Number of value of $'a'$ for which the system of equations,$A^2 x + (2 – a) y = 4 + a^2$ $a x + (2 a – 1) y = a^5 – 2$ possess no solution is

Let $A$ be a $3 \times 3$ real matrix such that $A \left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right) ; A \left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right)$ and $A \left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)$. If $X =\left( x _{1}, x _{2}, x _{3}\right)^{ T }$ and $I$ is an identity matrix of order $3$ , then the system $( A -2 I ) X =\left(\begin{array}{l}4 \\ 1 \\ 1\end{array}\right)$ has

If the system of linear equations, $x+y+z =6$ ; $x+2 y+3 z =10$ ; $3 x+2 y+\lambda z =\mu$ has more two solutions, then $\mu-\lambda^{2}$ is equal to

Let $A =$$\left[ {\begin{array}{*{20}{c}}1&2&3\\2&0&5\\0&2&1\end{array}} \right]$ and $b =$$\left[ {\begin{array}{*{20}{r}}0\\{ – 3}\\1\end{array}} \right]$ . Which of the following is true?

Solve the system of the following equations  $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$ ; $\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$  ; $\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$