Question

Three students run on a racing track such that their speeds add up to 6 km/h. However, double the speed of the third runner added to the speed of the first results in 7 km/h. If thrice the speed of the first runner is added to the original speeds of the other two, the result is 12 km/h. Using the matrix method, find the original speed of each runner.

Hint:

To solve a system of linear equations using the matrix method, express the system in the form \( A \cdot X = B \), where \( A \) is the coefficient matrix, \( X \) is the vector of unknowns, and \( B \) is the constants vector. Then, use matrix inversion or Gaussian elimination to solve for \( X \).

Answer 1

The speeds of the three students are denoted as follows: \( x \) km/h for the first student, \( y \) km/h for the second student, and \( z \) km/h for the third student. The given conditions are: 1. The sum of their speeds is 6 km/h: \[ x + y + z = 6 \quad (1) \] 2. Twice the speed of the third student plus the speed of the first student equals 7 km/h: \[ x + 2z = 7 \quad (2) \] 3. Three times the speed of the first student plus the speeds of the other two students equals 12 km/h: \[ 3x + y + z = 12 \quad (3) \] This system of linear equations can be represented in matrix form: \[ \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 6 \\ 7 \\ 12 \\ \end{pmatrix} \] Solving this system using matrix methods yields the following results: \( x = 3 \), \( y = 1 \), \( z = 2 \). Therefore, the speeds of the students are: - First student: \( 3 \, \text{km/h} \) - Second student: \( 1 \, \text{km/h} \) - Third student: \( 2 \, \text{km/h} \)