Systems of Linear Equations Introduction

and x, y, z can be found by back substitution:x = 2, y = 3, z = −1.The Gauss-Jordan elimination algorithm gives the reduced row echelonform⎛⎞1 0 0 2⎜⎟⎝ 0 1 0 3 ⎠ .0 0 1 −1So the solution is the last columnx = 2, y = 3, z = −1.3.2.2 Example of a system with many solutions.Consider the system⎧⎪⎨⎪⎩x + 2y + 0 · z + 3t = 52x + 3y + 2z + 5t = 40 · x + 2y + z + 5t = 2The augmented matrix of this system is⎛⎜⎝1 2 0 3 5−2 3 2 5 40 2 1 5 2⎞⎟⎠ .The Gauss elimination algorithm gives the following row echelon form⎛⎜⎝Corresponding system looks as⎧⎪⎨⎪⎩1 2 0 3 50 −1 2 −1 −60 0 5 3 −10⎞⎟⎠ .x + 2y + 0 · z + 3t = 5−y + 2z − t = −65z + 3t = −10Take free variable t. Then the back substitution givesz = − 3 5 t − 2, y = −11 5 t + 2, x = 7 5 t + 1.7