106I got (C)
the final expression was coming as |A-I|..
I took the matrix \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}
for considering the answer
A is an orthogonal matrix of odd ordeer such that |A|(x2+x+1) > 0, x belongs R.
If I is a unit matrix of the same order as of A then value of |A(I+A2)-(I+A)(I+A2-A)| is equal to
A) 1 B) -1
C) 0 D) 2
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4 Answers
106
1culd u post full solution............
after cnsidering I = as u mentioned wat to do next ............[7][7]
106for orthogonal matrix A2=A
=> |A| = 0,±1
but as |A|(x2+x+1) > 0
=> |A| = 1
the exp. given is
|A(I+A) - (I+A)(I+A-A)| (as A2=A)
= | A + A2 - I - A|
= |A-I|
then i took A = I (as I is also orthogonal with |A| = 1)
we get |I-I| = 0
but m not entirely convinced abt my solution