I've stumbled upon this question, and I have to say that I'm completely lost. I have no idea how one can find the amount of solutions of the system Ax = 0 when this is the case:

Let A be a 3 × 3 matrix and suppose that 2a_1 + a_2 − 4a_3 = 0.

(a) How many solutions will the system Ax = 0 have? Explain.

I also do not know how to answer this question, I know that a matrix is nonsingular when for a certain n x n matrix A, there exists a matrix A^-1 such that A * A^-1 = I_n

(b) Is A nonsingular? Explain.

Thanks in advance.


There are many equivalent statements one can make regarding the invertibility of a square matrix.

In all of the following, $A$ and $B$ are $n\times n$ matrices. The following are all equivalent statements.

  • The matrix $A$ is invertible
  • The rows of $A$ are linearly independent
  • The columns of $A$ are linearly independent
  • The determinant of $A$ is nonzero
  • Zero is not an eigenvalue of $A$
  • There exists a matrix $B$ such that $AB=I$
  • There exists a matrix $B$ such that $BA=I$
  • $Ax=b$ has only one solution
  • $rank(A)=n$
  • The reduced row echelon form of $A$ is the identity
  • $col(A)=\Bbb R^n$
  • $col(A^T)=\Bbb R^n$
  • $\ker(A)=\{0\}$

and probably more still that I am forgetting.

By "they are all equivalent statements" I mean that if a single one of these items on the list is true, they must all be true. If a single one of these items on the list is false, they must all be false.

Your problem statement says something about the linear dependence of the columns of $A$.


Infinitely many solution. Note that one of the solution is already given to you, namely $(2,1,-4)$ because $2a_1+a_2-4a_3=0$ is same as $A\begin{bmatrix} 2 \\ 1 \\ -4\end{bmatrix}=0$. Now note that any multiple of this will also be the solution for $Ax=0$. Regarding your second question, note that $Ax=0$ has non trivial solution, so its columns are not independent and thus there can not exist any inverse for it.


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