Find the characteristic equation and the eigenvalues (and corresponding eigenvectors) of the matrix. [2−2303−20−12] a) The characteristic equation λ3−7λ2+14λ−8=0 b) The eigenvalues (λ1, λ2, λ3)=(1, 2, 4) The corresponding eigenvectors x1=−7, 1, 1? x2=1, 0, 0 x3=13, −4, 2?

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Find the characteristic equation and the eigenvalues (and corresponding eigenvectors) of the matrix.  [2−2303−20−12]  a) The characteristic equation  λ3−7λ2+14λ−8=0  b) The eigenvalues  (λ1, λ2, λ3)=(1, 2, 4)  The corresponding eigenvectors  x1=−7, 1, 1?  x2=1, 0, 0  x3=13, −4, 2?

Deufemiak7 · Accepted Answer

Step 1 Explanation: Given that, [2−2303−20−12] a. Characteritic equation for given matrix λ3−7λ2+14λ−8=0 b. Eigenvalues are: λ3−7λ2+14λ−8=0 (λ−1)(λ−2)(λ−4)=0 λ−1=0 λ=1 λ−2=0 λ=2 λ−4=0 λ=4 λ1=1 λ2=2 λ3=4 Step 2 c) Find the eigenvectors To find the eigenvectors, we will set up the equation: Solve (A−λI) A−λ1I=[2−2303−20−12]−1[100010001] A−1I=[2−2303−20−12]−[100010001]=[1−2302−20−11] Now reduce this matrix using row echelon form. Reduce [1−2302−20−11] R3⇒R3+R22 =[1−2302−2000] R2⇒R22 =[1−2301−1000] R1⇒R1+2R2 =[10101−1000] (A−!I)[xyz]=[10101−1000][xyz]=0 x+z=0⇒x=−z

Mollie Nash · Accepted Answer

A=[2−2303−20−12] |A−λI|=0 implies [2−λ−2303−λ−20−12−λ]=0⇒(2−λ)[(3−λ)(2−λ)−2]=0 ⇒(2−λ)[6−5λ+λ2−2]=0⇒(2−λ)(λ2−5λ+4)=0 ⇒(2−λ)(1−λ)(4−λ)=0 or −λ3+7λ2−14λ+8=0 or λ3−7λ2+14λ−8=0 The eisen valus are λ=1, 2, 4 (λ1 λ2 λ3)=(1, 2, 4) i) for λ=1, (A−I)x=0 [1−2302−20−11][xy2]=[000]⇒y=z 2−2y+z2=0⇒x=−z ⇒x1=[−111] ii) λ=2, (A−2I)x=0 ⇒[0−2301−20−10][xyz]=[00]⇒y=0z=0x=k x2=[1  0  0]T iii) for λ=4, (A−4I)x=0 ⇒[−2−230−1−20−1−2][xyz]=[000] ⇒y=2

Find the characteristic equation and the eigenvalues (and corresponding eigenvectors)

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