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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 _______
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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 $\lambda_{3} = 3$

$\color{blue}{\det(A) = \lambda_{1}\lambda_{2}\lambda_{3} = (2+i)*(2-i)*3 = (2^2 - i^2)*3 = 5*3 = 15}$
by Boss (28.6k points)
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we "can have". Not "will have" :)
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Why Can ?
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So, it's not compulsory?
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I think will is correct, because there will always be 3 roots / eigen values. which may be repeated.
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again pooved that NAT has a hidden concept to catch.
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,

by Loyal (9.7k points)
edited
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no1 is telling if the answers are correct or not.atlst upvotes.. :(
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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 -
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Determinant = product of diagonal elements

Above is valid only for triangular matrix or diagonal matrix .

Does this hold for any type of square matix??
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Satyajeet Singh valid only for triangular matrix or diagonal matrix.

for other square matrix product of eigenvalue =determinant of matrix

https://math.stackexchange.com/questions/1243237/prove-the-determinant-is-the-product-of-its-diagonal-entries

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dear friend

for ANY matrix,

PRODUCT OF EIGEN VALUES = DETERMINANT OF THAT MATRIX ITSELF

any way in diagonal/triangular matrices determinant is product of the elements in principle diagonal therefore they are same as product of eigen values.
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what is the meaning of real matrix here?
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means there is no imaginary value in the matrix.

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.

by Loyal (9.6k points)

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.

by (149 points)
+1 vote

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

by Boss (11.4k points)