`|(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 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)|:}

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)|

|(x,a,x+a),(y,b,y+b),(z,c,z+c)| = 0

If [[ p , q-y , r-z],[ p-x , q , r-z ],[ p-x , q-y , r ]]=0 , then (p)/(x)+(q)/(y)+(r)/(z)=

[[ 0 , p-q , p-r ],[q-p , 0 , q-r],[ r-p , r-q , 0 ]] =

Let A= [[a,b,c],[p,q,r],[x,y,z]] and suppose then det (A) = 2, then det (B) equals, where B = [[4x,2a,-p],[4y, 2b, -q],[4z, 2c, -r]]

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

If [[ a , b , c],[ m , n , p],[ x , y , z ]] =k , then [[ 6 a , 2 b , 2 c ],[3 m , n , p],[ 3 x , y , z ]] =

Using the property of determinants and without expanding {:|( x,a,x+a),( y,b,y+b),(z,c,z+c)|:} =0

If a, b, c are in A.P., a, x, b are in G.P., and b, y, c are in G.P., then (x, y) lies on :