Question

# If A –2B = and 2A – 3B = , then matrix B is equal to–

## The correct answer is:

We have

B = (2A –3B) –2 (A –2B) = 2

### Related Questions to study

### The value of x for which the matrix A = is inverse of B = is

### The value of x for which the matrix A = is inverse of B = is

The greatest possible difference between two of the roots if [0, 2] is

The greatest possible difference between two of the roots if [0, 2] is

### Statement I : is a diagonal matrix.

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

### 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 A^{2} = I.

Statement-I :If AI and A– I, then det A= – 1

Statement-II : If A I and A – I then tr(A)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 A^{2} = 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

### 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

### 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^{–1}B

### 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

### 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) .