1. ## matrix product

The product of two square matrices of same order equals to zero implies,

1. Atleast one of the matrix is nonsingular
2.Either one is null matrix
3. Both are non singular matrix
4.Both are singular matrix.

Answer given is 2 or 4.

Option two looks obvious.but how 4 can be inferred from the given data.

2. Originally Posted by kumaran5555
The product of two square matrices of same order equals to zero implies,

1. Atleast one of the matrix is nonsingular
2.Either one is null matrix
3. Both are non singular matrix
4.Both are singular matrix.

Answer given is 2 or 4.

Option two looks obvious.but how 4 can be inferred from the given data.
The correct answer is one. Taking the matrices $\displaystyle A=\begin{pmatrix}0 &0\\ 0& 1\end{pmatrix}$ and $\displaystyle I=\begin{pmatrix}1& 1\\0 &0\end{pmatrix}$ is a counterexample to 2 and $\displaystyle A=\begin{pmatrix}0 &0\\ 0 & 0\end{pmatrix}$ and $\displaystyle B=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}$ is a counterexample to 4.

3. I don't understand your explanation.

The first example with A and I which is counter to 2 but it goes with point number 4. i.e both are singular.

similarly the second example is counter to 4 but satisfies condition two. i.e one of them is null matrix.

So the answer is still 2 or 4 only. Any one of this is true in your examples.

4. So we want to prove
$\displaystyle AB=0$ implies either:

2. At least one of the matrices is 0, OR
4. Both matrices are singular

I like to do questions with an "or" conclusion by showing that, if one of the options does not hold, then the other option MUST hold.

So, lets say we have two such matrices, and that #2 does NOT hold (that is, neither $\displaystyle A$ nor $\displaystyle B$ is 0).

Let's show both matrices are singular (i.e., neither is invertible).

It's easy to see that if $\displaystyle AB=0$ and $\displaystyle A$ is invertible, then

$\displaystyle AB=0\Rightarrow A^{-1}AB=A^{-1}0\Rightarrow B=0$,

which is clearly a contradiction to out assumption. Similar if $\displaystyle B$ is invertible.