Problem 14 - College Algebra

Question

Use Cramer's rule to solve each system or to determine that the system is inconsistent or contains dependent equations. $$ \begin{aligned}&x-2 y=5\\\&5 x-y=-2\end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system of equations is \( x = -9/11 \) and \( y = -27/11 \).

Step 1: Compute the Determinant of the Coefficient Matrix

The coefficient matrix of the system is given by the coefficients of x and y in our equations. Therefore, our 2x2 matrix is \(\begin{bmatrix}1 & -2\ 5 & -1\end{bmatrix}\). The determinant of a 2x2 matrix \(\begin{bmatrix}a & b\ c & d\end{bmatrix}\) is given by \(ad - bc\). Therefore, the determinant for our matrix, denoted as D, is \( (1*-1) - (5*-2) = 1 + 10 = 11 \).

Step 2: Compute the Determinant of X

We replace the coefficients of x in our original matrix with the solution vector, which consists of the solutions on the right-hand side of our system of equations (5 and -2). Now we have: \(\begin{bmatrix}5 & -2\ -2 & -1\end{bmatrix}\). The determinant of this matrix, denoted as Dx, is \( (5*-1) - (-2*-2) = -5 - 4 = -9 \).

Step 3: Compute the Determinant of Y

This time we replace the coefficients of y with the solution vector. The result is: \(\begin{bmatrix}1 & 5\ 5 & -2\end{bmatrix}\). The determinant of this matrix, denoted as Dy, is \( (1*-2) - (5*5) = -2 - 25 = -27 \).

Step 4: Apply Cramer's Rule

The x-value is computed as the determinant of X divided by the determinant of the coefficient matrix. The y-value is computed likewise. Therefore, we get \( x = Dx/D = -9/11 \) and \( y = Dy/D = -27/11 \).

Key Concepts

Determinant of a matrix

In linear algebra, the determinant is a scalar value that can be computed from the elements of a square matrix. It gives us important information about the matrix, including whether it is invertible, which is crucial in solving systems of linear equations using Cramer's rule. For a 2x2 matrix, the determinant is calculated by the formula \( ad - bc \) when our matrix is represented as \( \begin{bmatrix}a & b\ c & d\end{bmatrix} \).

In the context of our exercise, the coefficient matrix is \( \begin{bmatrix}1 & -2\ 5 & -1\end{bmatrix} \), and its determinant, denoted as D, can be calculated as follows: \( D = (1 \cdot -1) - (5 \cdot -2) = 1 + 10 = 11 \). This value plays a pivotal role in utilizing Cramer's rule to solve the system of equations. The determinant being non-zero also indicates that the system has a unique solution.

System of linear equations

A system of linear equations consists of two or more linear equations with the same set of variables. Our goal is to find the values of these variables that satisfy all equations in the system simultaneously. In our exercise, we're dealing with a system of two equations:

\[x - 2y = 5\]
\[5x - y = -2\]

When we represent these equations in a matrix form, where each row corresponds to each equation and each column corresponds to a coefficient of a variable, it allows for efficient computational techniques like Cramer's rule to be applied. This rule uses determinants to find the variable values (in this case, x and y) by evaluating the ratios of the determinant of the modified coefficient matrix to the determinant of the original coefficient matrix.

Coefficient matrix

The coefficient matrix is a matrix derived from the system of linear equations, containing only the coefficients of the variables. For our example:

\[\begin{bmatrix}1 & -2\ 5 & -1\end{bmatrix}\]

This matrix holds the coefficients of x and y from each equation of our system. When using Cramer's rule, we create modified versions of this matrix to find the determinant corresponding to each variable. For instance, to find Dx (determinant for x), we replace the first column (which corresponds to x) with the constants from the right side of the equations, resulting in \( \begin{bmatrix}5 & -2\ -2 & -1\end{bmatrix} \). A non-zero determinant of the coefficient matrix is a prerequisite for Cramer's rule to be applicable and indicates that the system of equations has a unique solution.