Question

यदि `[x -5 -1][(1,0,2), (0,2,1), (2,0,3)][(x),(4),(1)] = 0` है तो x का मान ज्ञात कीजिए।

Answer 1

यहाँ, `[x -5 -1][(1,0,2), (0,2,1), (2,0,3)][(x),(4),(1)] = [0]`

∴ `[x -5 -1][(x + 0 + 2), (0 + 8 + 1), (2x + 0 + 3)] = [0]`

⇒ [x -5 -1]`[(x + 2), (9), (2x + 3)] = [0]`

⇒ [x(x + 2) - 45 - (2x + 3)] = 0

⇒ x² + 2x - 45 - (2x + 3) = 0

⇒ x² + 2x - 45 - 2x - 3 = 0

⇒ x² - 48 = 0

⇒ x² = 48

⇒ x = ±`sqrt48 = 4sqrt3`