Isn't 100000 such a huge number (Part-10)?

Algebra Level 4

Consider the matrix Y = ( i 0 2 i 0 100000 0 2 i 0 i ) \color{#69047E}{\large {Y=\begin{pmatrix} i&0&-2i\\0&100000&0\\2i&0&-i\end{pmatrix}}}

If the sum of the eigenvalues can be represented as a + b i a+bi and the product of the eigenvalues can be represented as c + d i c+di .

Find the value of b + d c a \color{#3D99F6}{{\dfrac{b+d-c}{a}}}


The answer is 3.

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3 solutions

Pranjal Jain
Apr 6, 2015

Sum of Eigen values is the trace of matrix, thus, a + i b = i + 100000 i = 100000 + 0 i a+ib=i+100000-i=100000+0i

Product of eigen values is determinant. d e t ( Y ) = i 0 2 i 0 100000 0 2 i 0 i = 3 i 0 2 i 0 100000 0 0 0 i det(Y)=\begin{vmatrix} i&0&-2i\\0&100000&0\\2i&0&-i\end{vmatrix}\\=\begin{vmatrix} -3i&0&-2i\\0&100000&0\\0&0&-i\end{vmatrix}

Now determinant of diagonal matrix is just the product of diagonal elements, thus, d e t ( Y ) = 300000 + 0 i det(Y)=-300000+0i

Substituting values, ( 300000 ) 100000 = 3 \dfrac{-(-300000)}{100000}=3

a 400 points problem!

Aman Gautam - 6 years, 2 months ago

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Yea! Just find trace of matrix and its determinant and you get 400 point! This must be probably easiest 400 points problem!

Kunal Joshi - 6 years, 1 month ago
Sridhar Sri
Feb 26, 2016

Mudit Bansal
Feb 9, 2015

sum of eigen values is Tr(Y) and product of eigen values is determinant of Y. Enjoy!!

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