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

The system of linear equations $x+y+z=6, x+2y+3z=10$ and $x+2y+az=b$ has no solutions when

Use the product $\left[\begin{array}{lll}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{lll}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$ to solve the system of equations:
$x-y+2z=1$
$2y-3z=1$
$3x-2y+4z=2$

The sum of three numbers is $6$. If we multiply the third number by $3$ and add the second number to it,we get $11$. By adding the first and third numbers,we get double the second number. Represent this algebraically and find the numbers using the matrix method.

Let the system of equations $x+2y+3z=5$,$2x+3y+z=9$,and $4x+3y+\lambda z=\mu$ have an infinite number of solutions. Then $\lambda+2\mu$ is equal to:

If the system $\begin{bmatrix} 2 & 8 \\ 3 & 7 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = k \begin{bmatrix} a \\ b \end{bmatrix}$ has a non-trivial solution,then the positive value of $k$ and a solution of the system for that value of $k$ are: