Problem 44 - Algebra and Trigonometry

Question

\(29-44\) Use Cramer's Rule to solve the system. $$ \left\\{\begin{array}{l}{x+y=1} \\ {y+z=2} \\ {z+w=3} \\\ {w-x=4}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \((x, y, z, w) = (0, -1, -1, -1)\).

Step 1: Identify the Coefficient Matrix

First, identify the coefficient matrix from the given system of equations. This system is:\[\begin{align*}x + y &= 1 \y + z &= 2 \z + w &= 3 \w - x &= 4 \end{align*}\]The coefficient matrix A is:\[A = \begin{bmatrix}1 & 1 & 0 & 0 \0 & 1 & 1 & 0 \0 & 0 & 1 & 1 \-1 & 0 & 0 & 1\end{bmatrix}\]

Step 2: Form the Determinant of the Coefficient Matrix

Calculate the determinant of the matrix \(A\). To do this, make sure \(A\) is square and calculate its determinant using any applicable method, such as expansion by minors or row reduction. Here, we'll simplify the computation by seeing that swapping rows and columns preserves the properties:\[det(A) = \begin{vmatrix}1 & 1 & 0 & 0 \0 & 1 & 1 & 0 \0 & 0 & 1 & 1 \-1 & 0 & 0 & 1\end{vmatrix} = 1\]because operations required for simplicity leave one term in this permutable identity matrix form.

Step 3: Apply Cramer's Rule to Solve for x

To find \(x\), replace the first column of \(A\) with the constants from the right side to form \(A_x\):\[A_x = \begin{bmatrix}1 & 1 & 0 & 0 \2 & 1 & 1 & 0 \3 & 0 & 1 & 1 \4 & 0 & 0 & 1 \end{bmatrix}\]Then calculate the determinant of \(A_x\).\[det(A_x) = 1 \times 1 - (-3) = 0\]Now solve for \(x\):\[ x = \frac{det(A_x)}{det(A)} = \frac{0}{1} = 0\]

Step 4: Apply Cramer's Rule to Solve for y

Replace the second column of \(A\) with the constants to form \(A_y\):\[A_y = \begin{bmatrix}1 & 1 & 0 & 0 \0 & 2 & 1 & 0 \0 & 3 & 1 & 1 \-1 & 4 & 0 & 1 \end{bmatrix}\]Calculate \(det(A_y)\).\[det(A_y) = 1 \times (-1) - 0 = -1\]Thus, solve for \(y\):\[ y = \frac{det(A_y)}{det(A)} = \frac{-1}{1} = -1\]

Step 5: Apply Cramer's Rule to Solve for z

Replace the third column of \(A\) with the constants to form \(A_z\):\[A_z = \begin{bmatrix}1 & 1 & 1 & 0 \0 & 1 & 2 & 0 \0 & 0 & 3 & 1 \-1 & 0 & 4 & 1\end{bmatrix}\]Calculate \(det(A_z)\).\[det(A_z) = 1 \times 1 - 0 = -1\]Thus, solve for \(z\):\[ z = \frac{det(A_z)}{det(A)} = \frac{-1}{1} = -1\]

Step 6: Apply Cramer's Rule to Solve for w

Replace the fourth column of \(A\) with the constants to form \(A_w\):\[A_w = \begin{bmatrix}1 & 1 & 0 & 1 \0 & 1 & 1 & 2 \0 & 0 & 1 & 3 \-1 & 0 & 0 & 4\end{bmatrix}\]Calculate \(det(A_w)\).\[det(A_w) = -1 \times (-2) = -1\]Thus, solve for \(w\):\[ w = \frac{det(A_w)}{det(A)} = \frac{-1}{1} = -1\]

Key Concepts

System of Linear Equations

A system of linear equations is a collection of linear equations involving the same set of variables. For example, in the given set like:\[ \begin{align*} x + y &= 1 \ y + z &= 2 \ z + w &= 3 \ w - x &= 4 \end{align*} \]These equations represent relationships among four variables \(x, y, z, \) and \(w\). The goal is to find values for these variables that satisfy all the equations simultaneously.
Linear systems can be solved using several methods, such as substitution, elimination, or matrix techniques like Cramer's Rule when applicable, which are particularly useful for more complex systems with multiple variables and equations.
This structured approach is essential to find consistent solutions and understand the interactions among variables.

Determinant Calculation

The determinant is a special number that can be calculated from a square matrix. It provides important properties about the matrix, such as whether it is invertible. For matrices involved in a system of linear equations, the determinant can tell us about the existence of a unique solution.
In Cramer's Rule, we calculate the determinant of the coefficient matrix as well as various altered matrices to find the values of the variables. Calculating the determinant can be done using several methods, such as row reduction or expansion by minors.
For a 4x4 matrix \(A\), as seen in this exercise, the determinant calculation is vital:\[ det(A) = \begin{vmatrix} 1 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 1 \ -1 & 0 & 0 & 1 \end{vmatrix} = 1 \] This determinant being non-zero confirms that the system has a unique solution.

Coefficient Matrix

The coefficient matrix is derived from the system of equations and contains all the coefficients of the variables. For instance, the system:\[ \begin{align*} x + y &= 1 \ y + z &= 2 \ z + w &= 3 \ w - x &= 4 \end{align*} \]translates into the coefficient matrix \(A\) as:\[A = \begin{bmatrix} 1 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 1 \ -1 & 0 & 0 & 1 \end{bmatrix}\]This matrix captures how each variable coefficient across all equations is structured and is critical for applications like determinant calculations and Cramer's Rule. The structure of the coefficient matrix directly influences how we solve the system, as it affects the properties like row operations and determinants.

Solving Linear Systems

Solving linear systems involves finding the values of the variables that satisfy all equations in the system simultaneously. Cramer's Rule offers a systematic way to do this when the number of equations matches the number of variables, making the coefficient matrix a square one.
Starting with the coefficient matrix, we determine if a unique solution exists by ensuring its determinant is non-zero. Then for each variable, a new matrix is formed by replacing the corresponding column with the constants from the right-hand side of the equations.
Applying Cramer's Rule, each variable \(x, y, z, \) and \(w\) is calculated. After calculating the determinants for each of these matrices, solutions are derived by dividing the determinant of each altered matrix by the determinant of the original coefficient matrix. This step-by-step rule is particularly advantageous in scenarios where variable solutions need to be derived algebraically from well-defined equations.