# Solving a system of linear equations with parameters

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?

• Ooops! you changed the system! :D Oct 24, 2015 at 14:45
• @HR I have made the edit and changed one tnery in the matrix. I did not change the original system. If the matrix correspond to the system where the first equation is $(4-\lambda)x_1-2x_1-x_3=1$, then the entire is the first row must be $4-\lambda$, $-2$, $-1$, $1$. Originally, the first row contained $1$ in the third place. Oct 24, 2015 at 14:50
• @HR Feel free to rollback my edit, if you think this is more confusing than the original post. But the we will have a post where the given system and and the matrix written int the post are different. (I hope the OP will clarify what is the correct version.) Oct 24, 2015 at 14:51
• @MartinSleziak: Yeah! I think that was a typo from OP that I didn't notice! Any way I changed my answer regarding to your edit! :) Oct 24, 2015 at 14:54
• I just asked my professor and apparently there was a mistake made in the book. It's an Israeli math textbook from the Open University, so it's not really something you can look up online. sigh thanks for the help, everyone. Oct 26, 2015 at 18:00

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.

• Thanks! I'll do the arithmetic on the elementary row operations again but what you're saying does seem to make sense. Thanks again! Oct 24, 2015 at 18:14

Without doing any further computations, you can notice that by addition the first and the third equation you get $$(3-\lambda)x_1+(3-\lambda)x_3=2.$$ For $\lambda=3$ this is equation $0=2$. So you know that for $\lambda=3$ the system has no solution.

If you prefer working with matrices instead, you can add the first row of the matrix to the third row and new matrix will contain the row $\begin{pmatrix} 3-\lambda & 0 & 3-\lambda & 2 \end{pmatrix}$, which leads exactly to the same conclusion.

We also know that a system $Ax=b$ has unique solution whenever determinant of $A$ is non-zero. (In fact, in such case the unique solution can be even found using Cramer's rule, although this way of solving the equation is usually impractical, unless computation of the determinant is particularly simple.)

WolframAlpha says that in your case determinant is $(\lambda-3)(\lambda^2-6\lambda-3)$. So we know that the system has unique solution for all values which are not roots of this polynomial, i.e for $\lambda \ne 3, 3\pm2\sqrt3$. (The determinant by itself is not sufficient information to tell us whether we will have zero or infinitely many solutions for these values. Which is not surprising, since this computation did not involve the RHS at all. But at least we know that for $\lambda$ equal to any of these three values, it is not possible that the system has exactly one solution.)

Notice that if you change $+2x_3$ to $-2x_3$ in the second equation then the determinant is equal to $-(1-\lambda)(3-\lambda)(5-\lambda)$. So we know that for $\lambda\ne1,3,5$ the equation has ony one solution. (Notice the numbers $1$ and $5$, similar to numbers $-1$ and $5$ which appeared in the formulation problem.)

Or if we change the system to $$(4-\lambda)x_1+2x_2-x_3=1\\ 2x_1+(1-\lambda)x_2+2x_3=2\\ -x_1+2x_2+(4-\lambda)x_3=1$$ (i.e. the only changes are $-2x_1$ to $2x_1$ in the second equation and $2x_2$ instead of $-2x_2$ in the first one), then the determinant is $-(\lambda-5)^2(\lambda+1)$ and the system has unique solution for $\lambda\ne-1,5$.

I should say that calculating determinant is not much easier than doing Gauss-Jordan elimination. (Determinant can be calculating using elementary row operations. If we calculate it this way, then basically the only difference is that here we ignored RHS.) But we can check our manual computations using some software which is capable to solve such thing (like WolframAlpha). You can even ask WolframAlpha to solve the system from your question or the modification I have suggested or the second modification I suggested.

So I think that there is a typo in the book and it is supposed to be the system I suggested; or something similar which leads to a nice solution.

Other possibility is that you are not expected to solve the system completely for every value of the parameter $\lambda$. Maybe the authors expect you simply to say that there is no solution for $\lambda=3$. (Which can be discovered in the way I explained above. And which is sufficient to answer both questions.)

To me the first explanation seems more probable. (Since I do not see reason to mention $\lambda\ne-1,5$ if this is the second case.)

Now you have confirmed that there was a typo. So we no longer have to guess.