નિશ્ચાયકની કિમત મેળવો  : $\left|\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|$

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Let $A=\left[\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right]$

By expanding along the first column, we have:

$|A|=2\left|\begin{array}{cc}2 & -1 \\ -5 & 0\end{array}\right|-0\left|\begin{array}{cc}-1 & -2 \\ -5 & 0\end{array}\right|+3\left|\begin{array}{cc}-1 & -2 \\ 2 & -1\end{array}\right|$

$=2(0-5)-0+3(1+4)$

$=-10+15=5$

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જો $-9 $ એ સમીકરણ $\left| {\,\begin{array}{*{20}{c}}x&3&7\\2&x&2\\7&6&x\end{array}\,} \right| = 0$ નું બીજ હોય તો બાકી ના બે બીજ મેળવો.

  • [IIT 1983]

જો $A = \left| {\,\begin{array}{*{20}{c}}{ - 1}&2&4\\3&1&0\\{ - 2}&4&2\end{array}\,} \right|$અને $B = \left| {\,\begin{array}{*{20}{c}}{ - 2}&4&2\\6&2&0\\{ - 2}&4&8\end{array}\,} \right|$, તો $B =$

જો $\omega $ એ એકનું ઘનમૂળ હોય તો સમીકરણ $\left| {\begin{array}{*{20}{c}}
  {x + 2}&\omega &{{\omega ^2}} \\ 
  \omega &{x + 1 + {\omega ^2}}&1 \\ 
  {{\omega ^2}}&1&{x + 1 + \omega } 
\end{array}} \right| = 0$ નું બીજ મેળવો.

ધારો કે $D = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ અને $D' = \left| {\,\begin{array}{*{20}{c}}{{a_1} + p{b_1}}&{{b_1} + q{c_1}}&{{c_1} + r{a_1}}\\{{a_2} + p{b_2}}&{{b_2} + q{c_2}}&{{c_2} + r{a_2}}\\{{a_3} + p{b_3}}&{{b_3} + q{c_3}}&{{c_3} + r{a_3}}\end{array}\,} \right|$, તો . . .

સમીકરણની સંહતિ ${x_1} + 2{x_2} + 3{x_3} = a2{x_1} + 3{x_2} + {x_3} = $ $b3{x_1} + {x_2} + 2{x_3} = c$ ને . . . ઉકેલ છે.