BITSAT 2015 Paper

We know that, M (adj M) = |M| I
Replacing M by adj M, we get
adj M [adj (adj M) = det (adj M) I
= det (M) M–1 [adj (adj M) = α2l
[Since‌M−1=

1

|M|

‌adj(M)]
⇒ α M–1 [adj (adj M)] = α2I
⇒ M–1 [adj (adj M)] = α I
But M–1 [adj (adj M)] = KI
Hence, K = α