Let A = [abcd]
i.e [abcd] [012−1] = [2−110]
→[a(0)+b(2)a(1)+b(−1)c(0)+d(2)c(1)+d(−1)] = [2−110]
→[2ba−b2dc−d] = [2−110]
By comparing 2b = 2 a - b = -1 2d = 1 and c - d = 0 ∴ b = 2∕2 = 1 a - b = -1 ⇒ a - 1 = -1 ∴ a = 0 ∴ d = 1∕2 ⇒ c = d ∴ c = 1∕2
∴The matrice A = [011/21/2]