Answer 1

`|(4 + x, 4 - x, 4 - x),(4 - x, 4 + x, 4 - x),(4 - x, 4 - x, 4 + x)|` = 0

Applying C₁ → C₁ + C₂ + C₃, we get

`|(12 -  x, 4 - x, 4 - x),(12 - x, 4 + x, 4 - x),(12 - x, 4 - x, 4 + x)|` = 0

Taking (12 – x) common from C₁, we get

`(12 - x)|(1, 4 - x, 4 - x),(1, 4 + x, 4 - x),(1, 4 - x, 4 + x)|` = 0

Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₁ , we get

`(12 - x)|(1, 4 - x, 4 - x),(0, 2x, 0),(0, 0, 2x)|` = 0

∴ (12 – x)[1(4x² – 0) – (4 – x)(0 – 0) + (4 – x)(0 – 0)] = 0

∴ (12 – x)(4x²) = 0

∴ x² (12 – x) = 0

∴ x = 0 or 12 – x = 0

∴ x = 0 or x = 12