Could anyone help me to understand the following cases

$1.$ CASE (2.1), why rank of co-ef matrix is $2$ and augmented matrix is $3$? I can explain from the fact that if rank of co-ef matrix is $1$ and rank of augmented matrix is $3$ then it is case that all the three planes are parallel. And if rank of both is $3$ then it is case $1$, so what is the geometric explanation of case $2.1$?

$2.$ case $4.1$, shouldn't it be rank of augmented matrix is $3$? because three planes have different distances from the origin?

$3.$ case $3.1$ I also want the geometric and algebraic explanation.


Some of the ideas:

Generically (which to a mathematician means something like "most of the time"), any two planes will intersect at a line. In fact two planes either intersect at a line or are parallel and we know that two rows have to be the same (in the non-augmented matrix) for the two planes described by those rows to be parallel.

The rank $r$ is the number of planes that are linearly independent (in a sense). You have to try imagine adding planes together to get another plane in a way very similar to how you add vectors.

enter image description here

Then only if $r=3$ will you have a unique solution. This is the case where all three planes are linearly independent (again, in a sense).

This also should help you understand that if $r=1$ then all $3$ planes will be parallel.

If $r'>r$ then there will be no solution. I can't think of a geometric explanation (at least one in terms of planes) for this, but algebraically its because the target vector $b$ is outside the image of the transformation $x\mapsto Ax$.

Also, algebraically, $r'$ can't be more than $1$ higher than $r$ because you're only adding one more column. That column might be linearly independent of the other columns ($r'=r+1$) or linearly dependent on them ($r'=r$) but those are the only choices.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.