Menu Close

Category: Matrices and Determinants

A-1-2-2-5-B-3-1-4-2-determine-A-B-

Question Number 2564 by Rasheed Soomro last updated on 22/Nov/15 $${A}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{2}}&{\mathrm{5}}\end{bmatrix},{B}=\begin{bmatrix}{\mathrm{3}}&{−\mathrm{1}}\\{\mathrm{4}}&{+\mathrm{2}}\end{bmatrix} \\ $$$${determine}\:{A}^{{B}} \\ $$ Answered by Yozzi last updated on 22/Nov/15 $$\mid{A}\mid=\mathrm{5}−\mathrm{4}=\mathrm{1}\neq\mathrm{0}\:\Rightarrow{A}\:{is}\:{non}−{singular}. \\ $$$${A}^{{B}}…

If-A-and-B-are-two-matrices-of-suitable-order-does-there-exist-definition-for-A-B-

Question Number 2546 by Rasheed Soomro last updated on 22/Nov/15 $$\boldsymbol{\mathrm{If}}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{two}\:\mathrm{matrices}\:\mathrm{of}\:\mathrm{suitable}\:\mathrm{order} \\ $$$$\mathrm{does}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{definition}\:\mathrm{for}\:\mathrm{A}^{\mathrm{B}} \:? \\ $$ Commented by Yozzi last updated on 22/Nov/15 $${A}^{{B}} ={e}^{\left({lnA}\right){B}}…

2x-y-2z-4-x-10y-3z-10-

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}…

Given-a-matrix-A-M-n-n-k-N-define-A-n-if-A-and-k-1-I-n-if-A-n-and-k-0-A-k-1-A-if-A-n-and-k-1-Prove-that-A-k-A-r-A-k-r-

Question Number 605 by magmarsenpai last updated on 10/Feb/15 $${Given}\:{a}\:{matrix}\:{A}\:\in\:\mathbb{M}_{{n}\:×\:{n}\:} \:\forall\:\:{k}\:\in\:\mathbb{N}\:{define}: \\ $$$${A}=\begin{cases}{\varnothing_{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:{A}=\varnothing\:\:{and}\:\:{k}\geqslant\mathrm{1}}\\{{I}_{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:{A}\neq\varnothing_{{n}} \:{and}\:{k}\neq\mathrm{0}}\\{{A}^{{k}−\mathrm{1}} {A}\:\:\:\:{if}\:{A}\neq\varnothing_{{n}} \:{and}\:{k}\geqslant\mathrm{1}}\end{cases} \\ $$$${Prove}\:{that}\:{A}^{{k}} {A}^{{r}} ={A}^{{k}+{r}} \:\forall\:{k},{r}\:\in\:\mathbb{N}. \\ $$…

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-th

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