When can we multiply matrices to both sides of a equation?

Question

So this question was recently asked in my exam

If A and B are two square matrices of same order such that $AB=A$ and $BA=B$ then $A^2=?$

Here's what I did

As $$AB=A$$ $$\implies A^{-1}AB=A^{-1}A $$ (pre multiplying by $A^{-1}$) $$ \implies I B = I $$ or $$B=I$$

Here I is identity matrix of same order as A or B.

Similarly we can have from second equation $$A=I$$

Thus it follows $$A^2 = I^2 =I$$

So$$A^2 = I$$

But this was considered wrong and the correct answer was given as A as: $$A^2 = AA$$ $$\implies A^2 = (AB)(A) $$ As $AB=A$ $$\implies A^2 = A(BA) $$ $$\implies A^2 = A B $$ As $BA=B$ $$\implies A^2 = A $$ As $AB=A$

So $$A^2 = A$$

My question:

Why was my response considered wrong when actually both the answers are equivalent as $A= I$ ? Are they correct in saying my answer is wrong?

@HrishabhNayal If it is, yes. But it need not be. What if it's the null matrix of that order? — Allawonder, Mar 28 '20 at 08:56
The answers aren’t equivalent: $A^2=A$ does not imply that $A=I$. Every projection satisfies this identity. — amd, Mar 28 '20 at 19:53

Siong Thye Goh · Accepted Answer · 2020-03-28T08:45:32.927

2

What you have shown is that if $A$ is invertible, then $A=I$.

However, $A$ need not be invertible.

For example, it is possible that $A=B=0$. So yes, they are right to say that your answer is wrong.

You can multiply matrices of compatible size to both size if it exists.

edited Mar 28 '20 at 08:45

answered Mar 28 '20 at 08:39

Siong Thye Goh

149,520
20
88
149

Can you give an example of non invertible matrix where my answer would fail? And thanks for the answer. – Hrishabh Nayal Mar 28 '20 at 09:17
$A=B=0$ is an example – Siong Thye Goh Mar 28 '20 at 09:18
Any non null matrix? – Hrishabh Nayal Mar 28 '20 at 09:19
1

Let $A=\begin{bmatrix} 1 & 0 \ 0 & 0\end{bmatrix}=B$ – Siong Thye Goh Mar 28 '20 at 09:20

When can we multiply matrices to both sides of a equation?

1 Answers1