Answer to Question #125836 in Linear Algebra for hitendra GARG

Question

a) Check whether the matrices A and B are diagonalisable. Diagonalise those
matrices which are diagonalisable.
A = 1 0 0 B = 2 0 0
1 5 -3 -2 2 -1
2 8 -5 -1 0 1
b) Find inverse of the matrix B in part a) of the question by using
Cayley-Hamiltion theorem.

c) Find the inverse of the matrix A in part a) of the question by fnding its
adjoint.

Answer 1

a) Find eigenvalues from the characteristic polynomial of A:

$\begin{vmatrix} 1-\lambda & 0 & 0 \\ 1 & 5-\lambda & -3 \\ 2 & 8 & -5-\lambda \end{vmatrix}$ $=−λ^3+λ^2+λ−1=−(λ−1)×(λ^2−1)=−(λ−1)×(λ−1)×(λ+1)$

Hence, Eigenvalues are 1, 1, -1.

Corresponding to eigenvalue 1, G.M. = $3 - \rho(A-I)$

= $3 - \rho \begin{bmatrix} 0 & 0 & 0 \\ 1 & 4 & -3 \\ 2 & 8 & -6 \end{bmatrix} = 3-1 = 2$ .

Corresponding to eigenvalue -1, G.M. = 1.

Hence, A matrix is diagonlizable.

Characteristic polynomial of B is

$\begin{vmatrix} 2-\lambda & 0 & 0 \\ -2 & 2-\lambda & -1 \\ -1 & 0 & 1-\lambda \end{vmatrix}=0$

$\implies -\lambda^3+5 \lambda^2-8\lambda+4 = 0 \\ =-(\lambda-1)(\lambda^2-4\lambda+4)=-(\lambda-1)(\lambda-2)(\lambda-2) = 0$

Hence, eigenvalues of B are 1,2,2.

Corresponding to eigenvalue 2, G.M. = $3-\rho(A-2I) = 3 - \rho \begin{bmatrix} 0 & 0 & 0 \\ -2 & 0 & -1 \\ -1 & 0 & -1 \end{bmatrix}$

= 3 - 1 = 2.

Hence, B matrix is not diagonalizable.

b) According to Cayley-Hamilton theorem, $Ch(B)=0$ .

$-B^3+5 B^2-8B+4I = 0 \\ \implies 4I = B^3 - 5B^2+8B \\ \implies B^{-1} = \frac{1}{4} [B^2-5B+8I]$

So, $4 B^{-1} = \begin{bmatrix} 2 & 0 & 0 \\ -2 & 2 & -1 \\ -1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 & 0 \\ -2 & 2 & -1 \\ -1 & 0 & 1 \end{bmatrix} - 5 \begin{bmatrix} 2 & 0 & 0 \\ -2 & 2 & -1 \\ -1 & 0 & 1 \end{bmatrix} + 8 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\implies B^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{-1}{2} \\ \frac{1}{2} & 0 & 1 \end{bmatrix}$ .

c) $|A| = -25+24 = -1$ ,

Cofactors of A are $C_{11} = -2, C_{12} =-1, C_{13} = -2, C_{21} = 0, C_{22} = -5, \\ C_{23} =-8, C_{31} = 0, C_{32} = 3, C_{33} = 5$

$adj(A) = \begin{bmatrix} -1 & -1 & -2 \\ 0 & -5 & -8 \\ 0 & 3 & 5 \end{bmatrix}^T = \begin{bmatrix} -1 & 0 & 0 \\ -1 & -5 & 3 \\ -2 & -8 & 5 \end{bmatrix}$

$A^{-1} =\frac{1}{|A|}adj(A) = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 5 & - 3 \\ 2 & 8 & -5 \end{bmatrix}$ .