નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરી અને વિસ્તરણ કર્યા સિવાય સાબિત કરો : $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$

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$\left|\begin{array}{ccc}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=\left|\begin{array}{ccc}x & a & x \\ y & b & y \\ z & c & z\end{array}\right|+\left|\begin{array}{ccc}x & a & a \\ y & b & b \\ z & c & c\end{array}\right|$

Clearly, the two determinants have two identical columns. Thus,

$=0+0=0$

Similar Questions

$\left| {\,\begin{array}{*{20}{c}}{31}&{37}&{92}\\{31}&{58}&{71}\\{31}&{105}&{24}\end{array}\,} \right|$ = . . ..

જો $\Delta = \left| {\,\begin{array}{*{20}{c}}a&b&c\\x&y&z\\p&q&r\end{array}\,} \right|$, તો $\left| {\,\begin{array}{*{20}{c}}{ka}&{kb}&{kc}\\{kx}&{ky}&{kz}\\{kp}&{kq}&{kr}\end{array}\,} \right|$=

જો $a+x=b+y=c+z+1,$ જ્યાં $a, b, c, x, y, z$ એ શૂન્યેતર ભિન્ન વાસ્તવિક સંખ્યાઓ હોય તો $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ ની કિમત શોધો 

  • [JEE MAIN 2020]

જો $\left| {\begin{array}{*{20}{c}}   {a - b}&{b - c}&{c - a} \\    {b - c}&{c - a}&{a - b} \\    {c - a + 1}&{a - b}&{b - c}  \end{array}} \right| = 0$ ,$\left( {a,b,c \in R - \left\{ 0 \right\}} \right),$ 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.

  • [IIT 1999]