If `[(1,a,2),(1,2,5),(2,1,1)]` is non invertible then a= (A) 2 (B) 1 (C) 0 (D) -1

If A=[(1,lambda,2),(1,2,5),(2,1,1)] is not invertible then lambda =?

A=[{:(1,0,k),(2, 1,3),(k,0,1):}] is invertible for

Let A=[(1,-1,1),(2,1,-3),(1,1,1)] and B=[(4,2,2),(-5,0,alpha),(1,-2,3)] If B is the inverse of matrix A , then alpha= (A) 2 (B) -2 (C) 5 (D) -5

The matrix [[1, a, 2], [1, 2, 5], [2, 1, 1]] is not invertible, if a=

If A=[(1,2,2),(2,1,2),(2,2,1)] then (A) A^-1= 1/5(A-4I_3) (B) A^2-4A-5I_3=0 (C) A^2 is invertible (D) A^3 is non invertible

If |(a,a^2,1+a^3),(b,b^2,1+b^3),(c,c^2,1+c^2)|=0 and vectors (1,a,a^2),(1,b,b^2) and (1,c,c^2) are hon coplanar then the product abc equals (A) 2 (B) -1 (C) 1 (D) 0

Matrix [(a,b,(a alpha-b)),(b,c,(balpha-c)),(2,1,0)] is non invertible if

If A+B=[(1,0),(1,1)] and A-2B=[(-1,1),(0,-1)] then A= (A) 1/3 [(1,1),(2,1)] (B) 1/3 [(2,1),(1,2)] (C) [(1,1),(2,1)] (D) none of these

If =(0,1)B=(1,0),C=(1,2),D=(2,1) ,prove that vec AB=vec CD

If A^-1 = [(1,-1,0),(0,-2,1),(0,0,1)] then (A) A is non-singular (B) A is skew symmetric (C) |A|=2 (D) AdjA= [(1/2,- 1/2,0),(0,-1,1/2),(0,0,- 1/2)]