Let $\left| {\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}} \right| = x + iy$,then

One of the roots of the given equation $\left| \begin{array}{ccc} x+a & b & c \\ b & x+c & a \\ c & a & x+b \end{array} \right| = 0$ is

In a square matrix $A$ of order $3$,$a_{ii}$ are the sum of the roots of the equation $x^2 - (a + b)x + ab = 0$; $a_{i, i+1}$ are the product of the roots,$a_{i, i-1}$ are all unity,and the rest of the elements are all zero. The value of the determinant of $A$ is equal to

$A$ determinant is chosen at random from the set of all determinants of order $2 \times 2$ with elements $0$ or $1$ only. The probability that the determinant chosen is non-zero is .........

If ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$ such that $\left| {\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}} \right| = \lambda $,then the value of $\lambda $ is

The value of the determinant $\left| \begin{array}{ccc} 1 & \cos(\alpha - \beta) & \cos \alpha \\ \cos(\alpha - \beta) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \end{array} \right|$ is