If A = bmatrix 3 & 1 -1 & 2 bmatrix , C = bmatrix 7 & 3 0 & 6 bmatrix and AB = C , then the inverse of matrix B is

Step-by-step solution

Given AB = C Taking inverse on both sides, we get (AB)⁻¹ = C⁻¹ B⁻¹A⁻¹ = C⁻¹ Multiplying both sides by A on the right, we get B⁻¹ = C⁻¹A We have C = bmatrix 7 & 3 0 & 6 bmatrix |C| = 7(6) - 3(0) = 42 C⁻¹ = 1 42 bmatrix 6 & -3 0 & 7 bmatrix Now, B⁻¹ = 1 42 bmatrix 6 & -3 0 & 7 bmatrix bmatrix 3 & 1 -1 & 2 bmatrix B⁻¹ = 1 42 bmatrix 18+3 & 6-6 0-7 & 0+14 bmatrix B⁻¹ = 1 42 bmatrix 21 & 0 -7 & 14 bmatrix B⁻¹ = 7 42 bmatrix 3 & 0 -1 & 2 bmatrix = 1 6 bmatrix 3 & 0 -1 & 2 bmatrix Answer: 1 6 bmatrix 3 & 0 -1 & 2 bmatrix