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Matrices and DeterminantsQuestion and Answers: Page 1

Question Number 213435 Answers: 0 Comments: 0

A ∈ M_(2×2) ,and ,det (A)≠0 : A^3 = A^2 + A ⇒ det ( A −2I )=?

$$ \\ $$$$\:\:\:\:\:\:\:{A}\:\in\:\mathrm{M}_{\mathrm{2}×\mathrm{2}} \:\:,{and}\:,{det}\:\left({A}\right)\neq\mathrm{0}\::\:\:\:{A}^{\mathrm{3}} \:=\:{A}^{\mathrm{2}} \:+\:{A} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:\:{det}\:\left(\:{A}\:−\mathrm{2}{I}\:\right)=? \\ $$$$\:\:\:\:\:\: \\ $$

Question Number 212844 Answers: 2 Comments: 1

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$$\:\:\:\cancel{\downharpoonleft}\underline{\:} \\ $$

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Question Number 205586 Answers: 1 Comments: 0

Givem that the matrix A = ((3,1,5),(2,3,5),(5,1,6) ). If Adj. A = (((13),(-1),(-10)),((13),(-7),(-5)),((-13),2,7) ) (i) find A^(−1) (ii) Use the result in (i) to find the values of x, y and z that will satisfy the equations: 3x + y + 5z = 8 2x +3y + 5z = 0 5x + y + 6z = 13

$${Givem}\:{that}\:{the}\:{matrix}\:{A}\:=\:\begin{pmatrix}{\mathrm{3}}&{\mathrm{1}}&{\mathrm{5}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{5}}\\{\mathrm{5}}&{\mathrm{1}}&{\mathrm{6}}\end{pmatrix}.\: \\ $$$${If}\:{Adj}.\:{A}\:=\:\begin{pmatrix}{\mathrm{13}}&{-\mathrm{1}}&{-\mathrm{10}}\\{\mathrm{13}}&{-\mathrm{7}}&{-\mathrm{5}}\\{-\mathrm{13}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix} \\ $$$$\left({i}\right)\:{find}\:{A}^{−\mathrm{1}} \\ $$$$\left({ii}\right)\:{Use}\:{the}\:{result}\:{in}\:\left({i}\right)\:{to}\:{find}\:{the} \\ $$$${values}\:{of}\:{x},\:{y}\:{and}\:{z}\:{that}\:{will}\:{satisfy}\:{the} \\ $$$${equations}: \\ $$$$\mathrm{3}{x}\:+\:{y}\:+\:\mathrm{5}{z}\:=\:\mathrm{8} \\ $$$$\mathrm{2}{x}\:+\mathrm{3}{y}\:+\:\mathrm{5}{z}\:=\:\mathrm{0} \\ $$$$\mathrm{5}{x}\:+\:{y}\:+\:\mathrm{6}{z}\:=\:\mathrm{13} \\ $$

Question Number 205306 Answers: 1 Comments: 0

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Question Number 202371 Answers: 2 Comments: 0

If A ∈ M_(2×2) , det(A )≠ 0 , A^( 3) = A^2 +A ⇒ Find the values of det (2A −I )

$$ \\ $$$$\:\:\:{If}\:\:\:\:{A}\:\in\:{M}_{\mathrm{2}×\mathrm{2}} \:,\:{det}\left({A}\:\right)\neq\:\mathrm{0} \\ $$$$\:\:\:\:,\:\:{A}^{\:\mathrm{3}} \:=\:{A}^{\mathrm{2}} \:+{A}\:\Rightarrow\:{Find}\:{the}\: \\ $$$$\:\:\:\:{values}\:{of}\:\:\:{det}\:\left(\mathrm{2}{A}\:−{I}\:\right) \\ $$$$ \\ $$

Question Number 199847 Answers: 0 Comments: 6

Find all possible value (a/(a+b+d )) +(b/(a+b+c)) + (c/(b+c+d))+(d/(a+c+d)) when a,b,c,d vary over positive reals

$$\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{value}\: \\ $$$$\:\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}+\mathrm{d}\:}\:+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:+\:\frac{\mathrm{c}}{\mathrm{b}+\mathrm{c}+\mathrm{d}}+\frac{\mathrm{d}}{\mathrm{a}+\mathrm{c}+\mathrm{d}}\: \\ $$$$\:\mathrm{when}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{vary}\:\mathrm{over}\:\mathrm{positive} \\ $$$$\:\mathrm{reals}\: \\ $$

Question Number 199164 Answers: 0 Comments: 0

Question Number 199155 Answers: 0 Comments: 1

$$\:\underbrace{ } \\ $$

Question Number 199001 Answers: 1 Comments: 0

If ,A ∈ M_(n×n) , A^( 2) = A ,1≠ k ∈R. Find ( I − kA )^( −1) = ?

$$ \\ $$$$\:\mathrm{I}{f}\:,\mathrm{A}\:\in\:\mathrm{M}_{{n}×{n}} \:\:\:,\:\:\mathrm{A}^{\:\mathrm{2}} \:=\:\mathrm{A}\:,\mathrm{1}\neq\:{k}\:\in\mathbb{R}. \\ $$$$\:\:\:\mathrm{F}{ind}\:\:\:\:\left(\:\:\:\mathrm{I}\:−\:{k}\mathrm{A}\:\right)^{\:−\mathrm{1}} \:=\:?\: \\ $$

Question Number 198743 Answers: 1 Comments: 0

Question Number 198156 Answers: 1 Comments: 0

Prove that ((2t−1)/(lnt−ln(1−t)))=∫^( 1) _( 0) t^x (1−t)^(1−x) dx and ∫^( 1) _( 0) ((2t−1)/(lnt−ln(1−t)))dt = (π/2)∫^( 1) _( 0) ((x(1−x))/(sin(πx)))dx

$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\frac{\mathrm{2t}−\mathrm{1}}{\mathrm{lnt}−\mathrm{ln}\left(\mathrm{1}−\mathrm{t}\right)}=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \mathrm{t}^{\mathrm{x}} \left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{1}−\mathrm{x}} \mathrm{dx} \\ $$$$\mathrm{and}\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\mathrm{2t}−\mathrm{1}}{\mathrm{lnt}−\mathrm{ln}\left(\mathrm{1}−\mathrm{t}\right)}\mathrm{dt}\:\:=\:\:\frac{\pi}{\mathrm{2}}\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{sin}\left(\pi\mathrm{x}\right)}\mathrm{dx} \\ $$

Question Number 195203 Answers: 1 Comments: 0

Calculer ∫^( +∞) _( 0) (dt/((e^t −e^(−t) )^2 +a^2 ))

$$\mathrm{Calculer}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \frac{\mathrm{dt}}{\left(\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} \right)^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} } \\ $$

Question Number 195185 Answers: 0 Comments: 0

1/ Montrer que ∫^( +∞) _( 0) (((1−x^2 )^(2p−1) )/(1−x^(4p) ))dx=((2^(2p−3) /p))π[1+2Σ_(k=1) ^(p−1) cos^(2p−1) (((kπ)/(2p)))] 2/ En de^ duire ∫^( 1) _( 0) (((1−x^2 )^(2p−1) )/(1−x^(4p) ))dx

$$\mathrm{1}/\:\:\mathrm{Montrer}\:\mathrm{que}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \:\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}{p}−\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{4}{p}} }{dx}=\left(\frac{\mathrm{2}^{\mathrm{2}{p}−\mathrm{3}} }{{p}}\right)\pi\left[\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{p}−\mathrm{1}} {\sum}}{cos}^{\mathrm{2}{p}−\mathrm{1}} \left(\frac{{k}\pi}{\mathrm{2}{p}}\right)\right] \\ $$$$\mathrm{2}/\:\:\:\:\mathrm{En}\:\mathrm{d}\acute {\mathrm{e}duire}\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}{p}−\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{4}{p}} }{dx} \\ $$

Question Number 194915 Answers: 2 Comments: 4

(3/(x−3))+(5/(x−5))+(7/(x−17))+((19)/(x−19))=x^2 −11x−4

$$ \\ $$$$\frac{\mathrm{3}}{{x}−\mathrm{3}}+\frac{\mathrm{5}}{{x}−\mathrm{5}}+\frac{\mathrm{7}}{{x}−\mathrm{17}}+\frac{\mathrm{19}}{{x}−\mathrm{19}}={x}^{\mathrm{2}} −\mathrm{11}{x}−\mathrm{4} \\ $$

Question Number 194642 Answers: 2 Comments: 0

If A= (((a b c)),((b c a)),((c a b)) ) and a,b,c >0 such that abc=1 and A^T .A=I find a^3 +b^3 +c^3 −3abc .

$$\:\mathrm{If}\:\mathrm{A}=\begin{pmatrix}{\mathrm{a}\:\:\:\:\mathrm{b}\:\:\:\:\:\:\mathrm{c}}\\{\mathrm{b}\:\:\:\:\mathrm{c}\:\:\:\:\:\:\mathrm{a}}\\{\mathrm{c}\:\:\:\:\:\mathrm{a}\:\:\:\:\:\:\mathrm{b}}\end{pmatrix}\:\mathrm{and}\:\mathrm{a},\mathrm{b},\mathrm{c}\:>\mathrm{0} \\ $$$$\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{abc}=\mathrm{1}\:\mathrm{and}\:\mathrm{A}^{\mathrm{T}} .\mathrm{A}=\mathrm{I} \\ $$$$\:\mathrm{find}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} −\mathrm{3abc}\:. \\ $$

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Question Number 190266 Answers: 1 Comments: 0

a ball is thrown vertically upward from a point 0.5m above the ground with speed u = 7m/s find the height reached above ground g = 10m/s^2

$${a}\:{ball}\:{is}\:{thrown}\:{vertically}\:{upward} \\ $$$${from}\:{a}\:{point}\:\mathrm{0}.\mathrm{5}{m}\:{above}\:{the}\:{ground}\:{with} \\ $$$${speed}\:{u}\:=\:\mathrm{7}{m}/{s} \\ $$$${find}\:{the}\:{height}\:{reached}\:{above}\:{ground} \\ $$$${g}\:=\:\mathrm{10}{m}/{s}^{\mathrm{2}} \\ $$

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