Question

# 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

- 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 true but (R) is false.

### AA^{T} = I

|A| |A^{T}| = 1 |A|^{2} = 1 |A| = ± 1

|A + I| = |A + AA^{T}| = |A|^{3} |I + A^{T}|

= A^{3} | I + A |

when |A| = – 1 |A + I| = – |A + I|

2|A + I| = 0 |A + I| = 0

when |A| = 1 |A + I| = |A + I|

Reason is false

Let A =

A + I =

AA^{T} == 1 |AA^{T}| = 1 |A| = ± 1 but |A| = – 1

|A + I| = |A + AA^{T}| = |A|^{3} |I + A^{T}| = – |A + I|

|A + I| = 0

So assertion is true.

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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) .