Question #198526

Find A and write the following linear system in the matrix equation form (λI2 − A)X = 0


−x + y = λx

3x + y = λy


For the systems above, find:


(10.1) The determinant (known as the characteristic equation).

(10.2) Solve for λ when det(λI2 − A) = 0.

(10.3) Substitute for each value of λ from (ii) into the equation (λI2 − A)X = 0 and solve the corresponding system for X = x

y


1
Expert's answer
2021-05-28T08:18:15-0400

Rewriting the equations,

x(1λ)+y=0x(-1-\lambda)+y=0

3x+y(1λ)=03x+y(1-\lambda)=0


(1) [(1λ)13(1λ)][xy]=[00]\begin{bmatrix} (-1-\lambda) & 1 \\ 3 & (1-\lambda) \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}


(1λ)(1λ)3=1+λλ+λ23(-1-\lambda)(1-\lambda)-3=-1+\lambda-\lambda+\lambda^2-3

=λ24=\lambda^2-4


(2) when det(λI2A)=0\det(\lambda I_2-A)=0

λ24=0\Rightarrow\lambda^2-4=0

λ=±2\Rightarrow\lambda=\pm2


(3) Substituting the value of (λI2A)X=0(\lambda I_2-A)X=0

For system to be consistent, λ=±2\lambda\cancel=\pm2

(λI2A)X=0(\lambda I_2-A)X=0

[(1λ)13(1λ)][xy]=[00]\begin{bmatrix} (-1-\lambda) & 1 \\ 3 & (1-\lambda) \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}


x(1λ)+y=0x(-1-\lambda)+y=0

x(12)+y=0x(-1-2)+y=0

3x+y=0-3x+y=0

3xy=03x-y=0


3x+y(1λ)=03x+y(1-\lambda)=0

3x+y(12)=03x+y(1-2)=0

3xy=03x-y=0


Since both the equations are parallel therefore the system has infinitely many solutions.


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