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Two eigenvalues of a $3 \times 3$ real matrix $P$ are $(2+\sqrt {-1})$ and $3$. The determinant of $P$ is _______

Given two eigen values are (2+i) and 3.. since it is a real matrix the 3rd eigen value is 2-i
determinant of P = product of eigen values.
Solving we get,

edited
no1 is telling if the answers are correct or not.atlst upvotes.. :(
sure u r right .. but the 3rd eigen value is 2 - i because is is 3x3 matrix nt because it is real matrix .
and the complex root are in pair + and -
Eigen values are roots of Characterstic equation $|A - \lambda I | = 0$

For a $3×3$ matrix, characterstic equation will be cubic, so will have $3$ roots. Two roots are given as: $2 + i$ and $3$ and We know that complex roots always occur in pairs so, if $2+i$ is a root of characterstic equation, then $2-i$ must be other root.

$\lambda_{1} = 2+i$, $\lambda_{2} = 2-i$ and $\lambda_{3} = 3$

$\color{blue}{\det(A) = \lambda_{1}\lambda_{2}\lambda_{3} = (2+i)*(2-i)*3 = (2^2 - i^2)*3 = 5*3 = 15}$
edited by
we "can have". Not "will have" :)
Why Can ?
So, it's not compulsory?
I think will is correct, because there will always be 3 roots / eigen values. which may be repeated.
+1 vote

The determinant of a real matrix can never be imaginary. So, if one eigen value is complex, the other eigen value has to be its conjugate.   So, the eigen values of the matrix will be 2+i, 2-i and 3.   Also, determinant is the product of all eigen values. So, the required answer is (2+i)*(2-i)*(3) = (4-i2)*(3) = (5)*(3) = 15.