Left| {,begin{array}{*{20}{c}}{a - b - c}&{2a}&{2a}{2b}&{b - c

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

hard

$\left| {\,\begin{array}{*{20}{c}}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}\,} \right| = $

${(a + b + c)^2}$

${(a + b + c)^3}$

$(a + b + c)(ab + bc + ca)$

એકપણ નહી.

Solution

(b) $\left| {\,\begin{array}{*{20}{c}}{a – b – c}&{2a}&{2a}\\{2b}&{b – c – a}&{2b}\\{2c}&{2c}&{c – a – b}\end{array}\,} \right|$

= $\left| {\,\begin{array}{*{20}{c}}{ – \Sigma a}&0&{2a}\\{\Sigma a}&{ – \Sigma a}&{2b}\\0&{\Sigma a}&{c – a – b}\end{array}\,} \right|$ , $\left( \begin{array}{l}{C_1} \to {C_1} – {C_2}\\{C_2} \to {C_2} – {C_3}\end{array} \right)$

= ${(\Sigma a)^2}\,\left| {\,\begin{array}{*{20}{c}}{ – 1}&0&{2a}\\1&{ – 1}&{2b}\\1&1&{c – a – b}\end{array}\,} \right| = {(\Sigma a)^3}$, (on expansion)

= ${(a + b + c)^3}$.

Standard 12

Mathematics

ધારો કે જો પૂર્ણાકો $a, b \in[-3,3]$ એવાં છે કે જેથી $a+b \neq 0$. તો શક્ય તમામ એવી જોડ $(a, b)$ ની સંખ્યા શોધો, કે જેના માટે $\left|\frac{z-a}{z+b}\right|=1$ અને $\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|$ $=1, z \in C$, જ્યાં $\omega$ અને $\omega^2$ એ $x^2+x+$ $1=0$, નાં બીજ છે.

hard

(JEE MAIN-2025)

નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^{3}$

hard

ધારો કે $a-2 b+c=1$ છે . જો $f(x)=\left|\begin{array}{lll}{x+a} & {x+2} & {x+1} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|,$ હોય તો . . .

hard

(JEE MAIN-2020)

જો $f(x) = \left| {\begin{array}{*{20}{c}}1&x&{x + 1}\\{2x}&{x(x – 1)}&{(x + 1)x}\\{3x(x – 1)}&{x(x – 1)(x – 2)}&{(x + 1)x(x – 1)}\end{array}} \right|$ તો $f(100)$ મેળવો.

hard

(IIT-1999)

$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ – 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$

તો $f(x)$ ની મહતમ કિમંત મેળવો.

hard

(JEE MAIN-2021)

$\left| {\,\begin{array}{*{20}{c}}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}\,} \right| = $

Solution

Similar Questions

નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^{3}$

ધારો કે $a-2 b+c=1$ છે . જો $f(x)=\left|\begin{array}{lll}{x+a} & {x+2} & {x+1} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|,$ હોય તો . . .

જો $f(x) = \left| {\begin{array}{*{20}{c}}1&x&{x + 1}\\{2x}&{x(x – 1)}&{(x + 1)x}\\{3x(x – 1)}&{x(x – 1)(x – 2)}&{(x + 1)x(x – 1)}\end{array}} \right|$ તો $f(100)$ મેળવો.

$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ – 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$ તો $f(x)$ ની મહતમ કિમંત મેળવો.

Start a Free Trial Now

$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ – 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$

તો $f(x)$ ની મહતમ કિમંત મેળવો.