Let $A = \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix}$. If $A^3 = 4A^2 - A - 21I$,where $I$ is the identity matrix of order $3 \times 3$,then $2a + 3b$ is equal to:

If the unique solution of the simultaneous linear equations $3x - 2y + z = 5k$, $2x + 3y - 2z = -5k$, and $x + 4y + 3z = k$ is $x = \alpha, y = \beta, z = 3$, then $k =$

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 $\alpha^2+\beta^2=$

The value of $\lambda$ for which the system of equations $2x - y - z = 12,$ $x - 2y + z = -4,$ and $x + y + \lambda z = 4$ has no solution is

Let $k_1$ and $k_2$ be the maximum and minimum values of $k$ for which the system of equations $x + ky = 1$,$kx + y = 2$,and $x + y = k$ are consistent. Then $k_1^2 + k_2^2$ is equal to:

If $A=\begin{bmatrix} x & y & y \\ y & x & y \\ y & y & x \end{bmatrix}$ is a matrix such that $5 A^{-1}=\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$, then $A^2-4 A=$