If $A$ is an invertible matrix, show that $\det(A)$ not equal $0$ and $\det(A^{-1})$ not equal $0$.

My attempt to solve it but I am not sure it is correct:

Given $A$ is an invertible matrix.

So $\det A$ not equal $0$.

Let the inverse of $A$ be denoted by $A^{-1}$.

By the definition of inverse $AA^{-1} =A^{-1}A = I$, the identity matrix.

Taking determinant on both sides we get

$$\det (AA^{-1}) = \det I.$$

So $\det A \det A^{-1} = 1$ (by the properties of determinants).

Hence $\det A^{-1} = 1 / \det A$.

  • $\begingroup$ But $\text{det}(A) = 1/\text{det}(A^{-1}).$ Why do you want to prove it? And the definition of an invertible square matrix is that the determinant is not zero. $\endgroup$ – Alex Silva Jan 5 '15 at 19:32
  • $\begingroup$ I try to prove that detA not equal zero by prove that detA=1/detA^1 , I think with this way I proved that detA not equal zero and also det A^-1 $\endgroup$ – user155971 Jan 5 '15 at 19:34
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    $\begingroup$ @Alex the definition of an invertible matrix is that it has a multiplicative inverse. The determinant criterion is a theorem characterizing the invertible elements. $\endgroup$ – Matt Samuel Jan 5 '15 at 19:39
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    $\begingroup$ @Alex can you show me a one line proof that if a matrix doesn't have an inverse then its determinant is 0? $\endgroup$ – Matt Samuel Jan 5 '15 at 19:43
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    $\begingroup$ @AlexSilva: There are many equivalent definitions of a singular/nonsingular matrix. You need to justify the statement you've made and then also justify that singular implies determinant zero. The difficulty of a student doing either of those steps depends entirely on which definition of singular they're working with. $\endgroup$ – Jim Jan 5 '15 at 20:08

Your proof is quite good. One part is strange namely "Given $A$ is an invertible matrix. So $\det A$ not equal $0$." You are asked to prove this, and you simply assert it. This is in general not good. Do not assert this at this point. Rather go straight to

"Let the inverse of $A$ be denoted by $A^{-1}$." and continue as you do. This is a good way to argue. Only at the end, add one additional remark. Namely:

"Hence $\det(A) \ne 0$, and further $\det A^{-1} = 1 / \det A$." Obeserve that beforehand you did not need the determinat was non-zero, and at that point you get it, since if it were $0$ its product with the determininat of the inverse matrix would be $0$ and yet you showed it must be $1$.


$$|AB|=|A||B|\\AA^{-1}=I\\|AA^{-1}|=|I|\\|A||A^{-1}|=1\\(REAL-NUMBER)*(REAL-NUBER)=1\\|A|\neq 0,|A^{-1}|\neq 0 \\$$if one of two real number was zero ,answer must be zero

so each one , can not be zero


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