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  1. If $A$ has row 1 + row 2 = row 3, show that $A$ is not invertible:

(a) Explain why $Ax=(1,0,0)$ cannot have a solution

My original solution was

$$Ax=\begin{bmatrix} r_1\\ r_2\\ r_3\\ \end{bmatrix}x\tag{1}$$

$$=\begin{bmatrix} r_1x\\ r_2x\\ r_3x \end{bmatrix}\tag{2}$$

$$=\begin{bmatrix} (r_2+r_3)x\\ r_2x\\ r_3x \end{bmatrix}\tag{3}$$

$$=\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\tag{4}$$

But then we have $r_2x=0$, $r_3x=0$, and $r_2x+r_3x=1$. But the latter is false since $r_2x+r_3x=0$. Thus there is no solution to this system.

The solution manual simply says

In $Ax=(1,0,0)$, equation 1 + equation 2 - equation 3 is 0=1

My interpretation of this solution is that from (2), we have

$$=\begin{bmatrix} r_1x\\ r_2x\\ r_3x \end{bmatrix}=\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$$

and if we add the first two rows and subtract the last row on both sides we get

$$r_1x+r_2x-r_3x=1$$

which is not true since $r_1x+r_2x=r_3x$.

Is this the correct interpretation of the terse solution manual solution?

More importantly, what is the geometric interpretation of the reasoning here (in particular considering the perspective of the rows as planes)? Ie, geometrically, why doesn't this system have a solution?

My attempt at this sort of interpretation is the following.

Rows 2 and 3 in the system $Ax=(1,0,0)$ represent two planes passing through the origin. They intercept each other on a line.

The fact that row 1 is the sum of rows 2 and 3 means the normal vector to the plane represented by row 1 is on the same plane as the normal vectors of the planes in rows 2 and 3. If $(1,0,0)$ were $(0,0,0)$ then the plane in row 1 would simply share the same common line with the other two planes, and there would be infinite solutions. But since the right side is a $1$, the first plane does not pass through the origin. But then it crosses each of the other two planes on other lines. So there is no point that is on all three planes at the same time.

Now, I feel like there is a missing piece above: how do I explain why it is that the off-origin plane doesn't cross the line at which planes 2 and 3 intersect?

I think the explanation is that we know the normal vector to the first plane is normal to the line (since it is on the same plane as two other vectors that are normal to the line). Therefore, every other line on the first plane is parallel to the line. Ok, at this point I think I've answered my own question.

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1 Answer 1

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If $Ax=\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}$, we will arrive at a contradiction: $$ \begin{aligned} 0&=\begin{bmatrix}0&0&0\end{bmatrix}x\\ &=\left(\begin{bmatrix}1&1&-1\end{bmatrix}A\right)x\\ &=\begin{bmatrix}1&1&-1\end{bmatrix}(Ax)\\ &=\begin{bmatrix}1&1&-1\end{bmatrix}\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}\\ &=1. \end{aligned} $$

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