Given two matrices $A$ and $B$. If $A = B$, is $A_{ij} = B_{ij}$ always true?

Question

My professor on college came up one day with an exercise where there was an equality of matrices.

But i noticed pretty instantly that the insertions on the matrices didn't match up.

Not only that, but if $A_{ij} = B_{ij}$ were to always be true, it would generate a contradiction.

$$ A = \begin{bmatrix}x+1& 10\\1.5& 1\end{bmatrix}\\ B = \begin{bmatrix}3& 2\\-1& 3\end{bmatrix}\\ A = B\\ \begin{bmatrix}x+1& 10\\1.5& 1\end{bmatrix} = \begin{bmatrix}3& 2\\-1& 3\end{bmatrix}$$

The professor said that we couldn't equal $x+1=3$, because that would incur in a contradiction, since 1 would be equal to 3 and so forth.

The way we should solve this in his eyes, was to take the determinant of both, so it would give us a simple linear system.

$$(x+1) -15 = 9 - (-2) \\ x = 25$$

So, is this equality even possible? If so, would that mean that if $A=B$, $A_{ij} = B_{ij}$ is not always true?

Or worse, $A = B$ iff $A_{ij} = B_{ij}$ is not true?

I searched for a response in some textbooks, but couldn't really find the answer.

Answer 1

In your example, $A=B$ is not true. Setting $x=25$ ensures that $\det(A)=\det(B)$, but that does not mean that $A=B$.

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As for your question in the title, the answer is yes. If $A=B$, then for every $i,j$, $A_{ij}=B_{ij}$ is true.

Answer 2

As @JohnHughes mentioned, your title question's answer is yes. I think what your professor meant to say, or should have said, is that whereas $x+1=3$ has a root, there is no simultaneous root of $x+1=3,\,10=2$ etc., so the matrices are unequal regardless of $x$. Alternatively, we could say $x+1=3$ is inconsistent with $\det A=\det B$, which is equivalent to $x-14=11$.

Answer 3

An $m\times n$ matrix $A=(a_{ij})$ over a field $K$ can be viewed as a mapping

$A:[m]\times[n]\rightarrow K:(i,j)\mapsto a_{ij}$.

Then two $m\times n$ matrices $A=(a_{ij})$ and $B=(b_{ij})$ are equals iff they are equal as mappings, i.e., for all $(i,j)$, $a_{ij}=b_{ij}$.

The equality of matrices is thus reduced to the equality of mapping.