Question

# 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^{–1}B

- If both (A) and (R) are true, and (R) is the correct explanation of (A).
- If both (A) and (R) are true but (R) is not the correct explanation of (A).
- If (A) is true but (R) is false.
- If (A) is false but (R) is true.

## The correct answer is: If (A) is false but (R) is true.

### Statement : I is false

As AB = 0 A = 0 or B = 0 or A & B both zero or A & B neither zero

Statement : II is true,

So, answer is (d)

### Related Questions to study

### Assertion (A): The inverse of the matrix does not exist.

Reason (R) : The matrix is singular. [ = 0, since R_{2} = 2R_{1}]

### Assertion (A): The inverse of the matrix does not exist.

Reason (R) : The matrix is singular. [ = 0, since R_{2} = 2R_{1}]

### 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.

### 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

### 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(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 : 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 : 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}

### 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) .

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) .