Why matrix is not invertible if A has row 1 + row 2 = row 3?

I'm interested in intuition so please elaborate on the answer please.

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  • It has to do with matrix rank and determinants. The row 3 does not add any new information but you still have the 3 x something matrix. You can imagine this as having more variables than equations, therefore the system of linear equations is unsolvable. – user2974951 Oct 11 at 12:19
  • Because it has a null space - If you imagine the matrix as an operator or a transformation on a vector, the fact that a matrix with linearly dependent rows or columns is able to send a set of vectors to $\vec 0,$ implies that once you have "killed" these vectors in the null space, there is no way of recovering the actual vector that was mapped to $\vec 0.$ – Antoni Parellada Oct 11 at 12:21
  • Because you lose information after you multiply with the matrix. That the rows can be expressed as each other in terms of some equations mean that there is redundancy and that we lose information. – mathreadler Oct 11 at 12:41

Suppose we have $$\left\{\begin{matrix}x+y&=&2 \\ 2x+2y&=&4\end{matrix}\right.$$ It looks like we have two equations in two unknowns, but we don't: the second equation is equivalent to the first, it doesn't, as @user2974951 already pointed out in a comment, add extra information.

Now, what ways are there for an equation to not add new information? It could be proportional to another equation, as in the example above. Or, it could be a sum of other equations, as below: $$\left\{\begin{matrix}x&=&2 \\ y&=&3 \\ x+y &=& 5\end{matrix}\right.$$ This looks like a system of three equations in two unknowns, but the third equation is entirely superfluous.

To solve for $n$ unknowns, you need n meaningfully distinct equations, or $n$ linearly independent equations in math jargon. In cases such as the one you describe, you have less than $n$ meaningfully distinct equations, so you will not find a unique solution.

Consider a system $Ax=b$. If $row_1 + row_2 = row_3$, then $b_1+b_2=b_3$. Since this does not happen for every $b$, we cannot have solutions of $Ax=b$ for every $b$ and so $A$ cannot be invertible.

Hi welcome to stack exchange.

Let A be the matrix that you are describing.

I think using the property $(A^{T})^{-1} = (A^{-1})^T$ Will make it make more sense intuitively.

Let $\mathbb{x}_{1},\mathbb{x}_{2}, \mathbb{x}_3$. Be the columns of $A^{T}$. Using what you have we know that:

$\mathbb{x}_{1}+\mathbb{x}_{2}- \mathbb{x}_3 = 0$

Extract the coefficients and place into a vector $\mathbb{w} = [1,1,-1]^T$

Now. $A^{T}\mathbb{w} = \mathbb{x}_{1}+\mathbb{x}_{2}- \mathbb{x}_3 = 0$

Suppose A is invertible. This means that $(A^{T})^{-1}A^{T} = I$

i.e. $(A^{T})^{-1}A^{T}\mathbb{w} = \mathbb{0} $

$I\mathbb{w} = \mathbb{w} \neq 0$ Therefore $A$ cannot be invertible$

This is true in the general case as an $n\times n$ matrix that isn't of full rank is non-invertible.

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