જો $a \ne b \ne c,$ તો સમીકરણ $\left| {\,\begin{array}{*{20}{c}}0&{x - a}&{x - b}\\{x + a}&0&{x - c}\\{x + b}&{x + c}&0\end{array}\,} \right| = 0$ નું સમાધાન કરે તેવી $x$ ની કિમત મેળવો.
A$x = 0$
B$x = a$
C$x = b$
D$x = c$
Easy
Download our app for free and get started
A$x = 0$
Obviously, on putting $x = 0$, we observe that the determinant becomes ${\Delta _{x = 0}} = \left| {\,\begin{array}{*{20}{c}}0&{ - a}&{ - b}\\a&0&{ - c}\\b&c&0\end{array}\,} \right| = a(bc) - b(ac) = 0$
$\therefore $ $x = 0$ is a root of the given equation.
Aliter : Expanding $\Delta $, we get
$\Delta \equiv - (x - a)\,[ - (x + b)(x - c)] + (x - b)\,[(x + a)\,(x + c)] = 0$
$ \Rightarrow \ 2{x^3} - (2\,\Sigma ab)x = 0$
$ \Rightarrow $ Either $x = 0$ or ${x^2} = \sum {ab} \ ($i.e., $x = \pm \sum {ab} )$
Again $x = 0$ satisfies the given equation.
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
જો $A=\left(\begin{array}{cc}0 & \sin \alpha \\ \sin \alpha & 0\end{array}\right)$ અને $\operatorname{det}\left(A^{2}-\frac{1}{2} I\right)=0,$ હોય તો $\alpha$ ની શક્ય કિમંત મેળવો.
ધારો કે $A$ અને $B$ એ કક્ષા $3$ ના એવા બે ચોરસ શ્રેણિકો છે કે જેથી $|A|=3$ અને $|B|=2$. તો $\left|\mathrm{A}^{\mathrm{T}} \mathrm{A}(\operatorname{adj}(2 \mathrm{~A}))^{-1}(\operatorname{adj}(4 \mathrm{~B}))(\operatorname{adj}(\mathrm{AB}))^{-1} \mathrm{AA}^{\mathrm{T}}\right|$= ..............