Question

If the system of linear equations: $x+y+z=6,x+2y+3z=10,\text{and}3x+2y+𝜆z=𝜇$ has more than two solutions, then $\mu -{\lambda }^2$ is equal to

Answer 1

Step 1: Determination of two solutions

Let the three given equations be 1, 2, and 3.

In equations 1 and 2, put $z=0$, we get

$x+y=6\dots \left(4\right)$

$x+2y=10\dots \left(5\right)$

Subtract $\left(4\right)$ from $\left(5\right)$, then

$y=4\text{and}x=2$

So, we get the first solution as $\left(2,4,0\right)$

For the second solution put $y=0$ in 1 and 2,

$x+z=6\dots \left(6\right)$

$x+3z=10\dots \left(7\right)$

Subtract $\left(6\right)$ from $\left(7\right)$ then

$z=2\text{and}x=4$

Therefore, the second solution is $\left(4,0,2\right)$.

Step 2: Determination of the value of $\mathit{\mu }\mathbf{-}{\mathit{\lambda }}^{\mathbf{2}}$

Apply the first solution in equation 3,

$\begin{array}{rcl}\mu & =& 3\left(2\right)+2\left(4\right)+\lambda \left(0\right)\\ & =& 6+8+0\\ & =& 14\end{array}$

Now, apply the second solution in equation 3,

$\begin{array}{rcl}\therefore 3x+2y+𝜆z& =& 𝜇\\ & \Rightarrow & 3\left(4\right)+2\left(0\right)+𝜆\left(2\right)=14\\ & \Rightarrow & 12+2𝜆=14\\ & \Rightarrow & 𝜆=1\end{array}$

Using these values,

$\begin{array}{rcl}\mu -{\lambda }^2& =& 14-1^2\\ & =& 13\end{array}$

Hence, the value of $\begin{array}{rcl}& & \mathit{\mu }\mathbf{-}{\mathit{\lambda }}^{\mathbf{2}}\end{array}$ is $\mathbf{13}$.