Question regarding trivial and non trivial solutions to a matrix.

Question

I would very much appreciate som explanations regarding trivial and non trivial solutions to a matrix (I am a beginner in studies of linear algebra).

Suppose that we have two matrices $A$ and $B$. The matrix $B$ is the row reduced form of matrix $A$ (by using a number of elementary row operations) and $B = A_k...A_2, A_1, A$, where $A_k...A_2, A_1$ are elementary matrices.

This means that the matrix $B$ is triangular and $det (B)$ is the product of the diagonal elements in the matrix $B$, the diagonal elements are either $1$ or $0$.

So far so good, Now the part which I don't really understand:

If any of the diagonal elements in $B$ is $0$ then the last row of $B$ must be $0$ (By shifting rows, so that the $0$ row is last?).

This means that the system $BX = 0$ has non trivial solutions (Why is that so? An explanation would be very much appreciated!).

This is also true for the equivalent system $AX=0$ and this means that $A$ is non invertible (An explanation how they make this conclusion would also be much appreciated).

Thank you kindly for your help, anything is helpful since I am trying to supplement my textbook to understand this better.

Thank you!

Answer 1

1st question: This means that the system $Bx=0$ has non trivial solutions (Why is that so? An explanation would be very much appreciated!).

If one of the rows of the matrix $B$ consists of all zeros then in fact you will have infinitely many solutions to the system $Bx=0$. As a simple case consider the matrix $M=\left(\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$. Then the system $Mx=0$ has infinitely many solutions, namely all points on the line $x+y=0$.

2nd question: This is also true for the equivalent system $Ax=0$ and this means that A is non invertible (An explanation how they make this conclusion would also be much appreciated).

Since the system $Ax=0$ is equivalent to the system $Bx=0$ which has non-trivial solutions, $A$ cannot be invertible. If it were then we could solve for $x$ by multiplying both sides of $Ax=0$ by $A^{-1}$ to get $x=0$, contradicting the fact that the system has non-trivial solutions.