Problem 26 - Intermediate Algebra

Question

For Problems \(11-30\), use Cramer's rule to find the solution set of each system. (Objective 2) $$ \left(\begin{array}{rr} 7 x-2 y+3 z= & -4 \\ 5 x+2 y-3 z= & 4 \\ -3 x-6 y+12 z= & -13 \end{array}\right) $$

Step-by-Step Solution

Answer

\((x, y, z) = (-\frac{8}{3}, \frac{5}{2}, -\frac{1}{3})\)

Step 1: Set Up Matrices

First, identify the coefficient matrix \(A\), and the constant matrix \(B\) from the given system of equations. Matrix \(A\) is:\[A = \begin{bmatrix}7 & -2 & 3 \5 & 2 & -3 \-3 & -6 & 12\end{bmatrix}\]Matrix \(B\) is:\[B = \begin{bmatrix}-4 \4 \-13\end{bmatrix}\]

Step 2: Calculate Determinant of Coefficient Matrix (\(\Delta\))

Calculate the determinant of the coefficient matrix \(A\):\[\Delta = \begin{vmatrix}7 & -2 & 3 \5 & 2 & -3 \-3 & -6 & 12\end{vmatrix}\]Compute this determinant using the cofactor expansion or any method:\[\Delta = 7(2 \cdot 12 - (-3) \cdot (-6)) + 2(5 \cdot 12 - (-3) \cdot (-3)) + 3(5 \cdot (-6) - 2 \cdot (-3))\]\[= 7(24 - 18) + 2(60 - 9) + 3(-30 + 6)\]\[= 42 + 102 - 72 = 72\]

Step 3: Calculate Determinant for \(x\) (\(\Delta_x\))

Replace the first column of \(A\) with \(B\) to find \(\Delta_x\):\[A_x = \begin{bmatrix}-4 & -2 & 3 \4 & 2 & -3 \-13 & -6 & 12\end{bmatrix}\]Calculate \(\Delta_x\):\[\Delta_x = \begin{vmatrix}-4 & -2 & 3 \4 & 2 & -3 \-13 & -6 & 12\end{vmatrix}\]\[= (-4)(2 \cdot 12 - (-3) \cdot (-6)) - (-2)(4 \cdot 12 - (-13) \cdot (-3)) + 3(4 \cdot (-6) - 2 \cdot (-13))\]\[= (-4)(24 - 18) + 2(48 + 39) + 3(-24 + 26)\]\[= -24 - 174 + 6 = -192\]

Step 4: Calculate Determinant for \(y\) (\(\Delta_y\))

Replace the second column of \(A\) with \(B\) to find \(\Delta_y\):\[A_y = \begin{bmatrix}7 & -4 & 3 \5 & 4 & -3 \-3 & -13 & 12\end{bmatrix}\]Calculate \(\Delta_y\):\[\Delta_y = \begin{vmatrix}7 & -4 & 3 \5 & 4 & -3 \-3 & -13 & 12\end{vmatrix}\]\[= 7(4 \cdot 12 - (-3) \cdot (-13)) + 4(5 \cdot 12 - (-3) \cdot 3) + 3(5 \cdot (-13) - 4 \cdot (-3))\]\[= 7(48 - 39) + 4(60 + 9) + 3(-65 + 12)\]\[= 63 + 276 - 159 = 180\]

Step 5: Calculate Determinant for \(z\) (\(\Delta_z\))

Replace the third column of \(A\) with \(B\) to find \(\Delta_z\):\[A_z = \begin{bmatrix}7 & -2 & -4 \5 & 2 & 4 \-3 & -6 & -13\end{bmatrix}\]Calculate \(\Delta_z\):\[\Delta_z = \begin{vmatrix}7 & -2 & -4 \5 & 2 & 4 \-3 & -6 & -13\end{vmatrix}\]\[= 7(2 \cdot (-13) - 4 \cdot (-6)) + 2(5 \cdot (-13) - (-3) \cdot 4) + (-4)(5 \cdot (-6) - 2 \cdot (-3))\]\[= 7(-26 + 24) + 2(-65 + 12) - 4(-30 + 6)\]\[= -14 - 106 + 96 = -24\]

Step 6: Calculate Solutions Using Cramer's Rule

From Cramer's Rule, the solution is given by:\[x = \frac{\Delta_x}{\Delta}\]\[y = \frac{\Delta_y}{\Delta}\]\[z = \frac{\Delta_z}{\Delta}\]Plug in the values:\[x = \frac{-192}{72} = -\frac{8}{3}\]\[y = \frac{180}{72} = \frac{5}{2}\]\[z = \frac{-24}{72} = -\frac{1}{3}\]

Step 7: Final Step: Write Down the Solution Set

The solution set for the system is:\[(x, y, z) = \left(-\frac{8}{3}, \frac{5}{2}, -\frac{1}{3}\right)\]

Key Concepts

Systems of Equations

When dealing with systems of equations, you aim to find values for the variables that satisfy all the equations simultaneously. In our example, we have three equations and three unknowns, specifically:

• \( 7x - 2y + 3z = -4 \)
• \( 5x + 2y - 3z = 4 \)
• \( -3x - 6y + 12z = -13 \)

To solve such a system using Cramer's Rule, it's important to understand the structure and relationship of these equations. Each equation represents a plane in a three-dimensional space.
The solution, often a point represented by \((x, y, z)\), is where all the planes intersect.
Thus, solving the system means finding the coordinates of this intersection point.

Determinants

Determinants play a vital role in matrix algebra, especially when using Cramer's Rule to solve systems of equations. A determinant is a special number calculated from a square matrix, giving information about the matrix properties such as invertibility.
In our problem, the determinant \(\Delta\) of the coefficient matrix \(A\) is calculated as follows:\[\Delta = \begin{vmatrix}7 & -2 & 3 \ 5 & 2 & -3 \ -3 & -6 & 12\end{vmatrix} = 72\]This value must be non-zero for a unique solution to exist.
If the determinant were zero, the system could either have no solutions or an infinite number of solutions.
Determinants are calculated using cofactor expansion and involve quite a bit of arithmetic.

Matrix Algebra

Matrix algebra simplifies the handling of multiple equations and unknowns. Matrices are rectilinear arrays of numbers that represent systems of equations. The key matrices in our problem are the coefficient matrix \(A\) and the constant matrix \(B\).

• **Matrix \(A\)** contains the coefficients of the variables:\[A = \begin{bmatrix}7 & -2 & 3 \ 5 & 2 & -3 \ -3 & -6 & 12\end{bmatrix}\]
• **Matrix \(B\)** represents the constants on the right side of each equation:\[B = \begin{bmatrix}-4 \ 4 \ -13\end{bmatrix}\]

By organizing the system in matrix form, Cramer's Rule is applied, and we substitute columns of \(A\) with \(B\) to calculate determinants necessary for each variable's solution.
This approach streamlines computations and clearly presents the solution procedure.

Solving Linear Systems

Solving linear systems using Cramer's Rule allows us to find definite solutions for variables unless the determinant is zero. Here's how it works:

• Calculate the determinant \(\Delta\) of the coefficient matrix \(A\).
• For each variable \(x, y, z\), replace the respective column in \(A\) with matrix \(B\) to get \(A_x, A_y, A_z\).
• Calculate \(\Delta_x, \Delta_y, \Delta_z\) using the determinants of these new matrices.
• Apply Cramer's Rule: \[x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}, \quad z = \frac{\Delta_z}{\Delta}\]

In this example, we found the solutions:

• \(x = -\frac{8}{3}\)
• \(y = \frac{5}{2}\)
• \(z = -\frac{1}{3}\)

Using Cramer's Rule ensures clarity and consistency while solving linear systems.