(i) $Adj (A)$

(ii) $24A^{-1}+2A-I$

where $A=\begin{bmatrix} 1 & 2 & 3 & -2 \\ 0 & 2 & 4 & 6 \\ 0 & 0 & 4 & -5 \\ 0 & 0 & 0 & 6 \end{bmatrix}$

Solution:

Given: $A=\begin{bmatrix} 1 & 2 & 3 & -2 \\ 0 & 2 & 4 & 6 \\ 0 & 0 & 4 & -5 \\ 0 & 0 & 0 & 6 \end{bmatrix}$

$|A|=48$

Eigen value,

$|A-\lambda I|=0$

$\begin{bmatrix} 1-\lambda & 2 & 3 & -2 \\ 0 & 2-\lambda & 4 & 6 \\ 0 & 0 & 4-\lambda & -5 \\ 0 & 0 & 0 & 6-\lambda \end{bmatrix}=0$

$(1-\lambda)(2-\lambda)(4-\lambda)(6-\lambda)=0$

$\therefore \lambda_{1}=4$

$\therefore \lambda_{2}=2$

$\therefore \lambda_{3}=4$

$\therefore \lambda_{4}=6$

i) Adj of (A) $=\cfrac{|A|}{\text{eigen value}}=\cfrac{|A|}{\lambda_{1}}, \cfrac{|A|}{\lambda_{2}}, \cfrac{|A|}{\lambda_{3}},\cfrac{|A|}{\lambda_{4}}= \cfrac{48}{1}, \cfrac{48}{2}, \cfrac{48}{3}, \cfrac{48}{4}=48, 24, 12, 8$

ii) Eigen value for $24A^{-1}+2A-I$

The eigen value of A are = 1, 2, 4, 6

$\therefore$ Eigen value for $A^{-1}= \cfrac{1}{1}, \cfrac{1}{2}, \cfrac{1}{4}, \cfrac{1}{6}$

$\therefore$ Eigen value for $24A^{-1}= 24 \cdot \left(\cfrac{1}{1}\right), 24 \cdot \left( \cfrac{1}{2}\right), 24\cdot \left( \cfrac{1}{4} \right), 24 \cdot \left( \cfrac{1}{6} \right)$

$\therefore$ Eigen value for $24A^{-1}=24, 12, 6, 4$

$\therefore$ Eigen value for $2A=2(1), 2(2), 2(4), 2(6)=2, 4, 8, 12$

Hence, the eigen values for $24A^{-1}+2A-I$

i) 24+2-1=25

ii) 12+4-1=15

iii) 6+8-1=13

iv) 4+12-1=15

$\therefore$ The eigen values $24A^{-1}+2A-I$ is 25, 15, 13, 15.