Question Number 42728 by RANGAM kumar last updated on 01/Sep/18
![If A= [(( cos x),(sin x)),((−sin x),(cos x)) ] and A adj A = k [(1,0),(0,1) ], then the value of k is](https://www.tinkutara.com/question/Q42728.png)
$$\mathrm{If}\:{A}=\begin{bmatrix}{\:\:\:\mathrm{cos}\:{x}}&{\mathrm{sin}\:{x}}\\{−\mathrm{sin}\:{x}}&{\mathrm{cos}\:{x}}\end{bmatrix}\:\mathrm{and}\: \\ $$$${A}\:\mathrm{adj}\:{A}\:=\:{k}\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{is} \\ $$
Answered by math1967 last updated on 01/Sep/18
![adjA= [((cosx),(sinx)),((−sinx),(cosx)) ]^t = [((cosx),(−sinx)),((sinx),(cosx)) ] ∴ [((cosx),(sinx)),((−sinx),(cosx)) ] [((cosx),(−sinx)),((sinx),(cosx)) ] [((cos^2 x+sin^2 x),0),(0,(sin^2 x+cos^2 x)) ] [(1,0),(0,1) ]=1. [(1,0),(0,1) ]∴k=1](https://www.tinkutara.com/question/Q42729.png)
$${adjA}=\begin{bmatrix}{{cosx}}&{{sinx}}\\{−{sinx}}&{{cosx}}\end{bmatrix}^{{t}} =\begin{bmatrix}{{cosx}}&{−{sinx}}\\{{sinx}}&{{cosx}}\end{bmatrix} \\ $$$$\therefore\begin{bmatrix}{{cosx}}&{{sinx}}\\{−{sinx}}&{{cosx}}\end{bmatrix}\begin{bmatrix}{{cosx}}&{−{sinx}}\\{{sinx}}&{{cosx}}\end{bmatrix} \\ $$$$\begin{bmatrix}{{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}}&{\mathrm{0}}\\{\mathrm{0}}&{{sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}=\mathrm{1}.\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\therefore{k}=\mathrm{1} \\ $$
Answered by RANGAM kumar last updated on 02/Sep/18
$$\mathrm{zza} \\ $$
Commented by math1967 last updated on 02/Sep/18
$$??????\:\:{what}\:{is}\:{wrong}? \\ $$