Category: Matrices and Determinants

Question Number 78670 by berket last updated on 19/Jan/20 $${let}\:{A}=\begin{bmatrix}{{a}\:\:{b}}\\{{c}\:\:{d}}\end{bmatrix}{use}\:{the}\:{augmented}\:{matrix}\left[{A}\:{I}\right]\:{and}\:{elementary}\:{row}\:{operation}\:{to}\:{show}\:{A}^{−\mathrm{1}} =\:\frac{\mathrm{1}}{{ad}\:{bc}}\begin{bmatrix}{{a}\:\:{b}}\\{{c}\:\:\:{d}}\end{bmatrix}{and}\:{show}\:{that}\:{det}\left({A}^{−\mathrm{1}} \right)=\frac{\mathrm{1}}{{det}\left({A}\right)} \\ $$$$ \\ $$ Commented by abdomathmax last updated on 20/Jan/20 $${Pc}\left({A}\right)={det}\left({A}−{xI}\right)\:=\begin{vmatrix}{{a}−{x}\:\:\:\:\:\:\:{b}}\\{{c}\:\:\:\:\:\:\:\:\:\:{d}−{x}}\end{vmatrix} \\…

Question Number 346 by 123456 last updated on 25/Jan/15 $$\Gamma\left(\theta\right)=\begin{bmatrix}{\mathrm{cos}\:\theta}&{\mathrm{sin}\:\theta}\\{−\mathrm{sin}\:\theta}&{\mathrm{cos}\:\theta}\end{bmatrix} \\ $$$$\Lambda\left(\theta,{t}\right)=\begin{bmatrix}{\mathrm{cos}\:\theta}&{\mathrm{sinh}\:{t}\:\mathrm{sin}\:\theta}\\{\mathrm{sin}\:\theta}&{\mathrm{cosh}\:{t}\:\mathrm{cos}\:\theta}\end{bmatrix} \\ $$$$\zeta\left(\theta,{t}\right)=\Gamma\left(\theta\right)×\Lambda\left(\theta,{t}\right)+\Lambda\left(\theta,{t}\right)×\Gamma\left(\theta\right) \\ $$$$\zeta\left(\theta,\mathrm{0}\right)=? \\ $$$$\mathrm{det}\:\zeta\left(\theta,\mathrm{0}\right)=? \\ $$ Answered by prakash jain last…

Question Number 15 by user1 last updated on 25/Jan/15 $$\mathrm{If}\:{A}=\begin{bmatrix}{\:\:\mathrm{3}}&{−\mathrm{5}}\\{−\mathrm{4}}&{\:\:\:\mathrm{2}}\end{bmatrix}, \\ $$$$\mathrm{show}\:\mathrm{that}\:{A}^{\mathrm{2}} −\mathrm{5}{A}−\mathrm{14}{I}=\mathrm{0} \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{We}\:\mathrm{have}: \\ $$$${A}^{\mathrm{2}}…

Question Number 13 by user1 last updated on 25/Jan/15 $$\mathrm{Expand}\:\mathrm{the}\:\mathrm{determnent}\: \\ $$$$\:\:\bigtriangleup=\begin{vmatrix}{{a}}&{{h}}&{{g}}\\{{h}}&{{b}}&{{f}}\\{{g}}&{{f}}&{{c}}\end{vmatrix} \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{Expanding}\:\mathrm{by}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{row},\:\mathrm{we}\:\mathrm{have}: \\ $$$$\bigtriangleup={a}\centerdot\begin{vmatrix}{{b}}&{{f}}\\{{f}}&{{c}}\end{vmatrix}−{h}\centerdot\begin{vmatrix}{{h}}&{{f}}\\{{g}}&{{c}}\end{vmatrix}+{g}\centerdot\begin{vmatrix}{{h}}&{{b}}\\{{g}}&{{f}}\end{vmatrix}…

Question Number 9 by user1 last updated on 25/Jan/15 $$\mathrm{Let}\:{A}=\begin{bmatrix}{\:\:\:\mathrm{0}}&{−\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}}\\{\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}}&{\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{bmatrix}\:\mathrm{and}\:{I}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{identity}\:\mathrm{matrix}\:\mathrm{of}\:\mathrm{order}\:\mathrm{2}.\:\mathrm{Show}\:\mathrm{that}\: \\ $$$$\left({I}+{A}\right)=\left({I}−{A}\right)\centerdot\begin{bmatrix}{\mathrm{cos}\:{x}}&{−\mathrm{sin}\:{x}}\\{\mathrm{sin}\:{x}}&{\:\:\:\:\mathrm{cos}\:{x}}\end{bmatrix}. \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{Let}\:\:\:\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}={t} \\…