer's Rule and Systems of Equations Study Guide

Overview of Cramer's Rule for Systems of Linear Equations

Cramer's Rule is a mathematical method used to solve a system of $n$ linear equations with $n$ variables by utilizing determinants. For a system of three equations with three variables ($x$, $y$, and $z$), the general form is:

$\begin{cases} a_1 x + b_1 y + c_1 z = d_1 \\ a_2 x + b_2 y + c_2 z = d_2 \\ a_3 x + b_3 y + c_3 z = d_3 \end{cases}$

The variables are calculated as follows:

$x = \frac{\Delta_x}{\Delta}$

$y = \frac{\Delta_y}{\Delta}$

$z = \frac{\Delta_z}{\Delta}$

Where:

• $\Delta$ is the determinant of the coefficient matrix:

$\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$

• $\Delta_x$ is the determinant formed by replacing the $x$-coefficients with the constants $d_1$, $d_2$, and $d_3$:

$\Delta_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}$

• $\Delta_y$ is the determinant formed by replacing the $y$-coefficients with the constants:

$\Delta_y = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}$

• $\Delta_z$ is the determinant formed by replacing the $z$-coefficients with the constants:

$\Delta_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}$

Detailed Analysis of Question 106

Question 106 asks to identify the correct response for the following system of equations using Cramer's Rule:

$x - y + 2z = -3$$x + 2y + 3z = 4$$2x + y + z = -3$

Step 1: Calculate the Main Determinant ($\Delta$)

$\Delta = \begin{vmatrix} 1 & -1 & 2 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix}$

Using the expansion by the first row: $\Delta = 1 \times (2 \times 1 - 3 \times 1) - (-1) \times (1 \times 1 - 3 \times 2) + 2 \times (1 \times 1 - 2 \times 2)$$\Delta = 1 \times (2 - 3) + 1 \times (1 - 6) + 2 \times (1 - 4)$$\Delta = 1 \times (-1) + 1 \times (-5) + 2 \times (-3)$$\Delta = -1 - 5 - 6 = -12$

Step 2: Calculate $\Delta_x$

$\Delta_x = \begin{vmatrix} -3 & -1 & 2 \\ 4 & 2 & 3 \\ -3 & 1 & 1 \end{vmatrix}$

$\Delta_x = -3 \times (2 - 3) - (-1) \times (4 - (-9)) + 2 \times (4 - (-6))$$\Delta_x = -3 \times (-1) + 1 \times (13) + 2 \times (10)$$\Delta_x = 3 + 13 + 20 = 36$$x = \frac{36}{-12} = -3$

Step 3: Calculate $\Delta_y$

$\Delta_y = \begin{vmatrix} 1 & -3 & 2 \\ 1 & 4 & 3 \\ 2 & -3 & 1 \end{vmatrix}$

$\Delta_y = 1 \times (4 - (-9)) - (-3) \times (1 - 6) + 2 \times (-3 - 8)$$\Delta_y = 1 \times (13) + 3 \times (-5) + 2 \times (-11)$$\Delta_y = 13 - 15 - 22 = -24$$y = \frac{-24}{-12} = 2$

Step 4: Calculate $\Delta_z$

$\Delta_z = \begin{vmatrix} 1 & -1 & -3 \\ 1 & 2 & 4 \\ 2 & 1 & -3 \end{vmatrix}$

$\Delta_z = 1 \times (-6 - 4) - (-1) \times (-3 - 8) + (-3) \times (1 - 4)$$\Delta_z = 1 \times (-10) + 1 \times (-11) - 3 \times (-3)$$\Delta_z = -10 - 11 + 9 = -12$$z = \frac{-12}{-12} = 1$

Conclusion for Question 106: The solution is $x = -3$, $y = 2$, and $z = 1$. This corresponds to Option B.

Detailed Analysis of Question 107

Question 107 asks for the solution to the following system using Cramer's Rule:

$2x - 3y + 5z = -8$$3x + 2y - z = -3$$4x + y + 6z = 4$

Step 1: Calculate the Main Determinant ($\Delta$)

$\Delta = \begin{vmatrix} 2 & -3 & 5 \\ 3 & 2 & -1 \\ 4 & 1 & 6 \end{vmatrix}$

$\Delta = 2 \times (12 - (-1)) - (-3) \times (18 - (-4)) + 5 \times (3 - 8)$$\Delta = 2 \times (13) + 3 \times (22) + 5 \times (-5)$$\Delta = 26 + 66 - 25 = 67$

Step 2: Calculate $\Delta_x$

$\Delta_x = \begin{vmatrix} -8 & -3 & 5 \\ -3 & 2 & -1 \\ 4 & 1 & 6 \end{vmatrix}$

$\Delta_x = -8 \times (12 - (-1)) - (-3) \times (-18 - (-4)) + 5 \times (-3 - 8)$$\Delta_x = -8 \times (13) + 3 \times (-14) + 5 \times (-11)$$\Delta_x = -104 - 42 - 55 = -201$$x = \frac{-201}{67} = -3$

Step 3: Calculate $\Delta_y$

$\Delta_y = \begin{vmatrix} 2 & -8 & 5 \\ 3 & -3 & -1 \\ 4 & 4 & 6 \end{vmatrix}$

$\Delta_y = 2 \times (-18 - (-4)) - (-8) \times (18 - (-4)) + 5 \times (12 - (-12))$$\Delta_y = 2 \times (-14) + 8 \times (22) + 5 \times (24)$$\Delta_y = -28 + 176 + 120 = 268$$y = \frac{268}{67} = 4$

Step 4: Calculate $\Delta_z$

$\Delta_z = \begin{vmatrix} 2 & -3 & -8 \\ 3 & 2 & -3 \\ 4 & 1 & 4 \end{vmatrix}$

$\Delta_z = 2 \times (8 - (-3)) - (-3) \times (12 - (-12)) + (-8) \times (3 - 8)$$\Delta_z = 2 \times (11) + 3 \times (24) - 8 \times (-5)$$\Delta_z = 22 + 72 + 40 = 134$$z = \frac{134}{67} = 2$

Conclusion for Question 107: The solution is $x = -3$, $y = 4$, and $z = 2$. This corresponds to Option A.

Inequality Analysis (Question 108)

Question 108 introduces the topic of linear inequalities. The transcript provides the starting line of a system of linear inequalities:

$2x + y \le 3$

A linear inequality in two variables describes a region of the Cartesian plane. To solve such a system, one typically graphs the boundary line (in this case, $2x + y = 3$) and shades the region that satisfies the inequality. Because the inequality symbol is $\le$ (less than or equal to), the boundary line itself is included in the solution set (represented by a solid line).

Institutional Context

These problems are sourced from a questionnaire (cuestionario) provided by the "UNIDAD EDUCATIVA 'OTAVALO'", located in the Republic of Ecuador. The institution covers multiple levels of education including:

• Inicial (Initial)
• Preparatoria (Preparatory)
• EGB (Educación General Básica: Elemental, Media, Superior)
• Bachillerato Técnico (Technical Baccalaureate)
• General Unificado (Unified General)

The document is identified by the code AMIE 10H00411 and refers to page 19 of 20 of the questionnaire material.