Short Quiz

The determinant \(\left|\begin{array}{lll}b^2-a b & b-c & b c-a c \\ a b-a^2 & a-b & b^2-a b \\ b c-a c & c-a & a b-a^2\end{array}\right|\) equals

Let \(\mathrm{A}=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]\) be a square matrix such that adj \(\mathrm{A}=\mathrm{A}\). Then, \((\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})\) is equal to :

Let \(\Delta=\left|\begin{array}{ccc}\mathrm{A} x & x^2 & 1 \\ \mathrm{~B} y & y^2 & 1 \\ \mathrm{C} z & z^2 & 1\end{array}\right|\) and \(\Delta_1=\left|\begin{array}{ccc}\mathrm{A} & \mathrm{B} & \mathrm{C} \\ x & y & z \\ z y & z x & x y\end{array}\right|\), then

If \(\mathrm{A}(\operatorname{adj} \mathrm{A})=\left[\begin{array}{ccc}2026 & 0 & 0 \\ 0 & 2026 & 0 \\ 0 & 0 & 2026\end{array}\right]\), then the value of \(|\operatorname{adj} \mathrm{A}|\) is equal to :