Let `P=(AB^(T)-BA^(T))`
`:. P^(T)=(AB^(T)-BA^(T))^(T)=(AB^(T))^(T)-(BA^(T))^(T)`
`=(B^(T))^(T) (A)^(T)-(A^(T))^(T)B^(T)=BA^(T)-AB^(T)`
`=-(AB^(T)-BA^(T))=-P`
Hence, `(AB^(T)-BA^(T))` is a skew-symmetric matric.
Correct Answer - B
Let `A=[a_("ij")]`. Since a is skew-symmetric, we have
`a_(ii)=0` and `a_("ij")=-a_("ij") (i ne j)`
A is symmetric as well, so `a_("ij")=a_("ji")` for all `i` and `j`.
`:....
Correct Answer - B
Let `C=BA`, then
`C^(T)=(BA)^(T)=A^(T)B^(T)`
`=-AB^(T)` (as A is skew-symmetric)
`=BA^(T)" "("as "AB^(T)+BA^(T)=O)`
`=B (-A)`
`=- BA=-C`
Therefore, C is skew-symmetric.
Correct Answer - C
`(A(BA))^(T)=(BA)^(T) A^(T)`
`=(A^(T) B^(T)).A^(T)`
`=(AB) A`
`=A(BA)`
Similarly `((AB)A)^(T)=(AB)A`
Clearly both statements are true but statement 2 is not a correct explanation of statement 1.
Correct Answer - A::C
`(a,c)` `A+B=AB`
`impliesI-(A+B-AB)=I`
`implies(I-A)(I-B)=I`
`implies|I-A||I-B|=I`
`implies |I-A|`, `|I-B|` are non zero
Also `(I-B)(I-A)=I`
`impliesI-B-A+BA=I`
`impliesA+B=B+A`
`impliesAB=BA`
`implies(a)` and `(c )` are correct
Correct Answer - D
`(d)` For symmetric matrix each place in upper triangle and leading diagonal can be filled in `7` ways. Then number of symmetric matrices are `7^(6)`.
For skew...