We know that E₁ = I₁Z₁₁ +I₂ Z₁₂ E₂= I₁Z₂₁+I₂Z₂₂ Consider KVL for the first loop  -E₁ + {i₁x2} + [{i₁ - i₂} x 4] – 10 E₁ = 0 Upon deducting, you will get  6i₁ - 4i₂ = 11E₁ KVL for Second loop:  E₂ + 4(i₂ - i₁) – 10 E₁ = 0 Upon deducting, you will get,  E₂ = 10E₁ - 4(i₂ - i₁) Considering the two equations We can write as follows: E₁ = I₁ (6/11) + I₂ ((-4)/11) E₂ = I₁ ((-16)/11) + I₂ ((-4)/11) Comparing this to above two equations, Z₁₁ = 6/11 Z₁₂ = (-4)/11 Z₂₁ = (-16)/11 Z₂₂ = (-4)/11