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{\bf Question}

Find the values of $\lambda$ and $\mu$ for which the following
systems of equations is consistent, and find the general solution
in all cases. \begin{eqnarray*} x_1 + 2x_2 + x_3 +x_4 & = & 13 \\
x_1 + \lambda x_2 + 4x_3 + 6x_4 & = & 18 \\ x_1 - 5x_3 -9x_4 & = &
\lambda \\ x_1 + 5x_2 + 10 x_3 + 16 x_4 & = & \mu \end{eqnarray*}

\vspace{.25in}

{\bf Answer}

\begin{eqnarray*} \left.\begin{array}{cccc} 1 & 2 & 1 & 1 \\ 1 &
\lambda & 4 & 6 \\ 1 & 0 & -5 & -9 \\ 1 & 5 & 10 & 16 \end{array}
\right| \begin{array}{c} 13 \\ 18 \\ \lambda \\ \mu \end{array} &
\rightarrow & \left.\begin{array}{cccc} 1 & 2 & 1 & 1 \\ 0 &
\lambda-2 & 3 & 5 \\ 0 & -2 & -6 & -10 \\ 0 & 3 & 9 & 15
\end{array} \right| \begin{array}{c} 13 \\ 5 \\ \lambda-13 \\ \mu-13
\end{array} \\ & \rightarrow & \left.\begin{array}{cccc} 1 & 2 & 1 & 1 \\ 0 &
\lambda-2 & 3 & 5 \\ 0 & 3 & 9 & 15 \\ 0 & 3 & 9 & 15
\end{array} \right| \begin{array}{c} 13 \\ 5 \\ -\frac{3}{2}(\lambda-13) \\ \mu-13
\end{array} \end{eqnarray*}
Must have $\mu -13 = -\frac{3}{2}(\lambda -13)$

i.e. $3\lambda + 2\mu -65 = 0$

Then we get rid of row 4 $$\left.\begin{array}{cccc} 1 & 2 & 1 & 1
\\ 0 & \lambda-2 & 3 & 5 \\ 0 & 3 & 9 & 15 \end{array} \right|
\begin{array}{c} 13 \\ 5 \\ -\frac{3}{2}(\lambda-13)
\end{array}$$

$\lambda = 3$ gives both sides of row 3 to be zero.

$\lambda \not=3$ gives $x_2 = \frac{1}{2}$

${}$

$\lambda = 3$ gives $\mu =28$

Thus $\left. \begin{array}{cccc} 1 & 2 & 1 & 1 \\ & 1 & 3 & 5
\end{array}\right| \begin{array}{c} 13 \\ 5 \end{array}$

Solution is $$\left(\begin{array}{c} 3 \\ 5 \\ 0 \\ 0 \end{array}
\right) + x_3\left(\begin{array}{c} 5 \\ -3 \\ 1 \\ 0 \end{array}
\right) + x_4\left(\begin{array}{c} 9 \\ -5 \\ 0 \\ 1 \end{array}
\right)$$

$\lambda \not=3$ gives $x_2 = \frac{1}{2}$ and solution $$\left(
\begin{array}{c} 10 + \frac{\lambda}{6} \\ \frac{1}{2} \\ 2 -
\frac{\lambda}{6} \\ 0 \end{array} \right) + x_4 \left(
\begin{array}{c} \frac{2}{3} \\ 0 \\ -\frac{5}{3} \\ 1 \end{array}
\right)$$

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