Question

The smallest and largest Eigen values of the following matrix are:

\(\left[ {\begin{array}{*{20}{c}} 3&{ - 2}&2\\ 4&{ - 4}&6\\ 2&{ - 3}&5 \end{array}} \right]\)

1. 1.5 and 2.5
2. 0.5 and 2.5
3. 1.0 and 3.0
4. 1.0 and 2.0

Answer 1

Correct Answer - Option 4 : 1.0 and 2.0

Concept:

Eigen Values

Let A be a square matrix of order ‘n’ and ‘λ’ be a scalar.

​\(\left| {A - \lambda I} \right| = 0\) is called the characteristic equation of matrix A.

The roots of the characteristic equation are called Eigenvalues.

Corresponding to each eigen value ‘λ’, there exists a non-zero vector ‘X’ such that AX = λX or (A -λI)X = 0

Calculation:

Given matrix is,

\(\left[ {\begin{array}{*{20}{c}} 3&{ - 2}&2\\ 4&{ - 4}&6\\ 2&{ - 3}&5 \end{array}} \right]\)

The characteristic eqaution for the given matrix is as follows

 \(\left| {A - \lambda I} \right| = 0\)

\( \Rightarrow \left| {\begin{array}{*{20}{c}} {3 - \lambda }&{ - 2}&2\\ 4&{ - 4 - \lambda }&6\\ 2&{ - 3}&{5 - \lambda } \end{array}} \right| = 0\)

⇒ (3 – λ) (– 20 + 4λ – 5λ + λ² + 18) + 2 (20 – 4λ – 12) + 2 (-12 + 8 + 2λ) = 0

⇒ λ³ – 4λ² + 5λ – 2 = 0

Now we can put values from the given options, and see which option satisfies the above equation

Only 1 and 2 satisfy this equation.

λ = 1, 1, 2

Hence, smallest Eigen value = 1 and Largest Eigen value = 2