Statement I : is a diagonal matrix.
Statement II : A square matrix A = (aij) is a diagonal matrix if aij = 0 .

Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity. Denote by tr(A), The sum of diagonal entries of A, Assume that A² = I.
Statement-I :If AI and A– I, then det  A= – 1
Statement-II : If A  I and A  – I then tr(A)0.

Suppose , let x be a 2×2 matrix such that AX = B
Statement-I : X is non singular & |x| = ±2
Statement-II : X is a singular matrix

If then is equal to

Statement-I : If A & B are two 3×3 matrices such that AB = 0, then A = 0 or B = 0
Statement-II : If A, B & X are three 3×3 matrices such that AX = B, |A| 0, then X = A^–1B

Assertion (A): The inverse of the matrix does not exist.
Reason (R) : The matrix is singular. [ = 0, since R₂ = 2R₁]

Assertion (A): is a diagonal matrix
Reason (R) : A square matrix A = (a_ij) is a diagonal matrix if a_ij = 0 for all i  j.

Consider = – 1, where a_i. a_j + b_i. b_j + c_i.c_j = and i, j = 1,2,3
Assertion(A) : The value of  is equal to zero
Reason(R) : If A be square matrix of odd order such that AA^T = I, then | A + I | = 0

Assertion(A) : The inverse of the matrix A = [Aij]_{n × n} where a_ij = 0, i  j is B = [aij^–1]_{n× n} 
Reason(R): The inverse of singular matrix does not exist

Assertion : The product of two diagonal matrices of order 3 × 3 is also a diagonal matrix
Reason : matrix multiplicationis non commutative

Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det  = det (–)

Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix
Reason : If A is non-singular then it commutes with I, adj A and A^–1

Statement-I The equation  has exactly one solution in [0, 2].

Statement-II For equations of type  to have real solutions in   should hold true.

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not , is same like assertion and reason. Here, Start solving first Statement and try to prove it . Then solve the Statement-II . Remember cos a cosb -sin a sinb = cos ( a + b ) and sin a cosb + cosa sinb = sin( a+ b) .

Assertion : If A is a skew symmetric matrix of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det   = det (–)

Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix

Reason : If A is non-singular then it commutes with I, adj A and A^–1

Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1
Assertion : The system of equations has no solution for k 3.and
Reason : The determinant  0, for k  3.