Question Number 89422 by naka3546 last updated on 17/Apr/20
Commented by jagoll last updated on 17/Apr/20
$${f}\left({x}\right)\:=\:\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right) \\ $$$${f}\left({A}\right)\:=\:\left({A}+\mathrm{2}{I}\right)\left({A}+{I}\right)\: \\ $$$$ \\ $$
Answered by som(math1967) last updated on 17/Apr/20
![f(A)=A^2 +3A+2I = [(1,(−2)),(2,(−1)) ] [(1,(−2)),(2,(−1)) ]+3 [(1,(−2)),(2,(−1)) ] +2 [(1,0),(0,1) ] = [((−3),0),(0,(−3)) ]+ [(3,(−6)),(6,(−3)) ]+ [(2,0),(0,2) ] = [((−3+3+2),(0−6+0)),((0+6+0),(−3−3+2)) ] = determinant ((2,(−6)),(6,(−4)))....ansB](https://www.tinkutara.com/question/Q89438.png)
$${f}\left({A}\right)={A}^{\mathrm{2}} +\mathrm{3}{A}+\mathrm{2}{I} \\ $$$$=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{2}}\\{\mathrm{2}}&{−\mathrm{1}}\end{bmatrix}\begin{bmatrix}{\mathrm{1}}&{−\mathrm{2}}\\{\mathrm{2}}&{−\mathrm{1}}\end{bmatrix}+\mathrm{3}\begin{bmatrix}{\mathrm{1}}&{−\mathrm{2}}\\{\mathrm{2}}&{−\mathrm{1}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$=\begin{bmatrix}{−\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{−\mathrm{3}}\end{bmatrix}+\begin{bmatrix}{\mathrm{3}}&{−\mathrm{6}}\\{\mathrm{6}}&{−\mathrm{3}}\end{bmatrix}+\begin{bmatrix}{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{2}}\end{bmatrix} \\ $$$$=\begin{bmatrix}{−\mathrm{3}+\mathrm{3}+\mathrm{2}}&{\mathrm{0}−\mathrm{6}+\mathrm{0}}\\{\mathrm{0}+\mathrm{6}+\mathrm{0}}&{−\mathrm{3}−\mathrm{3}+\mathrm{2}}\end{bmatrix} \\ $$$$=\begin{vmatrix}{\mathrm{2}}&{−\mathrm{6}}\\{\mathrm{6}}&{−\mathrm{4}}\end{vmatrix}….{ansB} \\ $$
Commented by naka3546 last updated on 17/Apr/20
$${thank}\:\:{you} \\ $$