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Having a bit of an issue answering this question. The answer seems simple but I am quite unsure. I think the answer is true, as it is a true/false question. The question is:

If a square matrix B is invertible, then its inverse is also invertible - True or False? This is all the information I have. I do not know how to go about this, i've just been trying to read laws on invertible matrices but I am really confused. Going to extra classes tomorrow for it, as I said I joined the class late and now I suddenly have an assignment to hand in, so I was just looking for an answer!

Please help me out! Thanks

Saadiq
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4 Answers4

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A matrix $B$ is invertible if there exists a matrix $X$ so that $BX=XB=I$; That matrix $X$ is then denoted $B^{-1}$.

Now can you find a matrix $X$ so that $B^{-1}X=XB^{-1}=I$?

celtschk
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  • I really am so confused, I haven't studied this section yet. Got admitted to the course late (doing correspondence) and have an assignment due on it today. I understand the concept of an invertible matrix. However, what you said doesn't make much sense to me. – Saadiq Aug 12 '16 at 08:38
  • Where exactly is your problem? That is, at which point does it stop making sense to you? – celtschk Aug 12 '16 at 08:40
  • It makes sense to me until the last line (now can you find...) – Saadiq Aug 12 '16 at 09:19
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    Well, your task is to find a matrix $X$ that fulfils the equation. Here's another hint: In the first sentence of the first paragraph, replace $X$ by $B^{-1}$. Do you now see it? – celtschk Aug 12 '16 at 10:09
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I will assume the definition that an $n \times n$ matrix $A$ is invertible if there exists an $n \times n$ matrix $B$ such that $AB = BA = I$. (As mentioned by Ryo, it is in fact sufficient that $AB = I$ to get $BA = I$ as a corallary, but you won't need this if you assume the definition with $AB = BA = I$, so just forget about it for now.)

The reason we call this matrix 'the' inverse of $A$ is because there can only be one matrix $B$ satisfying these conditions. Indeed: suppose $B_1$, $B_2$ satisfy $AB_1 = AB_2 = B_1A = B_2A = I$. Then we have $$ B_1 = IB_1 = B_2AB_1 = B_2I = B_2. $$ Therefore it makes sense to call a matrix $B$ 'the' inverse of $A$ as soon as it satisfies $AB = BA = I$. It is completely defined by just these two equalities!

Okay, so suppose $B$ is the inverse of $A$. To show that $B$ is invertible, we need to find a matrix $C$ such that $BC = CB = I$. But we already know such a matrix: $A$! That's why $B$ (the inverse of $A$) is invertible and in fact that the inverse of $B$ (i.e. the inverse of the inverse of $A$) is equal to $A$.

Bib-lost
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Alternatively (to the approach above), consider the equivalent expressions (By definition of a matrix inverse):

$B^{-1} * (B^{-1})^{-1} = I$

Because both sides are equivalent,

$B * (B^{-1} * (B^{-1})^{-1}) = B * I$

I'll give you the right half, which simplifies by the definition of the identity matrix.

$ ??? = B $

For the left half, what property of invertible matrices & matrix multiplication allows you to declare that

$ ??? = B $

Rdesmond
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  • Though I'm not sure that the "Because both sides are equivalent" works. I assumed that, since the right half works, because the objects are equal, that the left side is valid as well. – Rdesmond Aug 12 '16 at 08:59
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True.

Definition: Matrix $B$ is said to be invertible iff there exists a matrix $X$ such that $BX = XB = I$, and in this case, $X$ is called an inverse of $B$.

Now, let $B$ be a square matrix. Assume that there exists a square matrix $X$ such that $BX = I$. According to the the theorem proven here

Matrices: left inverse is also right inverse?

$XB = I$. Therefore, $B$ is invertible and $X$ is an inverse of it.

Now, from Definition, $X$ is invertible and $B$ is an inverse of it.

Therefore, an inverse is invertible, too.

(I haven't proved the uniqueness of inverse, so I kept saying "an inverse of", but uniqueness can be proved easily, I think.)

Community
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Ryo
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