$\left| {\begin{array}{ccc} 1 & 1+ac & 1+bc \\ 1 & 1+ad & 1+bd \\ 1 & 1+ae & 1+be \end{array}} \right| = $

If $A=\left|\begin{array}{ccc}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|$ and $B=\left|\begin{array}{ccc}c_{1} & c_{2} & c_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|$,then

$\left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}} \right| = $

The determinant $\left| {\begin{array}{*{20}{c}}{{b_1} + {c_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\\{{b_2} + {c_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\\{{b_3} + {c_3}}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} \right|$ is equal to:

The product of a matrix and its transpose is an identity matrix. The value of the determinant of this matrix is:

Let $a, b, c$ be such that $(b+c) \neq 0$ and $\left|\begin{array}{ccc} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{array}\right|+\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n-1} b & (-1)^n c \end{array}\right|=0$. Then the value of $n$ is