The following system of equations $x+y+z=9$, $2x+5y+7z=52$, $x+7y+11z=77$ has

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 $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$,find $A^{-1}$. Using $A^{-1}$,solve the system of equations: $2x - 3y + 5z = 11$,$3x + 2y - 4z = -5$,and $x + y - 2z = -3$.

Let $f(x) = 2x^2 + 5x + 1$. If we write $f(x)$ as $f(x) = a(x+1)(x-2) + b(x-2)(x-1) + c(x-1)(x+1)$ for real numbers $a, b, c$, then:

If the system of equations $2x + 3y - 3z = 3$,$x + 2y + \alpha z = 1$,and $2x - y + z = \beta$ has infinitely many solutions,then $\frac{\alpha}{\beta} - \frac{\beta}{\alpha} =$

If $\begin{bmatrix} 2 & 1 & -1 \\ -1 & 2 & 1 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \\ -2 \end{bmatrix}$, then the distance of point $P(a, b, c)$ from the plane $2x + y + 2z = 10$ is