Prove that `|(a+bx ,c+dx,p+qx),(ax + b, cx +d, px +q),(u,v,w)|= (1- x^3) |(a,c,p),(b,d,q),(u,v,w)|`

Prove that Delta ={:|( a+bx,c+dx,p+qx),( ax+b,cx+d,px+q),(u,v,w) |:}=( 1-x^(2)) {:|( a,c,p),(b,d,q),(u,v,w)|:}

Prove, using Properites of determinants, {:|(a+bx^2,c+dx^2,p+qx^2),(ax^2+b,cx^2+d,px^2+q),(u,v,w)|=(x^4-1){:|(b,d,q),(a,c,p),(u,v,w)|

|(b+c, q+r, y+z),(c+a,r+p,z +x),(a+b,p+q,x+y)|=2|(a,p,x),(b,q,y),(c,r,z)|

Using Properties of determinants, prove that: {:|(b+c,c+a,a+b),(q+r,r+p,p+q),(y+z,z+x,x+y)|=2{:|(a,b,c),(p,q,r),(x,y,z)|

Using the property of determinants and without expanding {:|( b+c,q+r,y+z),( c+a,r+p,z+x),( a+b,p+q,x+y) |:}=2 {:|(a,p,x),( b,q,y),(c,r,z)|:}

d/dx (ax+b)^(cx+d) =

Let f(x) = (ax+b)/(cx + d), then f(x) has

If P ={ a, b, c} and Q = {r} , find P xx Q .

The value of |(p,0,0,0),(a,q,0,0),(b,c,r,0),(d,e,f,s)| is :

Let f(x) = (ax+b)/(cx+d) . Then fof(x) = x , then