Understanding Solutions in Linear Algebra: The Case of m>n

[Miike012]

I don't remember exactly how the question on my test was phrased but I believe it was phrased

"Let A be an mxn matrix where m>n. Explain why in general there is not a solution to the equation Ax = b where b is a vector in Rm"

This question was confusing to me because to me the meaning of the question is saying

For all matrices A with dimension mxn, m>n, There does not exist a solution x such that b = Ax, where b is a vector in Rm.

Which is obviously a false statement because I can easily produce a mxn (m>n) matrix A and a solution x such that
Ax = b, b is a vector in Rm.

This is what I would like to know. I don't want to know what you think it means. Based on how it is worded I want to know what the statement is saying. (I hope that makes sense what I'm asking for)

 

[Mark44]

I don't remember exactly how the question on my test was phrased but I believe it was phrased

"Let A be an mxn matrix where m>n. Explain why in general there is not a solution to the equation Ax = b where b is a vector in Rm"

This question was confusing to me because to me the meaning of the question is saying

For all matrices A with dimension mxn, m>n, There does not exist a solution x such that b = Ax, where b is a vector in Rm.

Which is obviously a false statement because I can easily produce a mxn (m>n) matrix A and a solution x such that
Ax = b, b is a vector in Rm.

This is what I would like to know. I don't want to know what you think it means. Based on how it is worded I want to know what the statement is saying. (I hope that makes sense what I'm asking for)

When it says "in general there is not a solution" this doesn't mean that there is never a solution. It means that there can be a solution, but such solutions would be fairly rare.

To convince yourself of what the statement is saying, make up an m x n matrix A with m > n, say a 5 X 3 matrix. Here x would have to be a 3 X 1 column vector, and b would have to be a 5 X 1 column vector. Given that there are more rows in A than columns, row reduction would leave you with at least 2 rows of zeroes, and possibly more. If, during row reduction (using an augmented matrix), the column for b didn't end up with zeroes in the positions that correspond to the zero rows in the reduced matrix, there wouldn't be a solution.

The situation would be something like this:
$$\left[ \begin{array}{c c c c c} 1 & 0 & 0 & | & b_1 \\
0 & 1 & 0 & | & b_2 \\
0 & 0 & 1 & | & b_3 \\
0 & 0 & 0 & | & b_4 \\
0 & 0 & 0 & | & b_5 \\ \end{array} \right]$$

 

[Miike012]

When it says "in general there is not a solution" this doesn't mean that there is never a solution. It means that there can be a solution, but such solutions would be fairly rare.

To convince yourself of what the statement is saying, make up an m x n matrix A with m > n, say a 5 X 3 matrix. Here x would have to be a 3 X 1 column vector, and b would have to be a 5 X 1 column vector. Given that there are more rows in A than columns, row reduction would leave you with at least 2 rows of zeroes, and possibly more. If, during row reduction (using an augmented matrix), the column for b didn't end up with zeroes in the positions that correspond to the zero rows in the reduced matrix, there wouldn't be a solution.

The situation would be something like this:
$$\left[ \begin{array}{c c c c c} 1 & 0 & 0 & | & b_1 \\
0 & 1 & 0 & | & b_2 \\
0 & 0 & 1 & | & b_3 \\
0 & 0 & 0 & | & b_4 \\
0 & 0 & 0 & | & b_5 \\ \end{array} \right]$$

Thanks. I was wrong then. I though it was saying there will never be a solution. I was hopeing I would be able to get more point on my test ha.