# System of Linear Equations - how many solutions?

For which real values of t does the following system of linear equations:

$$\left\{ \begin{array}{c} tx_1 + x_2 + x_3 = 1 \\ x_1 + tx_2 + x_3 = 1 \\ x_1 + x_2 + tx_3 = 1 \end{array} \right.$$

Have:
a) a unique solution?
b) infinitely many solutions?
c) no solutions?

I haven't done linear algebra in almost a year, so I'm really rusty and could use some pushes in the right direction.

• Note that you can write this system as $\begin{bmatrix}t&1&1\\1&t&1\\1&1&t\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix}$ – flawr Sep 13 '14 at 22:30
• Start by trying some simple things. What happens if you put $t=1$? What happens if you put $t=0$? You will find that two kinds of nasty things can happen. You can get equations that contradict each other, or equations that are redundant. – almagest Sep 13 '14 at 22:32
• @almagest I've already solved the matrix for a variety of values of t. For example, for 1 I find that x1=1−x2−x3, and I assume that means there are infinitely many solutions. For all other values I tested except -2 (it was inconsistent), I seem to have a single answer. My issue is more that I don't know a concise way to solve this problem. I'm guessing I would put it in row-echelon form, but I don't know where to go from there. – Alex Sep 13 '14 at 22:38
• The standard jargon is "Cramer's rule". I have not looked through them carefully, but there are dozens of questions about it on this site. But it is dangerous just to turn a handle. It is much better to think about what is going on! :) – almagest Sep 13 '14 at 22:51
• – Martin Sleziak Feb 3 '15 at 17:13

$\begin{bmatrix}t&1&1\\1&t&1\\1&1&t\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix}$
This has now the form $Ax = b$ where $A$ is the matrix $x$ the unknown and $b$ the vector of ones. If it can be solved the solution would be $x=A^{-1}b$. Now I recommend (as the other commentors) determining whether you can solve this by consulting the determinant of $A$ or the gaussian algorithm.
• Thanks! Let's see if I understand what's going on here: I've found the formula for the determinant to be $t^3 -3t + 2$, so if the determinant is nonzero then it will have a unique solution (found by, as you said, $x=A^(-1)b). If the determinant IS zero, however, then it can have either no solution or infinitely many? Apologies, I've been really slow getting back into this course. – Alex Sep 13 '14 at 22:57 • @Alex What you are saying is correct. (In the case when the determinant is 0, you can use row reduction to see if there is a solution or not.) – user84413 Sep 13 '14 at 22:59 • @Alex Yes, but why not get some intuition for this by looking at a much simpler example? If we have a$1\times 1$matrix$(a)$, when does$ax=b$have a unique solution, infinitely many solutions, or no solutions? – Slade Sep 13 '14 at 23:03 • Perfect, I think I understand now. Thanks a lot! @you-sir-22433 it would have no solution when a = 0, because 0 =/= b (I believe we assume that the solution is non-zero, right?). A unique solution when a = b, and infinitely many when a is any other real number? I don't even think that's right. Man, even these simple questions are eluding me... – Alex Sep 13 '14 at 23:15$\text{I have used the Gauss elimination and then studied the rank of the coefficient matrix:}\text{1)If}\ t=1 \text{ the system reduces to just one equation, and it has}\ \infty^2 \text{solutions.}\text{2)If }\ t=-2 \text{ there are no solutions.}\text{3)If}\ t≠1,-2 \text{ there is a unique solution, depending on t. }$• Isn't rank just the number of nonzero rows of the matrix in row echelon form? How can you use that to determine how many solutions the system has? – Alex Sep 13 '14 at 23:16 • because of this theorem: en.wikipedia.org/wiki/Rouché–Capelli_theorem – Mosk Sep 13 '14 at 23:19 • The rank in this case depends on t, so you have to study what happens for the values of t that gives you another zero in the row echelon form, so the rank changes and the solutions too – Mosk Sep 13 '14 at 23:23 We can find rref for this matrix using Gauss-Jordan elimination: $$\begin{pmatrix} t & 1 & 1\\ 1 & t & 1\\ 1 & 1 & t \end{pmatrix}\sim \begin{pmatrix} t & 1 & 1\\ 1 & t & 1\\ t+2 & t+2 & t+2 \end{pmatrix}\overset{(1)}\sim \begin{pmatrix} t & 1 & 1\\ 1 & t & 1\\ 1 & 1 & 1 \end{pmatrix}\sim \begin{pmatrix} t-1 & 0 & 0\\ 0 & t-1 & 0\\ 1 & 1 & 1 \end{pmatrix}\overset{(2)}\sim \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 1 & 1 & 1 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}$$ It is important to notice that the step (1) is valid only if$t+2\ne0$and step (2) is valid only if$t-1\ne0$. So from the above computation we see that for$t\ne-2,1$this matrix is invertible. In this case the system has exactly one solution. (If you add the RHS to the above matrix and to the same manipulation, you should find out that the solution in this case is$(\frac1{t+2},\frac1{t+2},\frac1{t+2})$.) It should be easy to find the answer for the remaining cases$t=-2$and$t=1\$.