`|[a^(2)+1,ab,ac],[ab,b^(2)+1,bc],[ac,bc,c^(2)+1]|=1+a^(2)+b^(2)+c^(2)`

Prove the following by multiplication of determinants and power cofactor formula |{:(0,c,b),(c,0,a),(b,a,0):}|^(2)=|{:(b^(2)+v^(2),ab,ac),(ab,c^(2)+a^(2),bc),(ac,bc,a^(2)+b^(2)):}| =|{:(-a^(2),ab,ac),(ab,-b^(2),bc),(ac,bc,-c^(2)):}|=4a^(2)b^(2)c^(2)

Using properties of determinants, prove the following: |[a^2 + 1,ab, ac], [ab,b^2 + 1,b c],[ca, cb, c^2+1]|=1+a^2+b^2+c^2

Using properties of determinant prove that |(a^(2)+1, ab, ac),(ab, b^(2)+1, bc),(ca, cb,c^(2)+1)|=(1+a^(2)+b^(2)+c^(2)) .

Using properties of determinants, prove that : |{:(a^(2)+1,ab,ac),(ba,b^(2)+1,bc),(ca,cb,c^(2)+1):}|=a^(2)+b^(2)+c^(2)+1

|[x^2+a^2,ab,ac] , [ab,x^2+b^2,bc] , [ac,bc,x^2+c^2]|=

The determinant "det(A)"= |[ a^(2) +x,ab,ac], [ab, b^(2) +x,bc,] [ac,bc, c^(2) +x]| is divisible by a. x b. x^2 c. x^3 d. none of these

If a, b, c are non zero complex numbers satisfying a^(2) + b^(2) + c^(2) = 0 and |(b^(2) + c^(2),ab,ac),(ab,c^(2) + a^(2),bc),(ac,bc,a^(2) + b^(2))| = k a^(2) b^(2) c^(2) , then k is equal to

abs([-a^2,ab,ac],[ba,-b^2,bc],[ca,cb,-c^2]) = 4a^2.b^2.c^2

If A=[{:(0,c,-b),(-c,0,a),(b,-a,0):}] and B=[{:(a^(2),ab,ac),(ab,b^(2),bc),(ac,bc,c^(2)):}] , then (A+B)^(2)=

Using properties of determinants, prove the following |(a^2+1,ab,ac),(ab,b^2+1,bc),(ca,cb,c^2+1)|=1+a^2+b^2+c^2 .