Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 10228 by j.masanja06@gmail.com last updated on 30/Jan/17

B is the matrix ⌈ 3 1 −3⌉ ∣ 1 2a 1 ∣ ⌊ 0 2 a ⌋ i.find the value of a for which B is singular ii.determine whether or not ⌈x ⌉ ⌈ −3.5⌉ B ∣y ∣ = ∣ 5.5∣ has the ⌊ z ⌋ ⌊ 5 ⌋ solution for each of the value of obtai ned in (i) above

$$\:\:\:\:\:\:\:\:\mathrm{B}\:\mathrm{is}\:\mathrm{the}\:\mathrm{matrix}\: \\ $$$$\:\:\:\:\:\:\:\lceil\:\:\mathrm{3}\:\:\:\:\:\:\mathrm{1}\:\:−\mathrm{3}\rceil \\ $$$$\:\:\:\:\:\:\:\mid\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{2a}\:\:\:\mathrm{1}\:\:\mid \\ $$$$\:\:\:\:\:\:\:\lfloor\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\mathrm{a}\:\rfloor \\ $$$$\:\mathrm{i}.\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{B}\:\mathrm{is}\: \\ $$$$\:\:\:\:\mathrm{singular} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{ii}.\mathrm{determine}\:\mathrm{whether}\:\mathrm{or}\:\mathrm{not}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\lceil\mathrm{x}\:\rceil\:\:\:\:\:\:\:\:\:\lceil\:−\mathrm{3}.\mathrm{5}\rceil \\ $$$$\:\:\:\mathrm{B}\:\:\:\:\:\:\:\:\mid\mathrm{y}\:\:\mid\:\:=\:\:\:\:\mid\:\:\:\:\:\:\mathrm{5}.\mathrm{5}\mid\:\:\mathrm{has}\:\mathrm{the}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\lfloor\:\mathrm{z}\:\:\rfloor\:\:\:\:\:\:\:\:\lfloor\:\:\:\:\:\:\:\mathrm{5}\:\:\rfloor \\ $$$$\mathrm{solution}\:\mathrm{for}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{obtai} \\ $$$$\mathrm{ned}\:\mathrm{in}\:\left(\mathrm{i}\right)\:\mathrm{above} \\ $$

Commented by prakash jain last updated on 31/Jan/17

determinant ((3,1,(−3)),(1,(2a),1),(0,2,a)) =3(2a^2 −2)−1(a+6) =6a^2 −6−a−6 ∣B∣=0 6a^2 −a−12=0 a=((1±(√(1−4(6)(−12))))/(12))=((1±17)/(12)) a=(3/2),−(4/3)

$$\begin{vmatrix}{\mathrm{3}}&{\mathrm{1}}&{−\mathrm{3}}\\{\mathrm{1}}&{\mathrm{2}{a}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}&{{a}}\end{vmatrix} \\ $$$$=\mathrm{3}\left(\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{1}\left({a}+\mathrm{6}\right) \\ $$$$=\mathrm{6}{a}^{\mathrm{2}} −\mathrm{6}−{a}−\mathrm{6} \\ $$$$\mid{B}\mid=\mathrm{0} \\ $$$$\mathrm{6}{a}^{\mathrm{2}} −{a}−\mathrm{12}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}\left(\mathrm{6}\right)\left(−\mathrm{12}\right)}}{\mathrm{12}}=\frac{\mathrm{1}\pm\mathrm{17}}{\mathrm{12}} \\ $$$${a}=\frac{\mathrm{3}}{\mathrm{2}},−\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Commented by j.masanja06@gmail.com last updated on 31/Jan/17

thanks sir ! how about loman (ii)?

$$\mathrm{thanks}\:\mathrm{sir}\:! \\ $$$$\mathrm{how}\:\mathrm{about}\:\mathrm{loman}\:\left(\mathrm{ii}\right)? \\ $$

Commented by j.masanja06@gmail.com last updated on 01/Feb/17

how about loman (ii) sir?

$$\mathrm{how}\:\mathrm{about}\:\mathrm{loman}\:\left(\mathrm{ii}\right)\:\mathrm{sir}? \\ $$

Commented by prakash jain last updated on 01/Feb/17

What does loman mean?

$$\mathrm{What}\:\mathrm{does}\:\mathrm{loman}\:\mathrm{mean}? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com