If a2+λ2ab+cλca−bλab−cλb2+λ2bc+aλca+bλbc−aλc2+λ2λc−b−cλab−aλ=1+a2+b2+c23, then the value of λ is

We observe that the elements in the pre-factor are the cofactors of the corresponding elements of the post-factor. Hence,λc−b−cλab−aλ3=λλ2+a2+b2+c23=1+a2+b2+c23⇒ λ=1Alternate solution:Writing a = 0, b = 0, c = 0on both sides, we getλ6λ3=1 or λ=1