GO Classes DPP | GATE CS | Linear Algebra | Eigen Values

Answer 1

Since $Q$ is a real matrix and one eigenvalue is $(1+\sqrt{-1})$, its conjugate $(1-\sqrt{-1})$ must also be an eigenvalue.

So the three eigenvalues are $(1+\sqrt{-1})$, $(1-\sqrt{-1})$, and $4$.

Now, the determinant of a matrix is the product of its eigenvalues.

$\therefore \det(Q)=(1+\sqrt{-1})(1-\sqrt{-1})\cdot 4$.

Using $(a+b)(a-b)=a^2-b^2$,

we get, $(1+\sqrt{-1})(1-\sqrt{-1})=1-(-1)=2$.

$\therefore\det(Q)=2\cdot 4=8$.

So, the correct answer is $\boxed{8}$.

Answer 2

det(Q)=8
Ans : 8

Comments

Answer 8 , logic determinants=product of eigen values

Conjugate pair occurs in pairs.

Answer 3

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