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2 votes

Then conjugate of (2+i) is 2-i and it is the third value of diagonal 

We know that Product of diagonal elements is Value of Determinant

(2+i )*3*(2-i) = (4 -(-1))*3 = 15

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Complex number exists in pairs

 λ1 = 2+i, λ2=2−i and λ3=3

det(P) = λ1λ2λ3 = (2+i)(2-i)(3) = 15 {i^2 = -1}

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0 votes
If one Eigen value is (2+i) then another one is (2-i). And the last one is 3. So the determinant will be (2+i)*(2-i)*3

= 5*3 =15
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0 votes
here the given two eigen values are (2 + (-1)^½ ) and 3
we know determinant of a matrix is product of it's eigen values, so here determinant can't be imaginary number so there would be a value which is conjugate of the imganiry number so on multiplication it will give real number and the value will be  (2 - (-1)^½ )
so now  (2 + (-1)^½ ) * (2 - (-1)^½ ) ==  (2^2 - (-1)^½ ^2) == 4 – (-1) = 5
so final answer will be 5 * 3 = 15
|P| = 15
Answer:

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