Problem 68 - A Graphical Approach to College Algebra

Question

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{aligned}x+y+z &=4 \\\2 x-y+3 z &=4 \\\4 x+2 y-z &=-15\end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{33}{17}, y = \frac{51}{17}, z = 5\).

Step 1: Identify the System of Equations

The given system of equations is: \[ \begin{align*} x + y + z &= 4 \ 2x - y + 3z &= 4 \ 4x + 2y - z &= -15 \end{align*} \] We need to solve these equations using Cramer's rule.

Step 2: Write the Coefficient Matrix

The coefficient matrix \(A\) for the system is \[ A = \begin{bmatrix} 1 & 1 & 1 \ 2 & -1 & 3 \ 4 & 2 & -1 \end{bmatrix} \]

Step 3: Calculate the Determinant of the Coefficient Matrix

We find the determinant \(D\) of the matrix \(A\) using the formula for the determinant of a 3x3 matrix:\[ D = 1((-1)(-1) - 3(2)) - 1(2(-1) - (4)(3)) + 1(2(2) - 4(-1)) \]This simplifies to:\[ D = (1)(1 - 6) - (1)(-2 - 12) + (1)(4 + 4) \]\[ D = (1)(-5) + (1)(14) + (1)(8) \]\[ D = -5 + 14 + 8 = 17 \] So \(D = 17\).

Step 4: Calculate the Determinant for X-variable (Dx)

Replace the first column of \(A\) with the constants from the equations to get matrix \(A_x\):\[ A_x = \begin{bmatrix} 4 & 1 & 1 \ 4 & -1 & 3 \ -15 & 2 & -1 \end{bmatrix} \]Now, calculate the determinant \(D_x\):\[ D_x = 4((-1)(-1) - 3(2)) - 1(4(-1) - (3)(-15)) + 1(4(2) - 4(-1)) \]This simplifies to:\[ D_x = 4(1 - 6) - (1)(-4 + 45) + (1)(8 + 4) \]\[ D_x = 4(-5) + 41 + 12 \]\[ D_x = -20 + 41 + 12 = 33 \] So \(D_x = 33\).

Step 5: Calculate the Determinant for Y-variable (Dy)

Replace the second column of \(A\) with the constants from the equations to get matrix \(A_y\):\[ A_y = \begin{bmatrix} 1 & 4 & 1 \ 2 & 4 & 3 \ 4 & -15 & -1 \end{bmatrix} \]Now, calculate the determinant \(D_y\):\[ D_y = 1(4(-1) - 3(-15)) - 4(2(-1) - 3(4)) + 1(2(-15) - 4(4)) \]This simplifies to:\[ D_y = 1(-4 + 45) - 4(-2 - 12) + 1(-30 - 16) \]\[ D_y = 41 + 56 - 46 \]\[ D_y = 97 - 46 = 51 \] So \(D_y = 51\).

Step 6: Calculate the Determinant for Z-variable (Dz)

Replace the third column of \(A\) with the constants from the equations to get matrix \(A_z\):\[ A_z = \begin{bmatrix} 1 & 1 & 4 \ 2 & -1 & 4 \ 4 & 2 & -15 \end{bmatrix} \]Now, calculate the determinant \(D_z\):\[ D_z = 1((-1)(-15) - 2(4)) - 1(2(-15) - 4(4)) + 4(2(2) - 4(-1)) \]This simplifies to:\[ D_z = 1(15 - 8) - 1(-30 - 16) + 4(4 + 4) \]\[ D_z = 7 + 46 + 32 \]\[ D_z = 85 \] So \(D_z = 85\).

Step 7: Apply Cramer's Rule to Find Solutions

Using Cramer's rule:\[ x = \frac{D_x}{D} = \frac{33}{17} \] \[ y = \frac{D_y}{D} = \frac{51}{17} \] \[ z = \frac{D_z}{D} = \frac{85}{17} \]

Step 8: Simplify the Solutions

Finally, simplify the solutions as:\[ x = \frac{33}{17}, \quad y = \frac{51}{17}, \quad z = 5 \]

Key Concepts

Understanding Systems of Equations

A system of equations is a collection of equations that involve the same set of variables. In mathematical terms, these are equations with two or more unknowns that are related in such a way that the same variables appear in each one. The main objective is to find a common solution set that satisfies all the given equations at the same time.
In the original exercise, the system of equations is presented with three equations: \(x + y + z = 4\), \(2x - y + 3z = 4\), \(4x + 2y - z = -15\). Each equation is related since they all contain the same three variables (x, y, and z).

To tackle these, we often use methods like substitution, elimination, or more advanced techniques such as matrix algebra. Cramer's Rule is one such method that specifically applies when we have a square matrix derived from the coefficients of the equations.

The Role of the Determinant in Solving Equations

The determinant is a special number that can be calculated from a square matrix. It provides crucial information about the matrix, such as whether it has an inverse and whether a set of linear equations has a unique solution.
When using Cramer's Rule, the determinant of the coefficient matrix (matrix derived from just the coefficients of the variables in the equations) is key. If this determinant is non-zero, it guarantees a unique solution to the system of equations.
In the original problem, the coefficient matrix of the system of equations is given as \([1 \, 1 \, 1; \, 2 \, -1 \, 3; \, 4 \, 2 \, -1]\) and its determinant is computed to be 17. Since 17 is not zero, this tells us that the system has a unique solution that can be determined using Cramer's Rule.

Matrix Algebra and its Application

Matrix algebra allows us to work with arrays of numbers in a structured way. It simplifies complex systems of equations into more manageable components using matrices, which are rectangular arrays of numbers, symbols, or expressions arranged in rows and columns.
To solve a system of equations using Cramer's Rule, we construct matrices from the coefficients of the variables and the constants. The coefficient matrix is just the matrix made up of the coefficients of the variables, while the constant matrix is made up of the numbers on the other side of the equations.

Cramer's Rule then uses these matrices to find values for each variable independently by replacing one column of the coefficient matrix at a time with the constants and calculating the resulting determinants. This process involves several steps of matrix manipulation, such as calculating the determinant of 3x3 matrices which provides partial determinants for each variable using their respective modified matrices.

Finding the Solution of Equations using Cramer's Rule

Cramer's Rule is a straightforward method used for finding the solution of a system of linear equations, provided the coefficient matrix is square and has a non-zero determinant.
It works by calculating the determinant of the coefficient matrix, and then substituting each column of this matrix with the constants vector to find the determinants for each variable. These are referred to as partial determinants: \(D_x\), \(D_y\), and \(D_z\).
The values of the variables are found by dividing these partial determinants by the determinant of the coefficient matrix. In our example: \(x = \frac{D_x}{D}, \, y = \frac{D_y}{D}, \, z = \frac{D_z}{D}\).

Ultimately, for the given exercise, Cramer's Rule reveals the solutions: \(x = \frac{33}{17}\), \(y = \frac{51}{17}\), \(z = 5\), denoting that these values when substituted back into the original equations will satisfy all three equations simultaneously.