${C_2} \to {C_2} - {C_1},{C_3} \to {C_3} - {C_1}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&0&0\\
2&{b - 2}&{c - 2}\\
4&{{b^2} - 4}&{{c^2} - 4}
\end{array}} \right|$
$ = \left( {b - 2} \right)\left( {c - 2} \right)\left| {\begin{array}{*{20}{c}}
1&1\\
{b + 2}&{c + 2}
\end{array}} \right|$
$\left| A \right| = \left( {b - 2} \right)\left( {c - 2} \right)\left( {c - b} \right)$
$2,b,c$ are in $AP \Rightarrow 2,2 + d,2 + 2d$
$ \Rightarrow \left| A \right| = \left( d \right)\left( {2d} \right)\left( d \right) = 2{d^3} \in \left[ {2,16} \right]$
$ \Rightarrow {d^3} \in \left[ {1,8} \right]$
$ \Rightarrow 2d \in \left[ {2,4} \right]$
$ \Rightarrow 2 + 2d \in \left[ {4,6} \right]$
$f (\theta)=\left|\begin{array}{ccc}-\sin ^{2} \theta & -1-\sin ^{2} \theta & 1 \\ -\cos ^{2} \theta & -1-\cos ^{2} \theta & 1 \\ 12 & 10 & -2\end{array}\right|$ ની ન્યૂનતમ અને મહત્તમ કિમતો અનુક્રમે $m$ અને $M$ હોય તો $( m , M )$ ની કિમત શોધો
$A = \left[ {\begin{array}{*{20}{c}}
{{{10}^{30}} + 5}&{{{10}^{20}} + 4}&{{{10}^{20}} + 6}\\
{{{10}^4} + 2}&{{{10}^8} + 7}&{{{10}^{10}} + 2n}\\
{{{10}^4} + 8}&{{{10}^6} + 4}&{{{10}^{15}} + 9}
\end{array}} \right]$ , $n \in N$, હોય તો . ..