[" D.Fare affors far "],[[x+a,b,c],[a,x+b,c],[a,b,x+c]|=x^(2)(x+a+b+c)]

S.T abs[[x+a,b,c],[a,x+b,c],[a,b,x+c]]=x^2(x+a+b+c)

Using the properties of determinants, prove that : |[[x+a,b,c],[a,x+b,c],[a,b,x+c]]| = x^2(x+a+b+c)

Prove that : |{:(x+a,b,c),(a,x+b,c),(a,b,x+c):}|=x^(2)(x+a+b+c)

Prove that : |{:(x+a,b,c),(a,x+b,c),(a,b,x+c):}|=x^(2)(x+a+b+c)

The values of x for which the matrix [[x+a,b,c],[a,x+b,c],[a,b,x+c]] is non-singular are (A) R-{0} (B) R-{-(a+b+c)} (C) R-{0,-(a+b+c)} (D) none of these

If a+b+c=0, one root of [[a-x, c, b],[c, b-x, a],[ b, a, c-x]]=0 is a. x=1 b. x=2 c. x=a^2+b^2+c^2 d. x=0

If a+b+c=0, one root of [[a-x, c, b],[c, b-x, a],[ b, a, c-x]]=0 is a. x=1 b. x=2 c. x=a^2+b^2+c^2 d. x=0

If a+b+c=0, one root of [[a-x, c, b],[c, b-x, a],[ b, a, c-x]]=0 is a. x=1 b. x=2 c. x=a^2+b^2+c^2 d. x=0

If a+b+c=0, one root of [[a-x, c, b],[c, b-x, a],[ b, a, c-x]]=0 is a. x=1 b. x=2 c. x=a^2+b^2+c^2 d. x=0

a+b+c=0, solve for x : [[a-x,c,b],[c,b-x,a],[b,a,c-x]]=0