Question

Check whether the matrices A and B are diagonalisable. Diagonalise those matrices which are diagonalisable. i) A =  1 0 0
1 2 −3
1 1 −2  
ii) B =  −2 −4 −1
3 5 1
1 1 2 .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 finding its adjoint.

Answer 1

$$A=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 2 & -3 \\ 1 & 1 & -2 \\ \end{bmatrix} , B=\begin{bmatrix} -2 & -4 & -1 \\ 3 & 5 & 1 \\ 1 & 1 & 2 \\ \end{bmatrix}$$

We first determine the eigenvalues of the matrix A

$$|A|=1(2(-2)-(-3)(1))-0+0=-1$$ $$A-\lambda I=\begin{bmatrix} 1-\lambda & 0 & 0 \\ 1 & 2-\lambda & -3 \\ 1 & 1 & -2-\lambda \\ \end{bmatrix}$$

Characteristic equation

$$|A-\lambda I|=0$$ $$(1-\lambda)((2-\lambda)(-2-\lambda)-(-3)(1))-0+0=0$$ $$-(1-\lambda)^2(1+\lambda)=0$$ $$\lambda_1=1, \lambda_2=1, \lambda_3=-1.$$

These are eigenvalues.

Next, find the eigenvectors.

$$\lambda=1$$ $$\begin{bmatrix} 1-\lambda & 0 & 0 \\ 1 & 2-\lambda & -3 \\ 1 & 1 & -2-\lambda \\ \end{bmatrix} =\begin{bmatrix} 0 & 0 & 0 \\ 1 & 1 & -3 \\ 1 & 1 & -3 \\ \end{bmatrix}$$

Perform row operations to obtain the rref of the matrix:

$$\begin{bmatrix} 0 & 0 & 0 \\ 1 & 1 & -3 \\ 1 & 1 & -3 \\ \end{bmatrix} \to \begin{bmatrix} 1 & 1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

Now, solve the matrix equation

$$\begin{bmatrix} 1 & 1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$$

If we take

$$v_2=t, v_3=s, then\ v_1=3s-t.$$

Therefore

$$v=\begin{bmatrix} 3s-t \\ t \\ 3 \\ \end{bmatrix} =\begin{bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix}t +\begin{bmatrix} 3 \\ 0 \\ 1 \\ \end{bmatrix}s$$

$$\lambda=-1$$ $$\begin{bmatrix} 1-\lambda & 0 & 0 \\ 1 & 2-\lambda & -3 \\ 1 & 1 & -2-\lambda \\ \end{bmatrix} =\begin{bmatrix} 2 & 0 & 0 \\ 1 & 3 & -3 \\ 1 & 1 & -1 \\ \end{bmatrix}$$

Perform row operations to obtain the rref of the matrix:

$$\begin{bmatrix} 2 & 0 & 0 \\ 1 & 3 & -3 \\ 1 & 1 & -1 \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

Now, solve the matrix equation

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$$

If we take

$$v_3=t, then\ v_1=0, v_2=t.$$

Therefore

$$v=\begin{bmatrix} 0 \\ t \\ t \\ \end{bmatrix} =\begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix}t$$

Form the matrix P, whose i-th column is the i-th eigenvector:

$$P=\begin{bmatrix} -1 & 3 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ \end{bmatrix}$$

Form the diagonal matrix D, whose element at row i, column i is i-th eigenvalue:

$$D=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix}$$

These matrices have the property that 

$$A=PDP^{-1}$$

We first determine the eigenvalues of the matrix B

$$|B|=-2(5(2)-(1)(1))+4(3(2)-(1)(1))-1(3(1)-(1)(5))=4$$ $$B-\lambda I=\begin{bmatrix} -2-\lambda & -4 & -1 \\ 3 & 5-\lambda & 1 \\ 1 & 1 & 2-\lambda \\ \end{bmatrix}$$

Characteristic equation

$$|B-\lambda I|=0$$ $$(-2-\lambda)(5-\lambda)(2-\lambda)-3-4+(5-\lambda)-(-2-\lambda)+12(2-\lambda)=0$$ $$-(\lambda)^3+5(\lambda)^2-8\lambda+4=0$$ $$\lambda_1=1, \lambda_2=2, \lambda_3=2.$$

These are eigenvalues.

Next, find the eigenvectors.

$$\lambda=1$$ $$\begin{bmatrix} -2-\lambda & -4 & -1 \\ 3 & 5-\lambda & 1 \\ 1 & 1 & 2-\lambda \\ \end{bmatrix} =\begin{bmatrix} -3 & -4 & -1 \\ 3 & 4 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}$$

Perform row operations to obtain the rref of the matrix:

$$\begin{bmatrix} -3 & -4 & -1 \\ 3 & 4 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

Now, solve the matrix equation

$$\begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$$

If we take

$$v_3=t, then\ v_1=-3t, v_2=2t.$$

Therefore

$$v=\begin{bmatrix} -3t \\ 2t \\ t \\ \end{bmatrix} =\begin{bmatrix} -3 \\ 2 \\ 1 \\ \end{bmatrix}t$$

$$\lambda=2$$ $$\begin{bmatrix} -2-\lambda & -4 & -1 \\ 3 & 5-\lambda & 1 \\ 1 & 1 & 2-\lambda \\ \end{bmatrix} =\begin{bmatrix} -4 & -4 & -1 \\ 3 & 3 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix}$$

Perform row operations to obtain the rref of the matrix:

$$\begin{bmatrix} -4 & -4 & -1 \\ 3 & 3 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix} \to \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

Now, solve the matrix equation

$$\begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$$

If we take

$$v_2=t, then\ v_1=-t, v_3=0.$$

Therefore

$$v=\begin{bmatrix} -t \\ t \\ 0 \\ \end{bmatrix} =\begin{bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix}t$$

Since the number of eigenvectors is less than dimension of the matrix, then the matrix is not diagonalizable.

The matrix B is not diagonalizable.

b) Find inverse of the matrix B in part a) of the question by using Cayley-Hamiltion theorem.

$$B=\begin{bmatrix} -2 & -4 & -1 \\ 3 & 5 & 1 \\ 1 & 1 & 2 \\ \end{bmatrix}$$

$$|B|=-2(5(2)-(1)(1))+4(3(2)-(1)(1))-1(3(1)-(1)(5))=4\mathrlap{\,/}{=}0$$

We have

$$B-\lambda I=\begin{bmatrix} -2-\lambda & -4 & -1 \\ 3 & 5-\lambda & 1 \\ 1 & 1 & 2-\lambda \\ \end{bmatrix}$$

$$p(\lambda)=|B-\lambda I|=$$

$$=(-2-\lambda)(5-\lambda)(2-\lambda)-3-4+(5-\lambda)-(-2-\lambda)+12(2-\lambda)=$$

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

Thus, we have obtained the characteristic polynomial

$$p(\lambda)=-(\lambda)^3+5(\lambda)^2-8\lambda+4$$

of the matrix B.

The Cayley-Hamilton theorem yields that

$$0=p(B)=-B^3+5B^2-8B+4I$$

where O is the 3×3 zero matrix.

Rearranging terms, we have

$$B^3-5B^2+8B=4I$$ $$B({1 \over 4}(B^2-5B+8I))=I$$

Similarly, we have

$$({1 \over 4}(B^2-5B+8I))B=I$$

It follows from these two equalities that the matrix

$${1 \over 4}(B^2-5B+8I)$$

is the inverse matrix of B.

Therefore, we have

$$B^{-1}={1 \over 4}(B^2-5B+8I)$$

$$B^2=\begin{bmatrix} -2 & -4 & -1 \\ 3 & 5 & 1 \\ 1 & 1 & 2 \\ \end{bmatrix}\begin{bmatrix} -2 & -4 & -1 \\ 3 & 5 & 1 \\ 1 & 1 & 2 \\ \end{bmatrix}=$$

$$=\begin{bmatrix} -9 & -13 & -4 \\ 10 & 4 & 4 \\ 3 & 3 & 4 \\ \end{bmatrix}$$

$$B^2-5B+8I=\begin{bmatrix} -9 & -13 & -4 \\ 10 & 4 & 4 \\ 3 & 3 & 4 \\ \end{bmatrix}-5\begin{bmatrix} -2 & -4 & -1 \\ 3 & 5 & 1 \\ 1 & 1 & 2 \\ \end{bmatrix}+$$

$$+8\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}=\begin{bmatrix} 9 & 7 & 1 \\ -5 & -3 & -1 \\ -2 & -2 & 2 \\ \end{bmatrix}$$

Then the inverse matrix of B is

$$B^{-1}=\begin{bmatrix} 9/4 & 7/4 & 1/4 \\ -5/4 & -3/4 & -1/4 \\ -1/2 & -1/2 & 1/2 \\ \end{bmatrix}$$

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

$$B=\begin{bmatrix} -2 & -4 & -1 \\ 3 & 5 & 1 \\ 1 & 1 & 2 \\ \end{bmatrix}$$ $$|B|=-2(5(2)-(1)(1))+4(3(2)-(1)(1))-1(3(1)-(1)(5))=4\mathrlap{\,/}{=}0$$

$$C_{11}=\begin{vmatrix} 5 & 1 \\ 1 & 2 \end{vmatrix}=9, C_{12}=-\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix}=-5, C_{13}=\begin{vmatrix} 3 & 5 \\ 1 & 1 \end{vmatrix}=-2,$$

$$C_{21}=-\begin{vmatrix} -4 & -1 \\ 1 & 2 \end{vmatrix}=7, C_{22}=\begin{vmatrix} -2 & -1 \\ 1 & 2 \end{vmatrix}=-3, C_{23}=-\begin{vmatrix} -2 & -4 \\ 1 & 1 \end{vmatrix}=-2,$$

$$C_{31}=\begin{vmatrix} -4 & -1 \\ 5 & 1 \end{vmatrix}=1, C_{32}=-\begin{vmatrix} -2 & -1 \\ 3 & 1 \end{vmatrix}=-1, C_{33}=\begin{vmatrix} -2 & -4 \\ 3 & 5 \end{vmatrix}=2.$$

$$Adj(B)=C^T=\begin{bmatrix} 9 & 7 & 1 \\ -5 & -3 & -1 \\ -2 & -2 & 2 \\ \end{bmatrix}$$

$$B^{-1}={1 \over {|B|}}C^T=\begin{bmatrix} 9/4 & 7/4 & 1/4 \\ -5/4 & -3/4 & -1/4 \\ -1/2 & -1/2 & 1/2 \\ \end{bmatrix}$$

$$B^{-1}=\begin{bmatrix} 9/4 & 7/4 & 1/4 \\ -5/4 & -3/4 & -1/4 \\ -1/2 & -1/2 & 1/2 \\ \end{bmatrix}$$