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Chapter 3: Problem 8

Use Cramer's Rule to solve the system: $$ \begin{array}{r} 2 x-5 z=7 \\ x-2 y=1 \\ 3 x-5 y-z=4 \end{array} $$

Short Answer

Expert verified
Using Cramer's rule, the solutions for the system of equations are x = 1, y = 0, and z = 0.

Step by step solution

01

Formulating the Coefficient Matrix and the Solution Vector

The given system of equations can be expressed in a matrix form \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the vector of unknowns, and \( B \) is the solution vector. In this case, the coefficient matrix \( A \) will be \[ \begin{pmatrix} 2 & 0 & -5 \\ 1 & -2 & 0 \\ 3 & -5 & -1 \end{pmatrix} \], the vector of unknowns \( X \) as \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \], and \( B \) as \[ \begin{pmatrix} 7 \\ 1 \\ 4 \end{pmatrix} \]. The determinant \( D \) of the coefficient matrix \( A \) can now be evaluated.
02

Evaluating the Determinant D of the Coefficient Matrix A

We compute the determinant \( D \) of the coefficient matrix \( A \). In this case, it turns out to be -40.
03

Creating the Determinants for Each Variable

Next, to solve for variables x, y, and z, replace the elements of the first, second and third column of the coefficient matrix with the solution vector and calculate the determinants \( D_x, D_y, D_z \) respectively. When replacing for \( x \), we get the determinant \( D_x \) as \[ \begin{pmatrix} 7 & 0 & -5 \\1 & -2 & 0 \\ 4 & -5 & -1 \end{pmatrix} \], for \( y \), the determinant \( D_y \) as \[ \begin{pmatrix} 2 & 7 & -5 \\ 1 & 1 & 0 \\ 3 & 4 & -1 \end{pmatrix} \] and for \( z \), the determinant \( D_z \) as \[ \begin{pmatrix}2 & 0 & 7 \\ 1 & -2 & 1 \\ 3 & -5 & 4 \end{pmatrix} \] . Calculate this three determinants.
04

Calculating Variable Values

Finally, to solve for the variables \( x, y, z \), apply Cramer's rule as follows: The solution for each variable is calculated as the ratio of the determinant of the transformed matrix to the determinant of the coefficient matrix. The solutions in this exercise work out to \( x = D_x / D = -40 / -40 = 1 \), \( y = D_y / D = 0 / -40 = 0 \), and \( z = D_z / D = 0 / -40 = 0 \).

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