# Interpreting results from a linear system

$$\begin{pmatrix}t&1&1\\1&t&1\\1&1&t\end{pmatrix}\cdot\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}1\\1\\1\end{pmatrix}$$ Is a linear system. Find the values of t for which the system has a unique solution, infinitely many soluionts and no solution. Row reduced echelon form yields: $$\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\cdot\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}\frac{1}{t+2}\\\frac{1}{t+2}\\\frac{1}{t+2}\end{pmatrix}$$ I am not sure how to interpret this. I am unsure between 2 options: either the system has no solution at $t=-2$ and there are infinitely many unique solutions for $t\in\mathbb{R}$ (not ${-2}$). Or (the one I am more confident about) is that there are infinitely many solutions at $t=-2$, no solutions at $x_1=x_2=x_3=0$ and infinitely many unique solutions as before.

• Hm, what happens when $t = 1$? – Simon S Oct 31 '14 at 2:08
• $x_1=x_2=x_3=\frac{1}{3}$. Why? – George Oct 31 '14 at 2:13
• Or 1/2, 1/4, 1/4. Or 1, 0, 0. Or 2, -3, 2. Or ... – Simon S Oct 31 '14 at 2:14
• I don't get it. You get the general solution $x_1=x_2=x_3=\frac{1}{t+2}$, how do you get those answers from? – George Oct 31 '14 at 2:22
• You row-reduced correctly, but forgot that this required dividing by $t-1$. See my answer below. – Ben Grossmann Oct 31 '14 at 21:59

See the steps described below. Note that we divide twice in order to row reduce: once by $(t-1)$, and once by $-(t +2)$. Thus, this sequence of steps does not work when $t = -2$ or $t = 1$. So, we must handle those cases separately.

Verify that when $t = 1$, we have infinitely many solutions while when $t = -2$, we have no solutions.

In particular, when $t = 1$, the row reduction is simply $$\pmatrix{ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1 } \to \pmatrix{ 1&1&1&1\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ }$$ indicating infinitely many solutions.

## Row Reduction Steps:

The row reduction can be performed as \begin{align} \begin{bmatrix} t & 1 & 1 & 1 \\ 1 & t & 1 & 1 \\ 1 & 1 & t & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & t & 1 \\ 1 & t & 1 & 1 \\ t & 1 & 1 & 1 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 1 & t & 1 \\ 0 & t-1 & 1-t & 0 \\ 0 & 1-t & 1-t^2 & 1-t \end{bmatrix}\\[2ex] &\text{(1) for $t\ne1$}\\[2ex] &\to \begin{bmatrix} 1 & 1 & t & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 1-t & 1-t^2 & 1-t \end{bmatrix}\\ &\to \begin{bmatrix} 1 & 1 & t & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 2-t-t^2 & 1-t \end{bmatrix}\\[2ex] &\text{(2) for $t\ne-2$}\\[2ex] &\to \begin{bmatrix} 1 & 1 & t & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 1/(t+2) \end{bmatrix} \end{align}

Thus the system has surely unique solution for $t\ne1$ and $t\ne-2$ and the solution is readily seen to be $$\begin{bmatrix}1/(t+2)\\1/(t+2)\\1/(t+2)\end{bmatrix}$$ Note that this is just one solution because $t$ is supposed to be a fixed number. You have “infinitely many systems”, one for each value of $t$, but each one has one solution (when $t\ne1$ and $t\ne-2$).

If $t=-2$, the matrix before stating condition (2) becomes $$\begin{bmatrix} 1 & 1 & -2 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix}$$ and the system has no solution.

If $t=1$, the matrix before stating condition (1) becomes $$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ and the system has infinitely many solution.