Problem

2. Consider the system of linear equations \[ \begin{array}{r} 2 y+3 z=0 \\ x+2 y+z=0 \\ -2 x-2 y+z=0 \end{array} \] a) [6 marks Solve the system using Gaussian or Gauss-Jordan elimination. b) $[6$ marks $]$ Write the system in matrix form $A X=B$. Find det $A$ by cofactor expansion. c) [2 marks] Is the system consistent? Is the system homogeneous?

Solution

Step 1 :Given the system of linear equations: \[\begin{array}{r} 2 y+3 z=0 \\ x+2 y+z=0 \\ -2 x-2 y+z=0 \end{array}\]

Step 2 :Write this system in augmented matrix form: \[\begin{array}{ccc|c} 0 & 2 & 3 & 0 \\ 1 & 2 & 1 & 0 \\ -2 & -2 & 1 & 0 \end{array}\]

Step 3 :Perform row operations to bring it to reduced row-echelon form: \[\begin{array}{ccc|c} 1 & 0 & -2 & 0 \\ 0 & 1 & 3/2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\]

Step 4 :This corresponds to the system of equations: \[\begin{array}{r} x - 2z = 0 \\ y + 3/2z = 0 \\ 0 = 0 \end{array}\]

Step 5 :From the first equation, express x in terms of z: \(x = 2z\)

Step 6 :From the second equation, express y in terms of z: \(y = -3/2z\)

Step 7 :Since there is no equation for z, let z be a free variable

Step 8 :The solution to the system of equations is \(x = 2z\), \(y = -3/2z\), z is free

Step 9 :Final Answer: The solution to the system of equations is \(\boxed{x = 2z, y = -3/2z, z \text{ is free}}\)

From Solvely APP
Source: https://solvelyapp.com/problems/22969/

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