Let X & Y be 2 invertible square matrix, then which of the following is true

(A) $(AB)^{-1} = A^{-1} B^{-1}$

(B) $(AB)^{-1} = B^{-1} A^{-1}$

(C) $(AB)' = A' B'$

(D) $(AB)' = B' A'$

Choose the answer from the options given below

Which of the following are true for the given graph.

(A) Break even point is at x = 10

(B) Break even point is at x = 8

(C) Break even point is at x = 6

(D) Fixed cost is 20

(E) Fixed cost is 0

Choose the correct answer from the options :

If $A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix}, I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A^2 = 8A + kI$, the value of k is :

If $A = \begin{bmatrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{bmatrix}$ and $B = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$, then AB is equals to :

For a 3 $\times$ 3 matrix A = [a$_{ij}$], whose element a$_{ij}$ is given by $a_{ij} = \sqrt{i^2 + 3j}$ then what is element a$_{31}$?

If $A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 2 & 3 \end{bmatrix}$

then $(3A + 2B)'$ is :

Let $P = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix}$, $Q = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix}$, $R = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$, then the matrix S such that QS - RP = 0 will be :