The system of equations $x + ky - z = 0$,$3x - ky - z = 0$,and $x - 3y + z = 0$ has a non-zero solution for $k =$

Solve the system of linear equations using the matrix method:
$5x + 2y = 3$
$3x + 2y = 5$

The product of all real values of $b$ such that there is no solution to the system of equations $2x + 5y + z = 19$,$-4x + by + 6z = -42$,and $-3y - bz = 81$ is:

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 the system of equations $3x - 2y + z = 0$,$\lambda x - 14y + 15z = 0$,and $x + 2y + 3z = 0$ has a non-trivial solution,then $\lambda = $

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 0 & 5 \\ 0 & 2 & 1 \end{bmatrix}$ and $b = \begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}$. Which of the following is true?