Using properties of determinants, prove the following: `|1 1+p1+p+q2 3+2p1+3p+2q3 6+3p1+6p+3q|=1`

Using properties of determinants,prove the det[[1+p,1+p+q2,3+2p,1+3p+2q3,6+3p,1+6p+3q]]=1

Show that: |11+p1+p+q23+2p1+3p+2q36+3p1+6p+3q|=1

Using properties of determinants.Prove that det[[2,3+2p,1+3p+2q3,6+3p,10+6p+3q]]=1

Show that det[[1,1+p,1+p+q2,3+2p,1+3p+2q3,6+3p,1+6p+3q]]=

Using the property of determinants and without expanding prove that abs([p,q,r],[p^2,q^2,r^2],[p^3,q^3,r^3])=pqr(p-q)(q-r)(r-p)

Using truth tables, prove the following equivalences : (1) ~pvvq-=~(p^^q)to[~pvv(~pvvq)] (2) p""iffq-=(p^^q)vv(~p^^~q) (3) (p^^q)tor-=pto(qtor)

{:(2x + 3y = 9),((p + q)x + (2p - q)y = 3(p + q+ 1)):}

If one root is square of the other root of the equation x^(2)+px+q=0, then the relation between p and q is (2004,1M)p^(3)-(3p-1)q+q^(2)=0p^(3)-q(3p+1)+q^(2)=0p^(3)+q(3p-1)+q^(2)=0p^(3)+q(3p+1)+q^(2)=0

If one root is square of the other root of the equation x^(2)+px+q=0, then the relation between p and q is p^(3)-q(3p-1)+q^(2)=0p^(3)-q(3p+1)+q^(2)=0p^(3)+q(3p-1)+q^(2)=0p^(3)+q(3p+1)+q^(2)=0

In a G.P. if (2p)^(t h) term is q^2a n d(2q)^(t h) term is p^2 where q in N , then its (p+q^(t h)) term is p q b. p^2q^2 c.1/2p^3q^3 d. 1/4p^3q^3