[Math] Solving a system of linear equations with parameters

gaussian eliminationlinear algebrasystems of equations

I've been given the following system of equations:

$$
(4-\lambda)x_1-2x_2-x_3=1\\
-2x_1+(1-\lambda)x_2+2x_3=2\\
-x_1+2x_2+(4-\lambda)x_3=1
$$

The resulting coefficient matrix would be:

$$
\begin{bmatrix}
(4-\lambda)&-2&\ &-1&\ &1\\\\
-2&(1-\lambda)&\ &2&\ &2\\\\
-1&2&\ &(4-\lambda)&\ &1
\end{bmatrix}
$$

I've managed to reduce it to:

$$
\begin{bmatrix}
(4-\lambda)&-2&-1&\ &1\\\\
0&\frac{-6-\lambda}{4-\lambda}&(5-\lambda)&\ &\frac{3-\lambda}{4-\lambda}\\\\
0&(5-\lambda)&(\lambda-5)&\ &0
\end{bmatrix}
$$

There are two claims I'm being asked to answer either true or false on:

  1. If $\lambda\neq -1,5$ the system of equations has only one solution
  2. The system of equations has a solution for any value of $\lambda$

Is it possible to confirm or deny these claims without actually solving the system? Because I've honestly tried reducing it further and gotten a pretty messy resulting matrix. Can someone shed some light on this for me please?

Best Answer

Your strategy looks good. But the computation you carried out may have some errors. I did the lengthy elementary row operations for the augmented matrix in MAPLE to get

$$\left[ {\begin{array}{*{20}{c}} {4 - \lambda }&{ - 2}&{ - 1}&1\\ 0&\lambda &{ - \frac{{2\left( { - 3 + \lambda } \right)}}{{ - 5 + \lambda }}}&{ - 2}\\ 0&0&{\frac{{{\lambda ^3} - 9{\lambda ^2} + 15\lambda + 9}}{{ - 5 + \lambda }}}&{ - \lambda - 3} \end{array}} \right]$$

while during the process of elimination it is assumed that $\lambda\ne 0,4,5$. Now, if you look at the third row which is an equation you get

$$\left( {\frac{{{\lambda ^3} - 9{\lambda ^2} + 15\lambda + 9}}{{ - 5 + \lambda }}} \right){x_3} = - \lambda - 3$$

The system can have no solutions only when this equation can lead to a contradiction. This can happen when the coefficient of $x_3$ in LHS is zero and the RHS is nonzero. If this happens we can conclude that the system doesn't have a solution for specified values of $\lambda$. So we may write

$\left\{ \begin{array}{l} {\lambda ^3} - 9{\lambda ^2} + 15\lambda + 9 = 0\\ - \lambda - 3 \ne 0 \end{array} \right.\,\,\,\,\, \to \,\,\,\,\,\left\{ \begin{array}{l} \lambda = 3,3 \pm 2\sqrt 3 \\ \lambda \ne - 3 \end{array} \right.\,\,\,\, \to \,\,\,\,\lambda = 3,3 \pm 2\sqrt 3 $

Since this is consistent with our previous assumptions that $\lambda\ne 0,4,5$, we can say that for these values of $\lambda$ the system doesn't have a solution. Consequently, your second statement is false since we proved that the system has no solutions for some values of $\lambda$. Also, as we have found some values for $\lambda\ne -1,5$ that the system does not have a solution, your first statement is false too.

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