Statement 1: For a singular square matrix `A ,A B=A C B=Cdot` Statement 2; `|A|=0,t h e nA^(-1)` does not exist.

Statement-1 (Assertion and Statement- 2 (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below. Statement-1 For a singular matrix A , if AB = AC rArr B = C Statement-2 If abs(A) = 0, thhen A^(-1) does not exist.

If A is a non-singular square matrix such that |A|=10 , find |A^(-1)|

A square non-singular matrix A satisfies A^2-A+2I=0," then "A^(-1) =

If A is a square matrix of order 3 such that |A|=2,t h e n|(a d jA^(-1))^(-1)| is ___________.

If A is a square matrix of order 3 such that |A|=2,t h e n|(a d jA^(-1))^(-1)| is ___________.

Let A=[(0,0,-1),(0,-1,0),(-1,0,0)] Then only correct statement about the matrix A is (A) A is a zero matrix (B) A^2=I (C) A^-1 does not exist (D) A=(-1) I where I is a unit matrix

If A is a non-singular square matrix such that A^(-1)=[(5, 3),(-2,-1)] , then find (A^T)^(-1) .

Statement -1 : Determinant of a skew-symmetric matrix of order 3 is zero. Statement -2 : For any matrix A, Det (A) = "Det"(A^(T)) and "Det" (-A) = - "Det" (A) where Det (B) denotes the determinant of matrix B. Then,

For non-singular square matrix A ,\ B\ a n d\ C of the same order then, (A B^(-1)C)^(-1)= (a) A^(-1)B C^(-1) (b) C^(-1)B^(-1)A^(-1) (c) C B A^(-1) (d) C^(-1)\ B\ A^(-1)

Statement 1: if a ,b ,c ,d are real numbers and A=[a b c d]a n dA^3=O ,t h e nA^2=Odot Statement 2: For matrix A=[a b c d] we have A^2=(a+d)A+(a d-b c)I=Odot