Question

 ii) B =


−1 −3 0
2 4 0
−1 −1 2

.
b) Find inverse of the matrix B in part a) of the question by finding the adjoint as well
as using Cayley-Hamiltion

Answer 1

Answer on Question #81450 – Math – Linear Algebra

Question

ii) $B = \begin{pmatrix} -1 & -3 & 0 \\ 2 & 4 & 0 \\ -1 & -1 & 2 \end{pmatrix}$

b) Find inverse of the matrix $B$ in part a) of the question by finding the adjoint as well as using Cayley-Hamilton theorem.

Solution

$$B = \begin{pmatrix} -1 & -3 & 0 \\ 2 & 4 & 0 \\ -1 & -1 & 2 \end{pmatrix}$$ $$\det(B) = \begin{vmatrix} -1 & -3 & 0 \\ 2 & 4 & 0 \\ -1 & -1 & 2 \end{vmatrix} = 2 \begin{vmatrix} -1 & -3 \\ 2 & 4 \end{vmatrix} = 2(-4 + 6) = 4 \neq 0$$ $$C_{11} = \begin{vmatrix} 4 & 0 \\ -1 & 2 \end{vmatrix} = 8, \quad C_{12} = - \begin{vmatrix} 2 & 0 \\ -1 & 2 \end{vmatrix} = -4, \quad C_{13} = \begin{vmatrix} 2 & 4 \\ -1 & -1 \end{vmatrix} = 2$$ $$C_{21} = - \begin{vmatrix} -3 & 0 \\ -1 & 2 \end{vmatrix} = 6, \quad C_{22} = \begin{vmatrix} -1 & 0 \\ -1 & 2 \end{vmatrix} = -2, \quad C_{23} = - \begin{vmatrix} -1 & -3 \\ -1 & -1 \end{vmatrix} = 2$$ $$C_{31} = \begin{vmatrix} -3 & 0 \\ 4 & 0 \end{vmatrix} = 0, \quad C_{32} = - \begin{vmatrix} -1 & 0 \\ 2 & 0 \end{vmatrix} = 0, \quad C_{33} = \begin{vmatrix} -1 & -3 \\ 2 & 4 \end{vmatrix} = 2$$ $$Agj(B) = C^T = \begin{pmatrix} 8 & 6 & 0 \\ -4 & -2 & 0 \\ 2 & 2 & 2 \end{pmatrix}$$ $$B^{-1} = \frac{1}{\det(B)} Agj(B) = \frac{1}{4} \begin{pmatrix} 8 & 6 & 0 \\ -4 & -2 & 0 \\ 2 & 2 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 3/2 & 0 \\ -1 & -1/2 & 0 \\ 1/2 & 1/2 & 1/2 \end{pmatrix}$$

Use Cayley-Hamilton theorem

$$p(t) = \det(B - tI)$$ $$\det(B - tI) = \begin{vmatrix} -1 - t & -3 & 0 \\ 2 & 4 - t & 0 \\ -1 & -1 & 2 - t \end{vmatrix} = (2 - t) \begin{vmatrix} -1 - t & -3 \\ 2 & 4 - t \end{vmatrix} =$$ $$= (2 - t)(-4 + t - 4t + t^2 + 6) =$$ $$= 4 - 6t + 2t^2 - 2t + 3t^2 - t^3 = -t^3 + 5t^2 - 8t + 4$$ $$p(B) = 0 \Rightarrow -B^3 + 5B^2 - 8B + 4I = 0$$ $$I = \frac{1}{4}(B^3 - 5B^2 + 8B)$$ $$I = \frac{1}{4}B(B^2 - 5B + 8I)$$ $$B^{-1} = \frac{1}{4}(B^2 - 5B + 8I)$$ $$\begin{aligned} B^2 &= \begin{pmatrix} -1 & -3 & 0 \\ 2 & 4 & 0 \\ -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} -1 & -3 & 0 \\ 2 & 4 & 0 \\ -1 & -1 & 2 \end{pmatrix} \\ &= \begin{pmatrix} -1(-1) - 3(2) + 0 & -1(-3) - 3(4) + 0 & 0 \\ 2(-1) + 4(2) + 0 & 2(-3) + 4(4) + 0 & 0 \\ -1(-1) - 1(2) + 2(-1) & -1(-3) - 1(4) + 2(-1) & 0 + 0 + 2(2) \end{pmatrix} \\ &= \begin{pmatrix} -5 & -9 & 0 \\ 6 & 10 & 0 \\ -3 & -3 & 4 \end{pmatrix} \end{aligned}$$ $$\begin{aligned} B^2 - 5B + 8I &= \begin{pmatrix} -5 & -9 & 0 \\ 6 & 10 & 0 \\ -3 & -3 & 4 \end{pmatrix} + \begin{pmatrix} 5 & 15 & 0 \\ -10 & -20 & 0 \\ 5 & 5 & -10 \end{pmatrix} + \begin{pmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{pmatrix} \\ &= \begin{pmatrix} 8 & 6 & 0 \\ -4 & -2 & 0 \\ 2 & 2 & 2 \end{pmatrix} \end{aligned}$$ $$B^{-1} = \frac{1}{4} \begin{pmatrix} 8 & 6 & 0 \\ -4 & -2 & 0 \\ 2 & 2 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 3/2 & 0 \\ -1 & -1/2 & 0 \\ 1/2 & 1/2 & 1/2 \end{pmatrix}$$ $$B^{-1} = \begin{pmatrix} 2 & 3/2 & 0 \\ -1 & -1/2 & 0 \\ 1/2 & 1/2 & 1/2 \end{pmatrix}$$

Augment the matrix with identity matrix

$$\begin{pmatrix} -1 & -3 & 0 & 1 & 0 & 0 \\ 2 & 4 & 0 & 0 & 1 & 0 \\ -1 & -1 & 2 & 0 & 0 & 1 \end{pmatrix}$$ $$\begin{pmatrix} -1 & -3 & 0 & 1 & 0 & 0 \\ 2 & 4 & 0 & 0 & 1 & 0 \\ -1 & -1 & 2 & 0 & 0 & 1 \end{pmatrix} \xrightarrow{R_2 + (2)R_1} \begin{pmatrix} -1 & -3 & 0 & 1 & 0 & 0 \\ 0 & -2 & 0 & 2 & 1 & 0 \\ -1 & -1 & 2 & 0 & 0 & 1 \end{pmatrix}$$ $$\begin{pmatrix} -1 & -3 & 0 & 1 & 0 & 0 \\ 0 & -2 & 0 & 2 & 1 & 0 \\ -1 & -1 & 2 & 0 & 0 & 1 \end{pmatrix} \xrightarrow{R_3 - R_1} \begin{pmatrix} -1 & -3 & 0 & 1 & 0 & 0 \\ 0 & -2 & 0 & 2 & 1 & 0 \\ 0 & 2 & 2 & -1 & 0 & 1 \end{pmatrix}$$ $$\begin{pmatrix} -1 & -3 & 0 & 1 & 0 & 0 \\ 0 & -2 & 0 & 2 & 1 & 0 \\ 0 & 2 & 2 & -1 & 0 & 1 \end{pmatrix} \xrightarrow{(-1)R_1} \begin{pmatrix} 1 & 3 & 0 & -1 & 0 & 0 \\ 0 & -2 & 0 & 2 & 1 & 0 \\ 0 & 2 & 2 & -1 & 0 & 1 \end{pmatrix}$$ $$\left(\begin{array}{cccccc} 1 & 3 & 0 & -1 & 0 & 0 \\ 0 & -2 & 0 & 2 & 1 & 0 \\ 0 & 2 & 2 & -1 & 0 & 1 \end{array}\right) \xrightarrow{R_3 + R_2} \left(\begin{array}{cccccc} 1 & 3 & 0 & -1 & 0 & 0 \\ 0 & -2 & 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 & 1 & 1 \end{array}\right)$$ $$\left(\begin{array}{cccccc} 1 & 3 & 0 & -1 & 0 & 0 \\ 0 & -2 & 0 & 2 & 1 & 0 \\ 0 & 2 & 2 & -1 & 0 & 1 \end{array}\right) \xrightarrow{R_2 / (-2)} \left(\begin{array}{cccccc} 1 & 3 & 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & -1 & -1/2 & 0 \\ 0 & 0 & 2 & 1 & 1 & 1 \end{array}\right)$$ $$\left(\begin{array}{cccccc} 1 & 3 & 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & -1 & -1/2 & 0 \\ 0 & 0 & 2 & 1 & 1 & 1 \end{array}\right) \xrightarrow{R_1 - (3)R_2} \left(\begin{array}{cccccc} 1 & 0 & 0 & 2 & 3/2 & 0 \\ 0 & 1 & 0 & -1 & -1/2 & 0 \\ 0 & 0 & 2 & 1 & 1 & 1 \end{array}\right)$$ $$\left(\begin{array}{cccccc} 1 & 0 & 0 & 2 & 3/2 & 0 \\ 0 & 1 & 0 & -1 & -1/2 & 0 \\ 0 & 0 & 2 & 1 & 1 & 1 \end{array}\right) \xrightarrow{R_3/2} \left(\begin{array}{cccccc} 1 & 0 & 0 & 2 & 3/2 & 0 \\ 0 & 1 & 0 & -1 & -1/2 & 0 \\ 0 & 0 & 1 & 1/2 & 1/2 & 1/2 \end{array}\right)$$

We have obtained the identity matrix to the left. So, we are done.

$$B^{-1} = \left( \begin{array}{ccc} 2 & 3/2 & 0 \\ -1 & -1/2 & 0 \\ 1/2 & 1/2 & 1/2 \end{array} \right)$$

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