Answer 1

a_ij = i – 3j
∴ a₁₁ = 1 – 3(1) = 1 – 3 = – 2
a₁₂ = 1 – 3(2) = 1 – 6 = – 5
a₂₁ = 2 – 3(1 = 2 – 3 = – 1
a₂₂ = 2 – 3(2) = 2 – 6 = – 4
a₃₁ = 3 – 3(1) = 3 – 3 = 0,
a₃₂ = 3 – 3(2) = 3 – 6 = – 3

∴ A = `[(-2 , -5),(-1, -4),(0, -3)]`.