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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 _______

16 Answers

Best answer
158 158 votes
Eigen values are roots of Characteristic equation $|A - \lambda I | = 0.$

For a $3×3$ matrix, characteristic 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 characteristic equation, then $2-i$ must be other root.

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

$\color{blue}{\det(A) = \lambda_{1}\lambda_{2}\lambda_{3} = (2+i)*(2-i)*3 = (2^2 - i^2)*3 = 5*3 = 15}$
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50 50 votes
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,

Answer 15.
edited by
9 9 votes

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.

6 6 votes

We know that for any real matrix the determinant must be real.

Now the determinant is product of eigen values, and if one of the eigen value is imaginary then to make it real it can only be multiplied by its conjugate.

So, the other eigen value must be conjugate of the first imaginary eigen value.

Hence, we can easily find the third eigen value.

3 3 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

1 1 vote

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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