How to determine 2 unknown value in a 3 x 3 matrix

Question

How to solve the following question

2 2 0 X = 0

P 2 0 Y = 0

Q -1 -1 Z =0

Answer 1

Given this matrix (call it A) we are looking to solve for the null space of A...

A = | 2 2 0 |

| P 2 0 |

| Q -1 -1 |

We can use row-echelon formalism to reduce the matrix A into a upper triangular format for easy solution gathering...

1) P*R₁ - 2*R₂ --> R₂

A = | 2 2 0 |

| 0 (2P-4) 0 |

| Q -1 -1 |

2) Q*R₁ - 2*R₃ --> R₃

A = | 2 2 0 |

| 0 (2P - 4) 0 |

| 0 (2Q - 2) -1 |

3a) The final representation is to get 2Q - 2 = -(2P - 4). We can write one of the unknowns (at this time) in terms of the other.

2Q - 2 = 4 - 2P

2Q = 6 - 2P

Q = 3 - P

3b) R₂ + R₃ --> R₃

A = | 2 2 0 |

| 0 (2P - 4) 0 |

| 0 (2P+2Q - 6) -1 |

A = | 2 2 0 |

| 0 (2P - 4) 0 |

| 0 2(P + Q - 3) -1 |

With the constraint above (in bold) we finally get...

A = | 2 2 0 |

| 0 (2P - 4) 0 |

| 0 0 -1 |

Therefore from R3 we see that

-1 * Z = 0 --> Z = 0

From R2 we see that

(2P - 4)Y = 0 --> Y = 0 or (2P - 4) = 0, Since we need to solve for P, we requested that Y ≠ 0 which leads to P = 2 and therefore Q = 1

From R1 we see that

2X + 2Y = 0 --> since Y is arbitrary. we can define X in terms of Y --> X = -Y

Finally our matrix and solution vector can be seen as

A = | 2 2 0 |

| 2 2 0 |

| 1 -1 -1 |

With solution vector (parametrized) of s * (-1, 1, 0) where s can be any value.