The system of linear equations $x + y + z = 2, 2x + 3y + 2z = 5$,and $2x + 3y + (a^2 - 1)z = a + 1$ is:

If $\begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix} X = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}$,then $X =$

While solving a system of linear equations $AX=B$ using Cramer's rule with the usual notation,if $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 2 \\ -1 & 1 & 5\end{array}\right|$,$\Delta_1=\left|\begin{array}{ccc}5 & 1 & 1 \\ 4 & -1 & 2 \\ 11 & 1 & 5\end{array}\right|$ and $X=\left[\begin{array}{l}\alpha \\ 2 \\ \beta\end{array}\right]$,then $\alpha^2+\beta^2=$

If $A$ is a matrix such that $\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] A \left[\begin{array}{ll} 1 & 1 \end{array}\right] = \left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$, then $A$ is equal to

The system of equations $x - 2y + 3z = 5$,$2x - 2y + z = 0$,and $-x + 2y - 3z = 6$ has

If the system of equations $x + y + z = 5$, $x + 2y + 3z = 9$, $x + 3y + \lambda z = \mu$ has infinitely many solutions, then the value of $\lambda + \mu$ is: