Consider the system of equations:

$$x-2y+3z=-1$$

$$-x+y-2z=k$$

$$x-3y+4z=1$$

Statement - 1 : The system of equations has no solution for $$k\ne3$$.

and

Statement - 2 : The determinant $$\left| {\matrix{ 1 & 3 & { - 1} \cr { - 1} & { - 2} & k \cr 1 & 4 & 1 \cr } } \right| \ne 0$$, for $$k \ne 3$$.

$$A=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right]$$, if $$U_{1}, U_{2}$$ and $$U_{3}$$ are columns matrices satisfying. $$\mathrm{AU}_{1}=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right], \quad \mathrm{AU}_{2}=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right], \quad \mathrm{AU}_{3}=\left[\begin{array}{l}2 \\ 3 \\ 1\end{array}\right]$$ and $$\mathrm{U}$$ is $$3 \times 3$$ matrix whose columns are $$\mathrm{U}_{1}, \mathrm{U}_{2}, \mathrm{U}_{3}$$ then answer the following questions

Comprehension IV

$\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right]$, if $\mathrm{U}_1, \mathrm{U}_2$ and $\mathrm{U}_3$ are columns matrices satisfying. $\mathrm{AU}_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right], \mathrm{AU}_2=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right], \mathrm{AU}_3=\left[\begin{array}{l}2 \\ 3 \\ 1\end{array}\right]$ and U is $3 \times 3$ matrix whose columns are $\mathrm{U}_1, \mathrm{U}_2, \mathrm{U}_3$ then answer the following questions

The sum of the elements of $\mathrm{U}^{-1}$ is:

$\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right]$, if $\mathrm{U}_1, \mathrm{U}_2$ and $\mathrm{U}_3$ are columns matrices satisfying. $\mathrm{AU}_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right], \mathrm{AU}_2=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right], \mathrm{AU}_3=\left[\begin{array}{l}2 \\ 3 \\ 1\end{array}\right]$ and U is $3 \times 3$ matrix whose columns are $\mathrm{U}_1, \mathrm{U}_2, \mathrm{U}_3$ then answer the following questions

The value of $\left[\begin{array}{lll}3 & 2 & 0\end{array}\right] U\left[\begin{array}{l}3 \\ 2 \\ 0\end{array}\right]$ is :