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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) $\begin{array}{rlr}\mathrm{C}= & 1 & 2 & 3 \mid \\ \mid 2 & 1 & 3 \mid \\\ \mid 3 & 3 & 6\end{array} \mid$

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: $$λ₁ = 2$$, $$λ₂ = 3$$, $$λ₃ = 3$$
See the step by step solution

## 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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