Question

The smallest eigenvalue and the corresponding eigenvector of the matrix \(\begin{bmatrix} 2 & -2 \\\ -1 & 6 \end{bmatrix}\), respectively are
1. \(1.55 \ \rm and \ \left\lbrace \begin{matrix} 2.00 \\\ -0.45 \end{matrix} \right\rbrace\)
2. \(1.55 \ \rm and \ \left\lbrace \begin{matrix} 2.00 \\\ 0.45 \end{matrix} \right\rbrace\)
3. \(2.00 \ \rm and \ \left\lbrace \begin{matrix} 1.00 \\\ 1.00 \end{matrix} \right\rbrace\)
4. \(1.55 \ \rm and \ \left\lbrace \begin{matrix} -2.55 \\\ -0.45 \end{matrix} \right\rbrace\)

Answer 1

Correct Answer - Option 2 : \(1.55 \ \rm and \ \left\lbrace \begin{matrix} 2.00 \\\ 0.45 \end{matrix} \right\rbrace\)

Explanation:

Eigenvector (X) that corresponding to Eigenvalue (λ) satisfies the equation AX = λX.

Determining eigenvalues first by characteristic Equation 

Given matrix A = \(\begin{bmatrix} 2 & -2 \\\ -1 & 6 \end{bmatrix}\)

The characteristic equation for the given matrix is

  |A - λI| = 0

 \(M = \left| {\begin{array}{*{20}{c}} {2 - λ }&-2\\ -1&{6 - λ } \end{array}} \right| = 0\)

⇒ (2 - λ) × (6 - λ) - (-2)(-1) = 0 

⇒ 12 - 2 λ - 6 λ + λ² - 2 = 0 

⇒  λ2 - 8λ + 10 = 0

 \(λ = \frac{{ - 8 \pm √ {64 - 40} }}{2} = \frac{{ - 8 \pm √ {24} }}{2} = 4 \pm √ 6 \)

⇒ λ = 4 + √6 , 4 - √6 

∴ Eigen values are 4 + √6 & 4 - √6

Smallest Eigen Value = 4 - √6 = 1.55 

Now Determining Eigenvectors corresponding to the eigenvalue λ = 1.55.

For λ = 1.55 The corresponding Eigenvector is 

  \(\left[ {\begin{array}{*{20}{c}} {2 - 1.55}&-2\\ -1&{6 - 1.55} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\)

 \(\left[ {\begin{array}{*{20}{c}} {0.45}&-2\\ -1&{4.45} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\)

 0.45 x₁ - 2 x₂ = 0  -- (1)

  -x₁ + 4.45 x₂ = 0  -- (2)

 Solving (1) and (2) we get.

 x₁ = 2 & x₂ = 0.45 

 ∴ The eigenvector corresponding to 1.55 is \(\left[ {\begin{array}{*{20}{c}} { 2}\\ 0.45 \end{array}} \right]\)