System of Linear Equations - how many solutions?

Question

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.

Answer 1

Hint: You can write your system of equations in vector/matrix form:

$\begin{bmatrix}t&1&1\\1&t&1\\1&1&t\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix}$

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.

Answer 2

$\text{I have used the Gauss elimination and then studied the rank of the coefficient matrix:}$

$\text{1)If}\ t=1 \text{ the system reduces to just one equation, and it has}\ \infty^2 \text{solutions.}$

$\text{2)If }\ t=-2 \text{ there are no solutions.}$

$\text{3)If}\ t≠1,-2 \text{ there is a unique solution, depending on t. }$

Answer 3

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 $t=1$.