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Solve the following system of equations by matrix method: $3x - 2y + 3z = 8$,$2x + y - z = 1$,$4x - 3y + 2z = 4$.

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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:

If the system of equations $\begin{bmatrix} 1 & -2 & 5 \\ 2 & -1 & 1 \\ 11 & -7 & p \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ q \end{bmatrix}$ has infinitely many solutions,then:

The system of linear equations $\lambda x + 2y + 2z = 5$,$2\lambda x + 3y + 5z = 8$,and $4x + \lambda y + 6z = 10$ has:

If the system of equations
$2x + y - z = 5$
$2x - 5y + \lambda z = \mu$
$x + 2y - 5z = 7$
has infinitely many solutions,then $(\lambda + \mu)^2 + (\lambda - \mu)^2$ is equal to

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