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Question: Verify Cayley-Hamilton Theorem for the matrix A = [[2, 1, 2], [5, 3, 3], [- 1, 0, - 2]]...

Verify Cayley-Hamilton Theorem for the matrix A = [[2, 1, 2], [5, 3, 3], [- 1, 0, - 2]]

Answer

Verified

Explanation

Solution

To verify the Cayley-Hamilton Theorem for a matrix A, we need to show that the matrix A satisfies its own characteristic equation. The characteristic equation is given by ∣A−λI∣=0|A - \lambda I| = 0, where II is the identity matrix and λ\lambda is an eigenvalue.

Given matrix: A=[212533−10−2]A = \begin{bmatrix} 2 & 1 & 2 \\ 5 & 3 & 3 \\ -1 & 0 & -2 \end{bmatrix}

Step 1: Find the characteristic equation

First, form the matrix A−λIA - \lambda I: A−λI=[2−λ1253−λ3−10−2−λ]A - \lambda I = \begin{bmatrix} 2-\lambda & 1 & 2 \\ 5 & 3-\lambda & 3 \\ -1 & 0 & -2-\lambda \end{bmatrix}

Now, calculate the determinant ∣A−λI∣|A - \lambda I|: We will expand the determinant along the second column (C2) because it contains a zero, simplifying the calculation. ∣A−λI∣=−1⋅∣53−1−2−λ∣+(3−λ)⋅∣2−λ2−1−2−λ∣−0⋅∣2−λ153∣|A - \lambda I| = -1 \cdot \begin{vmatrix} 5 & 3 \\ -1 & -2-\lambda \end{vmatrix} + (3-\lambda) \cdot \begin{vmatrix} 2-\lambda & 2 \\ -1 & -2-\lambda \end{vmatrix} - 0 \cdot \begin{vmatrix} 2-\lambda & 1 \\ 5 & 3 \end{vmatrix}

=−1[5(−2−λ)−3(−1)]+(3−λ)[(2−λ)(−2−λ)−2(−1)]= -1 [5(-2-\lambda) - 3(-1)] + (3-\lambda) [(2-\lambda)(-2-\lambda) - 2(-1)] =−1[−10−5λ+3]+(3−λ)[−(2−λ)(2+λ)+2]= -1 [-10 - 5\lambda + 3] + (3-\lambda) [-(2-\lambda)(2+\lambda) + 2] =−1[−7−5λ]+(3−λ)[−(4−λ2)+2]= -1 [-7 - 5\lambda] + (3-\lambda) [-(4 - \lambda^2) + 2] =7+5λ+(3−λ)[λ2−4+2]= 7 + 5\lambda + (3-\lambda) [\lambda^2 - 4 + 2] =7+5λ+(3−λ)[λ2−2]= 7 + 5\lambda + (3-\lambda) [\lambda^2 - 2] =7+5λ+3λ2−6−λ3+2λ= 7 + 5\lambda + 3\lambda^2 - 6 - \lambda^3 + 2\lambda =−λ3+3λ2+(5+2)λ+(7−6)= -\lambda^3 + 3\lambda^2 + (5+2)\lambda + (7-6) =−λ3+3λ2+7λ+1= -\lambda^3 + 3\lambda^2 + 7\lambda + 1

Setting the determinant to zero, the characteristic equation is: −λ3+3λ2+7λ+1=0-\lambda^3 + 3\lambda^2 + 7\lambda + 1 = 0 Multiplying by -1, we get: λ3−3λ2−7λ−1=0\lambda^3 - 3\lambda^2 - 7\lambda - 1 = 0

Step 2: Verify P(A)=OP(A) = O

According to the Cayley-Hamilton Theorem, if P(λ)=λ3−3λ2−7λ−1P(\lambda) = \lambda^3 - 3\lambda^2 - 7\lambda - 1, then P(A)=A3−3A2−7A−IP(A) = A^3 - 3A^2 - 7A - I must be equal to the null matrix OO.

First, calculate A2A^2: A2=A⋅A=[212533−10−2][212533−10−2]A^2 = A \cdot A = \begin{bmatrix} 2 & 1 & 2 \\ 5 & 3 & 3 \\ -1 & 0 & -2 \end{bmatrix} \begin{bmatrix} 2 & 1 & 2 \\ 5 & 3 & 3 \\ -1 & 0 & -2 \end{bmatrix} A2=[(2)(2)+(1)(5)+(2)(−1)(2)(1)+(1)(3)+(2)(0)(2)(2)+(1)(3)+(2)(−2)(5)(2)+(3)(5)+(3)(−1)(5)(1)+(3)(3)+(3)(0)(5)(2)+(3)(3)+(3)(−2)(−1)(2)+(0)(5)+(−2)(−1)(−1)(1)+(0)(3)+(−2)(0)(−1)(2)+(0)(3)+(−2)(−2)]A^2 = \begin{bmatrix} (2)(2)+(1)(5)+(2)(-1) & (2)(1)+(1)(3)+(2)(0) & (2)(2)+(1)(3)+(2)(-2) \\ (5)(2)+(3)(5)+(3)(-1) & (5)(1)+(3)(3)+(3)(0) & (5)(2)+(3)(3)+(3)(-2) \\ (-1)(2)+(0)(5)+(-2)(-1) & (-1)(1)+(0)(3)+(-2)(0) & (-1)(2)+(0)(3)+(-2)(-2) \end{bmatrix} A2=[4+5−22+3+04+3−410+15−35+9+010+9−6−2+0+2−1+0+0−2+0+4]A^2 = \begin{bmatrix} 4+5-2 & 2+3+0 & 4+3-4 \\ 10+15-3 & 5+9+0 & 10+9-6 \\ -2+0+2 & -1+0+0 & -2+0+4 \end{bmatrix} A2=[7532214130−12]A^2 = \begin{bmatrix} 7 & 5 & 3 \\ 22 & 14 & 13 \\ 0 & -1 & 2 \end{bmatrix}

Next, calculate A3A^3: A3=A2⋅A=[7532214130−12][212533−10−2]A^3 = A^2 \cdot A = \begin{bmatrix} 7 & 5 & 3 \\ 22 & 14 & 13 \\ 0 & -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1 & 2 \\ 5 & 3 & 3 \\ -1 & 0 & -2 \end{bmatrix} A3=[(7)(2)+(5)(5)+(3)(−1)(7)(1)+(5)(3)+(3)(0)(7)(2)+(5)(3)+(3)(−2)(22)(2)+(14)(5)+(13)(−1)(22)(1)+(14)(3)+(13)(0)(22)(2)+(14)(3)+(13)(−2)(0)(2)+(−1)(5)+(2)(−1)(0)(1)+(−1)(3)+(2)(0)(0)(2)+(−1)(3)+(2)(−2)]A^3 = \begin{bmatrix} (7)(2)+(5)(5)+(3)(-1) & (7)(1)+(5)(3)+(3)(0) & (7)(2)+(5)(3)+(3)(-2) \\ (22)(2)+(14)(5)+(13)(-1) & (22)(1)+(14)(3)+(13)(0) & (22)(2)+(14)(3)+(13)(-2) \\ (0)(2)+(-1)(5)+(2)(-1) & (0)(1)+(-1)(3)+(2)(0) & (0)(2)+(-1)(3)+(2)(-2) \end{bmatrix} A3=[14+25−37+15+014+15−644+70−1322+42+044+42−260−5−20−3+00−3−4]A^3 = \begin{bmatrix} 14+25-3 & 7+15+0 & 14+15-6 \\ 44+70-13 & 22+42+0 & 44+42-26 \\ 0-5-2 & 0-3+0 & 0-3-4 \end{bmatrix} A3=[3622231016460−7−3−7]A^3 = \begin{bmatrix} 36 & 22 & 23 \\ 101 & 64 & 60 \\ -7 & -3 & -7 \end{bmatrix}

Now, substitute A3,A2,AA^3, A^2, A and II into the characteristic equation: A3−3A2−7A−IA^3 - 3A^2 - 7A - I =[3622231016460−7−3−7]−3[7532214130−12]−7[212533−10−2]−[100010001]= \begin{bmatrix} 36 & 22 & 23 \\ 101 & 64 & 60 \\ -7 & -3 & -7 \end{bmatrix} - 3 \begin{bmatrix} 7 & 5 & 3 \\ 22 & 14 & 13 \\ 0 & -1 & 2 \end{bmatrix} - 7 \begin{bmatrix} 2 & 1 & 2 \\ 5 & 3 & 3 \\ -1 & 0 & -2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

=[3622231016460−7−3−7]−[211596642390−36]−[14714352121−70−14]−[100010001]= \begin{bmatrix} 36 & 22 & 23 \\ 101 & 64 & 60 \\ -7 & -3 & -7 \end{bmatrix} - \begin{bmatrix} 21 & 15 & 9 \\ 66 & 42 & 39 \\ 0 & -3 & 6 \end{bmatrix} - \begin{bmatrix} 14 & 7 & 14 \\ 35 & 21 & 21 \\ -7 & 0 & -14 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

=[(36−21−14−1)(22−15−7−0)(23−9−14−0)(101−66−35−0)(64−42−21−1)(60−39−21−0)(−7−0−(−7)−0)(−3−(−3)−0−0)(−7−6−(−14)−1)]= \begin{bmatrix} (36-21-14-1) & (22-15-7-0) & (23-9-14-0) \\ (101-66-35-0) & (64-42-21-1) & (60-39-21-0) \\ (-7-0-(-7)-0) & (-3-(-3)-0-0) & (-7-6-(-14)-1) \end{bmatrix}

=[(15−14−1)(7−7)(14−14)(35−35)(22−21−1)(21−21)(−7+7)(−3+3)(−13+14−1)]= \begin{bmatrix} (15-14-1) & (7-7) & (14-14) \\ (35-35) & (22-21-1) & (21-21) \\ (-7+7) & (-3+3) & (-13+14-1) \end{bmatrix}

=[000000000]=O= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O

Since A3−3A2−7A−I=OA^3 - 3A^2 - 7A - I = O, the Cayley-Hamilton Theorem is verified for the given matrix A.