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

Let the speeds of the three students be: - \( x \) = speed of the first student (in km/h) - \( y \) = speed of the second student (in km/h) - \( z \) = speed of the third student (in km/h) We are given the following conditions: 1. The sum of their speeds is 6 km/h: \[ x + y + z = 6 \tag{1} \] 2. Double the speed of the third runner added to the speed of the first results in 7 km/h: \[ 2z + x = 7 \tag{2} \] 3. Thrice the speed of the first runner added to the original speeds of the other two results in 12 km/h: \[ 3x + y + z = 12 \tag{3} \] These equations can be written in matrix form as: \[ \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 6 \\ 7 \\ 12 \\ \end{pmatrix} \] Solving this system using matrix methods gives the solution: \[ x = 3, \quad y = 1, \quad z = 2 \] Thus, the original speeds of the students are: - First student: \( 3 \, \text{km/h} \) - Second student: \( 1 \, \text{km/h} \) - Third student: \( 2 \, \text{km/h} \)