Question #272373

Known Matrix:


A=[212122113]A=\begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \end{bmatrix}B=[30201a022a]B=\begin{bmatrix} 3 & 0 & 2 \\ 0 & 1 & a \\ 0 & 2 & 2a \end{bmatrix}


  1. Determine the characteristic equation det(A − λI) = 0 of the matrix above
  2. Determine the eigenvalues of the matrix and the basis of the eigenspace

Note: In matrix B, let the value of a be so that the eigenvalues and the basis of the eigenspace are dependent on a.


1
Expert's answer
2021-11-30T11:19:06-0500

a)


A=[212122113]A=\begin{bmatrix} 2& 1 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \end{bmatrix}

A−λI=[2−λ1212−λ2113−λ]A-\lambda I=\begin{bmatrix} 2 -\lambda & 1 & 2 \\ 1 & 2-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}

The characteristic equation


∣2−λ1212−λ2113−λ∣=0\begin{vmatrix} 2 -\lambda & 1 & 2 \\ 1 & 2-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{vmatrix}=0

(2−λ)∣2−λ213−λ∣−(1)∣1213−λ∣(2 -\lambda)\begin{vmatrix} 2 -\lambda & 2 \\ 1 & 3 -\lambda \end{vmatrix}-(1)\begin{vmatrix} 1 & 2 \\ 1 & 3 -\lambda \end{vmatrix}

+(2)∣12−λ11∣+(2)\begin{vmatrix} 1 & 2-\lambda \\ 1 & 1 \end{vmatrix}

=(2−λ)((2−λ)(3−λ)−2)−(3−λ−2)=(2-\lambda)((2-\lambda)(3-\lambda)-2)-(3-\lambda-2)

+2(1−(2−λ))+2(1-(2-\lambda))

=(2−λ)(4−5λ+λ2)−1+λ+2λ−2=(2-\lambda)(4-5\lambda+\lambda^2)-1+\lambda+2\lambda-2

=8−10λ+2λ2−4λ+5λ2−λ3+3λ−3=8-10\lambda+2\lambda^2-4\lambda+5\lambda^2-\lambda^3+3\lambda-3

=−λ3+7λ2−11λ+5=0=-\lambda^3+7\lambda^2-11\lambda+5=0

−λ2(λ−1)+6λ(λ−1)−5(λ−1)=0-\lambda^2(\lambda-1)+6\lambda(\lambda-1)-5(\lambda-1)=0

−(λ−1)(λ2−6λ+5)=0-(\lambda-1)(\lambda^2-6\lambda+5)=0

Matrix A.A. The characteristic equation is


−(λ−1)(λ−1)(λ−5)=0-(\lambda-1)(\lambda-1)(\lambda-5)=0


B=[30201a022a]B=\begin{bmatrix} 3& 0 & 2 \\ 0 & 1 & a \\ 0 & 2 & 2a \end{bmatrix}

B−λI=[3−λ0201−λa022a−λ]B-\lambda I=\begin{bmatrix} 3 -\lambda & 0 & 2 \\ 0 & 1-\lambda & a \\ 0 & 2 & 2a-\lambda \end{bmatrix}

The characteristic equation


∣3−λ0201−λa022a−λ∣=0\begin{vmatrix} 3 -\lambda & 0 & 2 \\ 0 & 1-\lambda & a \\ 0 & 2 & 2a-\lambda \end{vmatrix}=0

(3−λ)∣1−λa22a−λ∣−(0)∣0222a−λ∣(3 -\lambda)\begin{vmatrix} 1 -\lambda & a \\ 2 & 2a -\lambda \end{vmatrix}-(0)\begin{vmatrix} 0 & 2 \\ 2 & 2a -\lambda \end{vmatrix}

+(0)∣021−λa∣+(0)\begin{vmatrix} 0 & 2 \\ 1-\lambda & a \end{vmatrix}

=(3−λ)((1−λ)(2a−λ)−2a)=(3-\lambda)((1-\lambda)(2a-\lambda)-2a)

+2(1−(2−λ))+2(1-(2-\lambda))

=(3−λ)(2a−λ−2aλ+λ2−2a)=(3-\lambda)(2a-\lambda-2a\lambda+\lambda^2-2a)

=(3−λ)λ(λ−(1+2a))=0=(3-\lambda)\lambda(\lambda-(1+2a))=0

Matrix B.B. The characteristic equation is


(3−λ)λ(λ−(1+2a))=0(3-\lambda)\lambda(\lambda-(1+2a))=0

2.

Matrix A.A. The characteristic equation is


−(λ−1)(λ−1)(λ−5)=0-(\lambda-1)(\lambda-1)(\lambda-5)=0

The eigen values are λ1=5,λ2=1,λ3=1.\lambda_1=5, \lambda_2=1, \lambda_3=1.

λ1=5\lambda_1=5

[2−λ1212−λ2113−λ]=[−3121−3211−2]\begin{bmatrix} 2 -\lambda & 1 & 2 \\ 1 & 2-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}=\begin{bmatrix} -3 & 1 & 2 \\ 1 & -3 & 2 \\ 1 & 1 & -2 \end{bmatrix}

R1=R1/(−3)R_1=R_1/(-3)


[1−1/3−2/31−3211−2]\begin{bmatrix} 1 & -1/3 & -2/3 \\ 1 & -3 & 2 \\ 1 & 1 & -2 \end{bmatrix}

R2=R2−R1R_2=R_2-R_1


[1−1/3−2/30−8/38/311−2]\begin{bmatrix} 1 & -1/3 & -2/3 \\ 0 & -8/3 & 8/3 \\ 1 & 1 & -2 \end{bmatrix}

R3=R3−R1R_3=R_3-R_1


[1−1/3−2/30−8/38/304/3−4/3]\begin{bmatrix} 1 & -1/3 & -2/3 \\ 0 & -8/3 & 8/3 \\ 0 & 4/3 & -4/3 \end{bmatrix}

R2=−3R2/8R_2=-3R_2/8

[1−1/3−2/301−104/3−4/3]\begin{bmatrix} 1 & -1/3 & -2/3 \\ 0 & 1 & -1 \\ 0 & 4/3 & -4/3 \end{bmatrix}

R1=R1+R2/3R_1=R_1+R_2/3


[10−101−104/3−4/3]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 4/3 & -4/3 \end{bmatrix}

R3=R3−4R2/3R_3=R_3-4R_2/3


[10−101−1000]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}

Solve the matrix equation


[10−101−1000][x1x2x3]=[000]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

If we take x3=t,x_3=t, then x1=t,x2=t,x3=t.x_1=t, x_2=t, x_3=t.

The eigen vector is [111].\begin{bmatrix} 1 \\ 1\\ 1 \end{bmatrix}.


λ2=1\lambda_2=1

[2−λ1212−λ2113−λ]=[112112112]\begin{bmatrix} 2 -\lambda & 1 & 2 \\ 1 & 2-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}=\begin{bmatrix} 1 & 1 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \end{bmatrix}

R2=R2−R1R_2=R_2-R_1


[112000112]\begin{bmatrix} 1 & 1 & 2 \\ 0 & 0 & 0 \\ 1 & 1 & 2 \end{bmatrix}

R3=R3−R1R_3=R_3-R_1


[112000000]\begin{bmatrix} 1 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

Solve the matrix equation


[112000000][x1x2x3]=[000]\begin{bmatrix} 1 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

If we take x2=t,x3=s,x_2=t,x_3=s, then x1=−t−2s,x2=t,x3=s.x_1=-t-2s, x_2=t, x_3=s.

The eigen vectors are [−110],[−201]\begin{bmatrix} -1 \\ 1\\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0\\ 1 \end{bmatrix}



Matrix B.B. The characteristic equation is


(3−λ)λ(λ−(1+2a))=0(3-\lambda)\lambda(\lambda-(1+2a))=0

The eigen values are λ1=3,λ2=1+2a,λ3=0.\lambda_1=3, \lambda_2=1+2a, \lambda_3=0.

λ1=3\lambda_1=3


[3−λ0201−λa022a−λ]=[0020−2a022a−3]\begin{bmatrix} 3 -\lambda & 0 & 2 \\ 0 & 1-\lambda & a \\ 0 & 2 & 2a-\lambda \end{bmatrix}=\begin{bmatrix} 0 & 0 & 2 \\ 0 & -2 & a \\ 0 & 2 & 2a-3 \end{bmatrix}


Swap the rows 1and 2


[0−2a002022a−3]\begin{bmatrix} 0 & -2 & a \\ 0 & 0 & 2 \\ 0 & 2 & 2a-3 \end{bmatrix}


R1=R1/(−2)R_1=R_1/(-2)


[01−a/2002022a−3]\begin{bmatrix} 0 & 1 & -a/2 \\ 0 & 0 & 2 \\ 0 & 2 & 2a-3 \end{bmatrix}

R3=R3−2R1R_3=R_3-2R_1


[01−a/2002003a−3]\begin{bmatrix} 0 & 1 & -a/2 \\ 0 & 0 & 2 \\ 0 & 0& 3a-3 \end{bmatrix}

R2=R2/2R_2=R_2/2


[01−a/2001003a−3]\begin{bmatrix} 0 & 1 & -a/2 \\ 0 & 0 & 1 \\ 0 & 0& 3a-3 \end{bmatrix}

R1=R1+aR2/2R_1=R_1+aR_2/2


[010001003a−3]\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0& 3a-3 \end{bmatrix}

R3=R3−3(a−1)R2R_3=R_3-3(a-1)R_2


[010001000]\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0& 0 \end{bmatrix}

Solve the matrix equation


[010001000][x1x2x3]=[000]\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

If we take x1=t,x_1=t, then x1=t,x2=0,x3=0.x_1=t, x_2=0, x_3=0.

The eigen vector is [100].\begin{bmatrix} 1 \\ 0\\ 0 \end{bmatrix}.


λ2=2a+1\lambda_2=2a+1

[3−λ0201−λa022a−λ]=[2−2a020−2aa02−1]\begin{bmatrix} 3 -\lambda & 0 & 2 \\ 0 & 1-\lambda & a \\ 0 & 2 & 2a-\lambda \end{bmatrix}=\begin{bmatrix} 2-2a & 0 & 2 \\ 0 & -2a & a \\ 0 & 2 & -1 \end{bmatrix}

R1=R1/(2−2a),a≠1R_1=R_1/(2-2a), a\not=1

[101/(1−a)0−2aa02−1]\begin{bmatrix} 1 & 0 & 1/(1-a) \\ 0 & -2a & a \\ 0 & 2 & -1 \end{bmatrix}

R2=R2/(−2a)R_2=R_2/(-2a)

[101/(1−a)01−1/202−1]\begin{bmatrix} 1 & 0 & 1/(1-a) \\ 0 & 1 & -1/2 \\ 0 & 2 & -1 \end{bmatrix}

R3=R3−2R2R_3=R_3-2R_2

[101/(1−a)01−1/2000]\begin{bmatrix} 1 & 0 & 1/(1-a) \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \end{bmatrix}

Solve the matrix equation


[101/(1−a)01−1/2000]=[000]\begin{bmatrix} 1 & 0 & 1/(1-a) \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

If we take x3=t,x_3=t, then x1=1/(a−1),x2=1/2,x3=1.x_1=1/(a-1), x_2=1/2, x_3=1.

The eigen vector is [1/(a−1)1/21].\begin{bmatrix} 1 /(a-1)\\ 1/2\\ 1 \end{bmatrix}.


λ3=0\lambda_3=0

[3−λ0201−λa022a−λ]=[30201a022a]\begin{bmatrix} 3 -\lambda & 0 & 2 \\ 0 & 1-\lambda & a \\ 0 & 2 & 2a-\lambda \end{bmatrix}=\begin{bmatrix} 3 & 0 & 2 \\ 0 & 1 & a \\ 0 & 2 & 2a \end{bmatrix}

R1=R1/3R_1=R_1/3

[102/301a022a]\begin{bmatrix} 1 & 0 & 2/3 \\ 0 & 1 & a \\ 0 & 2 & 2a \end{bmatrix}

R3=R3−2R2R_3=R_3-2R_2


[102/301a000]\begin{bmatrix} 1 & 0 & 2/3 \\ 0 & 1 & a \\ 0 & 0 & 0 \end{bmatrix}

Solve the matrix equation


[102/301a000]=[000]\begin{bmatrix} 1 & 0 & 2/3 \\ 0 & 1 & a \\ 0 & 0 & 0 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

If we take x3=t,x_3=t, then x1=−2/3,x2=−a,x3=1.x_1=-2/3, x_2=-a, x_3=1.

The eigen vector is [−2/3−a1].\begin{bmatrix} -2/3\\ -a\\ 1 \end{bmatrix}.



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