# Why is this linear equation system inconsistent?

I have the following linear equation system: $$(S)=\left\{ \begin{array}{ll} a2008 + b = 34.564\\ a2009 + b = 35.104\\ a2010 + b = 35.481\\ a2011 + b = 36.891\\ a2012 + b = 37.331\\ a2013 + b = 38.061\\ a2016 + b = 39.071\\ a2018 + b = 39.345 \end{array} \right.$$ and the corresponding augmented matrix $$\mathbf{M} = \left[ \begin {array}{ccc} 2008&1& 34.564\\ {}2009&1& 35.104\\ {}2010&1& 35.481\\ {}2011&1& 36.891\\ {}2012&1& 37.331\\ {}2013&1& 38.061\\ {}2016&1& 39.071\\ {}2018&1& 39.345\end {array} \right]$$ After bringing this to reduced row-echelonform it is evident that rank $$\mathbf{M}$$ > rank $$\mathbf{A}$$ and that the linear equation system therefore is inconsistent - note that $$\mathbf{A}$$ is the coefficient matrix.

However, bringing the augmented matrix to RREF takes time, and for some reason I was able to figure that it was consistent before performing any row operations. I think it is because that there are all ones in the second column but it does not seem like a satisfactory argument/explanation.

So how would you argue that this linear equation system clearly has no solution (is inconsistent) without having the calculate anything?

• That is no equation system, it is a matrix. Maybe write down the equation? – mathreadler Jul 26 '19 at 10:35
• It is the augmented matrix of a linear equation system as I wrote? – npp Jul 26 '19 at 10:58
• But yea, I can also write the linear equation system in the question, even though it should not be necessary when you already have the augmented matrix for that system :)) – npp Jul 26 '19 at 10:59
• Without any matrices or any fancy stuff, just look at the differences of the different rows in the first equation. On the left-hand side you always get $a$ (or a multiple of $a$), but what do you get on the right-hand side? Based on this, the right-hand side difference should always be the same ... Just take a look at the first three rows, for example. – Matti P. Jul 26 '19 at 11:04
• In addition to what @MattiP. says: I generally don't use RREF reduction. You can use the fact that if there is a non-null minor $M$ of order $k$ and all minors of order $k+1$ obtained adding one row and one column to $M$ are null, then the rank of the matrix is precisely $k$. – dfnu Jul 26 '19 at 11:10

Since you have only two independent variables $$a$$ and $$b$$, any two linear equations in $$a$$ and $$b$$ will have a unique solution (as long as they are independent). So you can pick two pairs of equations from your list, solve for $$a$$ and $$b$$, and see if the solutions are the same.

In this case it is particular easy to see that the equations in $$S$$ are not consistent:

$$2008a + b = 34.564 \\ 2009a + b = 35.104 \\ \Rightarrow a = 35.104 - 34.564 = 0.54$$

$$2009a + b = 35.104 \\ 2010a + b = 35.481 \\ \Rightarrow a = 35.481 - 35.104 = 0.377$$

It really doesn't take much effort to reduce $$\mathbf M$$.

$$\left[ \begin {array}{ccc} 2008 & 1 & 34.564 \\ 2009 & 1 & 35.104 \\ 2010 & 1 & 35.481 \\ 2011 & 1 & 36.891 \\ 2012 & 1 & 37.331 \\ 2013 & 1 & 38.061 \\ 2016 & 1 & 39.071 \\ 2018 & 1 & 39.345 \end {array} \right]$$

The result of \begin{align} 2\mathbf M_2 &\to \mathbf M_2 \\ \mathbf M_2 - \mathbf M_1 &\to \mathbf M_2 \\ \mathbf M_2 - \mathbf M_3 &\to \mathbf M_2 \\ \dfrac{1}{0.163} \mathbf M_2 &\to \mathbf M_2 \\ \end{align}

is

$$\left[ \begin {array}{ccc} 2008 & 1 & 34.564 \\ 0 & 0 & 1.000 \\ 2010 & 1 & 35.481 \\ 2011 & 1 & 36.891 \\ 2012 & 1 & 37.331 \\ 2013 & 1 & 38.061 \\ 2016 & 1 & 39.071 \\ 2018 & 1 & 39.345 \end {array} \right]$$

And we will quickly get

$$\left[ \begin {array}{ccc} 2008 & 1 & 0 \\ 0 & 0 & 1 \\ 2010 & 1 & 0 \\ 2011 & 1 & 0 \\ 2012 & 1 & 0 \\ 2013 & 1 & 0 \\ 2016 & 1 & 0 \\ 2018 & 1 & 0 \end {array} \right]$$

$$\left[ \begin {array}{ccc} 2008 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 0 & 0 \\ 3 & 0 & 0 \\ 4 & 0 & 0 \\ 5 & 0 & 0 \\ 8 & 0 & 0 \\ 10 & 0 & 0 \end {array} \right]$$

$$\left[ \begin {array}{ccc} 2008 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end {array} \right]$$

$$\left[ \begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end {array} \right]$$