Question

दिए गए आव्यूह के व्युत्क्रम (जिनका अस्तित्व हो) ज्ञात कीजिए।

`[(1,0,0),(0,cos alpha, sin alpha),(0,sin alpha, -cos alpha)]`

Answer 1

A = `[(1, 0, 0), (0, cos alpha, sin alpha), (0,sin alpha, -cos alpha)]`

|A| = 1(−cos² α − sin² α)

= −1 ≠ 0

∴ A⁻¹ का अस्तित्व है।

C_ij = (−1)^{i + j} M_ij

C₁₁ = `(-1)^(1 + 1) |(cos alpha, sin alpha), (sin alpha, -cos alpha)|`

= −cos² α − sin ²α

= −1

C₁₂ = `(-1)^(1 + 2) |(0, sin alpha), (0, -cos alpha)|` = 0

C₁₃ = `(-1)^(1 + 3) |(0, cos alpha), (0, sin alpha)|` = 0

C₂₁ = `(-1)^(2 + 1) |(0, 0), (sin alpha, -cos alpha)|` = 0

C₂₂ = `(-1)^(2 + 2) |(1, 0), (0, -cos alpha)|` = −cos α

C₂₃ = `(-1)^(2 + 3) |(1, 0), (0, sin alpha)|` = −sin α

C₃₁ = `(-1)^(3 + 1) |(0, 0), (cos alpha, sin alpha)|` = 0

C₃₂ = `(-1)^(3 + 2) |(1, 0), (0, sin alpha)|` = −sin α

C₃₃ = `(-1)^(3 + 3) |(1, 0), (0, cos alpha)|` = cos α

अब, Adj A = `[(-1, 0, 0), (0, -cos alpha, -sin alpha), (0, -sin alpha, cos alpha)]`

A⁻¹ = `1/|A|` (adj A)

= `1/-1 [(-1, 0, 0), (0, -cos alpha, -sin alpha), (0, -sin alpha, cos alpha)]`

= `[(1, 0, 0), (0, cos alpha, sin alpha), (0, sin alpha, -cos alpha)]`