Let's say I have matrix A and its n x n and non-singular. Which of the following always true?
- The linear system Ax = 0 has only one non-trivial solution.
- A row echelon form of A has no non-pivot column
- det(A) = n
- rank(A) can be any non-zero integer value not more than n.
From what we've done in class, I think it should be 2.
It's not 1 as Ax=0 only has the trivial solution as its non-singular. Its not 3 as det(A) does not always equal n and since A is non-singular, it should be full rank.
Best Answer
You're right.
False, because if the matrix is non-singular, then the system $Ax=0$ has only the trivial solution (and consequently no non-trivial solutions). This is because the matrix being non-singular implies that every system $Ax=b$ has unique solution, and $x=0$ is always a solution to $Ax=0$, so it's unique in the case of $A$ being non-singular.
True (consecuence of the matrix having determinant different from $0$, and also with the fact said in point 4, because if it had a non-pivot column, then it would not have full rank and it would be a singular matrix).
False, the determinant can be anything different from $0$, but in general it's not equal to $n$ (take for example $I_2$, the $2\times 2$ identity matrix, then $|I_2|=1\neq 2$).
False. If the determinant is different from $0$, then the column vectors of $A$ are linearly independent, and then you conclude that $\text{rank}(A)=n$ (full rank).