The number of $3 \times 2$ matrices $A$, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^{T}A$ is $5$, is . . . . . . .

Let $A$ be a non-zero periodic matrix with period $4$ and $A^{12} + B = I$,where $I$ is the identity matrix and $B$ is any square matrix of the same order as $A$. The matrix product $AB$ is equal to:

Suppose $a_1, a_2, \dots$ are real numbers,with $a_1 \neq 0$. If $a_1, a_2, a_3, \dots$ are in $A.P.$,then:

If $\Delta _1 = \left| \begin{array}{ccc} b^5c^6(c^3 - b^3) & a^4c^6(a^3 - c^3) & a^4b^5(b^3 - a^3) \\ b^2c^3(b^6 - c^6) & ac^3(c^6 - a^6) & ab^2(a^6 - b^6) \\ b^2c^3(c^3 - b^3) & ac^3(a^3 - c^3) & ab^2(b^3 - a^3) \end{array} \right|$ and $\Delta _2 = \left| \begin{array}{ccc} a & b^2 & c^3 \\ a^4 & b^5 & c^6 \\ a^7 & b^8 & c^9 \end{array} \right|$,then $\Delta _1 \Delta _2$ is equal to

Let $A$ denote the matrix $\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$,where $i^2=-1$,and let $I$ denote the identity matrix $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$. Then,$I+A+A^2+\ldots+A^{2010}$ is

If $A$ is a square matrix of order $3$ with $|A| = 2$,then the value of $|(A - A^T)^5| + |(A^T - A)^3|$ is-