Question Number 1771 by hareem ali last updated on 19/Sep/15 $$\mathrm{2}{x}−{y}+\mathrm{2}{z}=\mathrm{4} \\ $$$${x}+\mathrm{10}{y}−\mathrm{3}{z}=\mathrm{10} \\ $$ Answered by 123456 last updated on 19/Sep/15 $$\begin{bmatrix}{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{10}}&{−\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}=\begin{bmatrix}{\mathrm{4}}\\{\mathrm{10}}\\{\mathrm{0}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{2}}&{\mathrm{4}}\\{\mathrm{1}}&{\mathrm{10}}&{−\mathrm{3}}&{\mathrm{10}}\end{bmatrix}…
Question Number 605 by magmarsenpai last updated on 10/Feb/15 $${Given}\:{a}\:{matrix}\:{A}\:\in\:\mathbb{M}_{{n}\:×\:{n}\:} \:\forall\:\:{k}\:\in\:\mathbb{N}\:{define}: \\ $$$${A}=\begin{cases}{\varnothing_{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:{A}=\varnothing\:\:{and}\:\:{k}\geqslant\mathrm{1}}\\{{I}_{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:{A}\neq\varnothing_{{n}} \:{and}\:{k}\neq\mathrm{0}}\\{{A}^{{k}−\mathrm{1}} {A}\:\:\:\:{if}\:{A}\neq\varnothing_{{n}} \:{and}\:{k}\geqslant\mathrm{1}}\end{cases} \\ $$$${Prove}\:{that}\:{A}^{{k}} {A}^{{r}} ={A}^{{k}+{r}} \:\forall\:{k},{r}\:\in\:\mathbb{N}. \\ $$…
Question Number 529 by 8905571695 last updated on 25/Jan/15 $$ \\ $$ Answered by prakash jain last updated on 25/Jan/15 $$\mathrm{blank} \\ $$ Terms of…
Question Number 131509 by snipers237 last updated on 05/Feb/21 $${Let}\:{a},{b},{c}\:\:{be}\:\:{no}\:{null}\:{integers}.\:{A}\:{ballot}\:{box}\:{contains}\:\:“{a}''\:{black}\:{bowls}\:{and}\:“{b}''{white}\:{bowls}. \\ $$$${After}\:{a}\:{print}\:{we}\:{put}\:{the}\:{bowl}\:{back}\:{in}\:{the}\:{ballot}\:{box}\:{with}\:“{c}''\:{another}\:{bowls}\:{of}\:{the}\:{same}\:{color}. \\ $$$${Prove}\:{that}\:{the}\:{probability}\:{to}\:{extract}\:{a}\:{green}\:{bowl}\:{at}\:{any}\:\:\:{print}\:{is}\:{always} \\ $$$${p}\:=\:\frac{{a}}{{a}+{b}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com