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Problem 223

Find the characteristic polynomials and the eigenvalues of the matrices. (i) \(\begin{array}{rl}\mathrm{A}= & 2 & 3 \\ & 1 & 4\end{array} \mid\) (ii) $\mathrm{B}=\mid \begin{array}{rr}\cos \alpha & \sin \alpha \\ \mid-\sin \alpha & \cos \alpha \mid\end{array}$ (iii) $\begin{array}{rlr}\mathrm{C}= & 1 & 2 & 3 \mid \\ \mid 2 & 1 & 3 \mid \\\ \mid 3 & 3 & 6\end{array} \mid$

Short Answer

Expert verified
The short answer for the eigenvalues of the matrices are as follows: - Matrix A: \(λ₁ = 1\), \(λ₂ = 5\) - Matrix B: \(λ₁ = \cos\alpha + \sin\alpha\), \(λ₂ = \cos\alpha - \sin\alpha\) - Matrix C: \(λ₁ = 2\), \(λ₂ = 3\), \(λ₃ = 3\)
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Step 1: Write down the matrix A

The given matrix A is: \(A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}\)

Step 2: Find the characteristic polynomial for matrix A

Subtract λI from matrix A and find the determinant: \(|A - λI| = \begin{vmatrix} 2-λ & 3 \\ 1 & 4-λ \end{vmatrix} = (2-λ)(4-λ) - (1)(3) = λ^2 -6λ +5\) Thus, the characteristic polynomial for matrix A is: \(P_A(λ) = λ^2 - 6λ + 5\)

Step 3: Find the eigenvalues for matrix A

The eigenvalues of matrix A are the roots of the characteristic polynomial. We need to solve the equation: \(P_A(λ) = λ^2 - 6λ + 5 = 0\) By factoring this quadratic equation, we get: \((λ-1)(λ-5)=0\) Thus, the eigenvalues for matrix A are: \(λ₁ = 1\) \(λ₂ = 5\) Matrix B:

Step 1: Write down the matrix B

The given matrix B is: \(B = \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix}\)

Step 2: Find the characteristic polynomial for matrix B

Subtract λI from matrix B and find the determinant: \(|B - λI| = \begin{vmatrix} \cos\alpha-λ & \sin\alpha \\ -\sin\alpha & \cos\alpha-λ \end{vmatrix} = (\cos\alpha-λ)^2 - (\sin\alpha)(-\sin\alpha)\) Using trigonometric identities, \(\sin^2\alpha+\cos^2\alpha=1\): \((\cos\alpha-λ)^2 - (1-\cos^2\alpha) = λ^2 -2\cos\alpha λ +1\) Thus, the characteristic polynomial for matrix B is: \(P_B(λ) = λ^2 - 2\cos\alpha λ + 1\)

Step 3: Find the eigenvalues for matrix B

The eigenvalues of matrix B are the roots of the characteristic polynomial. We need to solve the equation: \(P_B(λ) = λ^2 - 2\cos\alpha λ + 1 = 0\) Using the quadratic formula for λ: \(λ = \frac{2\cos\alpha \pm \sqrt{(-2\cos\alpha)^2 - 4(1)}}{2}\) As sin^2α < 1 (α cannot be \(±\pi/2\)): \(λ_{1,2} = \cos\alpha \pm \sqrt{\sin^2\alpha}\) Since \(\sin^2\alpha+\cos^2\alpha=1\): \(λ_{1,2} = \cos\alpha \pm \sin\alpha\) Thus, the eigenvalues for matrix B are: \(λ₁ = \cos\alpha + \sin\alpha\) \(λ₂ = \cos\alpha - \sin\alpha\) Matrix C:

Step 1: Write down the matrix C

The given matrix C is: \(C = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 3 & 6 \end{pmatrix}\)

Step 2: Find the characteristic polynomial for matrix C

Subtract λI from matrix C and find the determinant: \(|C - λI| = \begin{vmatrix} 1-λ & 2 & 3 \\ 2 & 1-λ & 3 \\ 3 & 3 & 6-λ \end{vmatrix}\) By the expanding the determinant by the first row, we get: \((1-λ)((1-λ)(6-λ) - (3)(3)) - (2)((2)(6-λ) - (3)(3)) + (3)((2)(3) - (1-λ)(3))\) \(= -(λ^3 - 8λ^2 + 21λ - 18)\) Thus, the characteristic polynomial for matrix C is: \(P_C(λ) = λ^3 - 8λ^2 + 21λ - 18\)

Step 3: Find the eigenvalues for matrix C

The eigenvalues of matrix C are the roots of the characteristic polynomial. We need to solve the equation: \(P_C(λ) = λ^3 - 8λ^2 + 21λ - 18 = 0\) By factoring or using numerical methods, we find the eigenvalues for matrix C: \(λ₁ = 2\) \(λ₂ = 3\) \(λ₃ = 3\) In conclusion, the eigenvalues of the given matrices are: - Matrix A: 1 and 5 - Matrix B: \(\cos\alpha + \sin\alpha\) and \(\cos\alpha - \sin\alpha\) - Matrix C: 2, 3, and 3

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Most popular questions from this chapter

Chapter 9

If $\mathrm{F}(\mathrm{X})=\begin{array}{ccccc}1 & 0 & 1 \mid & \mid 2 & 1 & 0 \mid \\ & \mid 2 & 1 & 1 \mid & -\mid-1 & 1 & 1 \\ & 1 & 1 & 1 \mid & \mid 0 & 1 & 0\end{array} \mid$ $\begin{array}{rllll} & \mid 1 & 1 & 1 \mid & & \\ \mathrm{X}+ & \mid 1 & 0 & 1 \mid \mathrm{X}^{2}, & \text { and } \mathrm{B}= & 1 & 1 \\ & \mid 0 & 1 & 0 \mid & 1 & 0 \\ & \text { find } \mathrm{F}_{\mathrm{L}}(\mathrm{B}) & \text { and } \mathrm{F}_{\mathrm{R}}(\mathrm{B}) .\end{array}$

Chapter 9

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Chapter 9

Let \(\mathrm{V}\) be the vector space \(\mathrm{R}^{2}\) and let \(\mathrm{T}\) be the operator defined by $\quad \mathrm{T}(\mathrm{xy})=(2 \mathrm{x}-\mathrm{y}, \mathrm{x}+\mathrm{y})$ Let \(\mathrm{f}(\mathrm{x})=2+3 \mathrm{x}\) and \(\mathrm{g}(\mathrm{x})=\mathrm{x}+\mathrm{x}^{2} .\) Find \(\mathrm{f}(\mathrm{T})\) and \(\mathrm{g}(\mathrm{T})\)

Chapter 9

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Chapter 9

Let \(\quad \varphi(\lambda)=-2-5 \lambda+3 \lambda^{2}\) $$ \begin{array}{rrrr}A= & \mid 1 & 2 \mid . & \text { Show that } & \varphi(A)=\mid 14 & 2 \mid \\ & \mid 3 & 1 \mid & & 13 & 14\end{array} $$

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