Question

The matrix A = −14 −6 6 28 12 −4 0 0 4 has characteristic polynomial

Answer 1

Answer:

Characteristic equation:

$p(\lambda) = -\lambda^3 + 2\lambda^2 + 8\lambda$

Eigen values:

$\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4$

Step-by-step explanation:

We are given the matrix:

$\displaystyle\left[\begin{array}{ccc}-14&-6&6\\28&12&-4\\0&0&4\end{array}\right]$

The characteristic equation can be calculated as:

$det(A-\lambda I) = 0\\|A-\lambda I| = 0$

We follow the following steps to calculate characteristic equation:

$=det\Bigg(\displaystyle\left[\begin{array}{ccc}-14&-6&6\\28&12&-4\\0&0&4\end{array}\right]-\lambda\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\Bigg)\\\\= det\Bigg(\displaystyle\left[\begin{array}{ccc}-14-\lambda&-6&6\\28&12-\lambda&-4\\0&0&4-\lambda\end{array}\right]\Bigg)\\\\=(-14-\lambda)[(12-\lambda)(4-\lambda)]+6[28(4-\lambda)]-6[(28)(0)-(12-\lambda)(0)]\\\\= -\lambda^3 + 2\lambda^2 + 8\lambda$

$p(\lambda)= -\lambda^3 + 2\lambda^2 + 8\lambda$

To obtain the eigen values, we equate the characteristic equation to 0:

$p(\lambda) = -\lambda^3 + 2\lambda^2 + 8\lambda = 0\\-\lambda(\lambda^2-2\lambda-8) = 0\\-\lambda(\lambda^2-4\lambda+2\lambda-8) = 0\\-\lambda[(\lambda(\lambda-4)+2(\lambda-4)] = 0\\-\lambda(\lambda+2)(\lambda-4) = 0 \\\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4$

We can arrange the eigen values as:

$\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4\\-2 < 0 < 4\\\lambda_2 < \lambda_1 < \lambda_3$