# How do you solve the system x + 2y -4z = 0, 2x + 3y + z = 1, 4x + 7y + lamda*z = mu?

Feb 23, 2017

x = (28-14mu+2 lambda)/(lambda + 7); y = (-16+9mu-lambda)/(lambda + 7) ;z = (mu-1)/(lambda+7)

#### Explanation:

This system can be solved a number of ways, such as substitution and elimination, matrix row operations, etc.

Using Matrix Row Operations we want to get: $\left(\begin{matrix}1 & 0 & 0 & x \\ 0 & 1 & 0 & y \\ 0 & 0 & 1 & z\end{matrix}\right)$

Since we have a 1 in the top left do the following row operations.
To row 2: $- 2 {R}_{1} + {R}_{2}$ and to row 3: $- 4 {R}_{1} + {R}_{3}$

((1,2,-4,0),(2,3,1,1),(4,7,lambda,mu)) => ((1,2,-4,0),(0,-1,9,1),(0,-1,16+lambda,mu))

To get a 1 in the 2nd column of row 2: $- {R}_{2}$
$\left(\begin{matrix}1 & 2 & - 4 & 0 \\ 0 & 1 & - 9 & - 1 \\ 0 & - 1 & 16 + \lambda & \mu\end{matrix}\right)$

To get (0, 1, 0) in the 2nd column:
To row 1: $- 2 {R}_{2} + {R}_{1}$ and to row 3: ${R}_{2} + {R}_{3}$
$\left(\begin{matrix}1 & 0 & 14 & 2 \\ 0 & 1 & - 9 & - 1 \\ 0 & 0 & 7 + \lambda & \mu - 1\end{matrix}\right)$

To get a 1 in the 3rd row, 3rd column: ${R}_{3} / \left(\lambda + 7\right)$
$\left(\begin{matrix}1 & 0 & 14 & 2 \\ 0 & 1 & - 9 & - 1 \\ 0 & 0 & 1 & \frac{\mu - 1}{\lambda + 7}\end{matrix}\right)$

To get (0,0,1) in the 3rd column:
To row 1: $- 14 {R}_{3} + {R}_{1}$ and to row 2: $9 {R}_{3} + {R}_{2}$
$\left(\begin{matrix}1 & 0 & 0 & - 14 \frac{\mu - 1}{\lambda + 7} + 2 \\ 0 & 1 & 0 & 9 \frac{\mu - 1}{\lambda + 7} - 1 \\ 0 & 0 & 1 & \frac{\mu - 1}{\lambda + 7}\end{matrix}\right) \implies \left(\begin{matrix}1 & 0 & 0 & \frac{28 - 14 \mu + 2 \lambda}{\lambda + 7} \\ 0 & 1 & 0 & \frac{9 \mu - \lambda - 16}{\lambda + 7} \\ 0 & 0 & 1 & \frac{\mu - 1}{\lambda + 7}\end{matrix}\right)$

So x = (28-14mu+2 lambda)/(lambda + 7); y = (-16+9mu-lambda)/(lambda + 7) ;z = (mu-1)/(lambda+7)