For square matrices A and B of the same order, we have adj(AB) = ? A.

1.	For square matrices A and B of the same order, we have adj(AB) = ? A. (adj A)(adj B) B. (adj B)(adj A) C. │AB│ D. none of these
Answer» We know that (AB)^-1 = adj(AB)/ \|AB\| adj (AB)= (AB)^-1 \|AB\| We also know that (AB)^-1 = B^-1. A^-1 \|AB\| = \|A\| \|B\| Putting them in 1 Adj (AB) = B^-1. A^-1. \|A\|.\|B\| = (A^-1. \|A\|) (B^-1 \|B\|) = adj(A) adj(B) Since, adj (A)= (A)^-1 \|A\| adj (B)= (B)^-1\|B\|

For square matrices A and B of the same order, we have adj(AB) = ? A. (adj A)(adj B) B. (adj B)(adj A) C. │AB│ D. none of these

Answer»

We know that (AB)^-1 = adj(AB)/ |AB|

adj (AB)= (AB)^-1 |AB|

We also know that (AB)^-1 = B^-1. A^-1

|AB| = |A| |B|

Putting them in 1

Adj (AB) = B^-1. A^-1. |A|.|B|

= (A^-1. |A|) (B^-1 |B|)

= adj(A) adj(B)

Since, adj (A)= (A)^-1 |A|

adj (B)= (B)^-1|B|