Question Number 44612 by rahul 19 last updated on 02/Oct/18
$${Prove}\:{that}\:\mathrm{One}\:\mathrm{factor}\:\mathrm{of}\begin{vmatrix}{{a}^{\mathrm{2}} +{x}}&{\:\:{ab}}&{\:\:{ac}}\\{\:\:{ab}}&{{b}^{\mathrm{2}} +{x}}&{\:\:{cb}}\\{\:\:{ca}}&{\:\:{cb}}&{{c}^{\mathrm{2}} +{x}}\end{vmatrix}\:\mathrm{is}\:{x}^{\mathrm{2}} . \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18
Commented by $@ty@m last updated on 02/Oct/18
$${it}\:{is}\:{correct}. \\ $$$${but}\:{it}\:{is}\:{expected}\:{to}\:{use}\:{properties} \\ $$$${of}\:{determinant}\:{to}\:{solve}\:{problems} \\ $$$${of}\:{determinant}\:{chapter}. \\ $$
Commented by rahul 19 last updated on 02/Oct/18
$${Yes}\:{sir},\:{i}\:{want}\:{this}\:{Q}.\:{to}\:{be}\:{solved}\:{by} \\ $$$${properties}. \\ $$$$\left({Although}\:{thanks}\:{for}\:{this}\:{also}\right). \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18
$${let}\:\bigtriangleup={x}^{\mathrm{2}} {f}\left({a},{b},{c}\right) \\ $$$${now}\:{using}\:{factor}\:{theorem} \\ $$$${p}\left({x}\right)=\left({x}−\alpha\right){q}\left({x}\right) \\ $$$${p}\left(\alpha\right)=\mathrm{0}\:\:{that}\:{means}\:{x}−\alpha\:{is}\:{s}\:{factor}\:{of}\:{p}\left({x}\right) \\ $$$${using}\:{same}\:{principle}\:{we}\:{put}\:{x}=\mathrm{0}\:{in}\:\bigtriangleup \\ $$$${nxt}\:{take}\:{common}\:{from}\:\mathrm{1}{st}\:{row}\: \\ $$$${b}\:{from}\:{second}\:{row} \\ $$$${c}\:{from}\:{third}\:{row}.. \\ $$$${it}\:{is}\:{clear}\:{that}\:{three}\:{row}\:{identical}\: \\ $$$${so}\:\bigtriangleup=\mathrm{0} \\ $$$${hence}\:{proved}… \\ $$
Answered by math1967 last updated on 02/Oct/18
$$\:\:{abc}\:\begin{vmatrix}{{a}+\frac{{x}}{{a}}}&{{b}}&{{c}}\\{{a}}&{{b}+\frac{{x}}{{b}}}&{{c}}\\{{a}}&{{b}}&{{c}+\frac{{x}}{{c}\:}\:}\end{vmatrix} \\ $$$$={abc}\begin{vmatrix}{\frac{{x}}{{a}}}&{−\frac{{x}}{{b}}\:\:}&{\mathrm{0}}\\{\mathrm{0}}&{\frac{{x}}{{b}}}&{−\frac{{x}}{{c}}}\\{{a}}&{{b}}&{{c}+\frac{{x}}{{c}}}\end{vmatrix}{R}_{\mathrm{1}} '\rightarrow{R}_{\mathrm{1}} −{R}_{\mathrm{2}} \:\:,{R}_{\mathrm{2}\:} '\rightarrow{R}_{\mathrm{2}\:} −{R}_{\mathrm{3}} \\ $$$$\:\:\:\:\:{abcx}^{\mathrm{2}} \begin{vmatrix}{\frac{\mathrm{1}}{{a}}}&{−\frac{\mathrm{1}}{{b}}}&{\mathrm{0}}\\{\mathrm{0}}&{\frac{\mathrm{1}}{{b}}}&{−\frac{\mathrm{1}}{{c}}}\\{{a}}&{{b}}&{{c}+\frac{{x}}{{c}}}\end{vmatrix} \\ $$$$\frac{{abcx}^{\mathrm{2}} }{{ab}}\begin{vmatrix}{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{−\frac{\mathrm{1}}{{c}}}\\{{a}^{\mathrm{2}} }&{{b}^{\mathrm{2}} }&{{c}+\frac{{x}}{{c}}}\end{vmatrix} \\ $$$${cx}^{\mathrm{2}} \begin{vmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{−\frac{\mathrm{1}}{{c}}}\\{{a}^{\mathrm{2}} }&{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }&{{c}+\frac{{x}}{{c}}}\end{vmatrix}{C}_{\mathrm{2}} '\rightarrow{C}_{\mathrm{1}} +{C}_{\mathrm{2}\:} \\ $$$${cx}^{\mathrm{2}} \left\{\mathrm{1}×\left({c}+\frac{{x}}{{c}}\right)+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{c}}\right\} \\ $$$$={x}^{\mathrm{2}} \left({x}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\therefore{x}^{\mathrm{2}} \:{is}\:{factor}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by rahul 19 last updated on 02/Oct/18
thanks sir ����.