Answer to Question #198526 in Linear Algebra for prince

Question

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

Answer 1

Rewriting the equations,

$x(-1-\lambda)+y=0$

$3x+y(1-\lambda)=0$

(1) $\begin{bmatrix} (-1-\lambda) & 1 \\ 3 & (1-\lambda) \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$

$(-1-\lambda)(1-\lambda)-3=-1+\lambda-\lambda+\lambda^2-3$

$=\lambda^2-4$

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

$\Rightarrow\lambda^2-4=0$

$\Rightarrow\lambda=\pm2$

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

For system to be consistent, $\lambda\cancel=\pm2$

$(\lambda I_2-A)X=0$

$\begin{bmatrix} (-1-\lambda) & 1 \\ 3 & (1-\lambda) \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$

$x(-1-\lambda)+y=0$

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

$-3x+y=0$

$3x-y=0$

$3x+y(1-\lambda)=0$

$3x+y(1-2)=0$

$3x-y=0$

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