This is not an assignment or anything from my book, just a pure interest. If I wanted to show this, wouldn't I have to say that the square matrix can be reduced into the identity square matrix and then say that the determinant is not zero. After I say that and use the fact that the only element in the kernel is the zero vector, could I conclude that the matrix is invertible?
[Math] Proving that a square matrix whose kernel is $\{0\}$ is invertible
linear algebra
Related Solutions
I found the answer in this book (in Section $6.4.14$, “Determinants, Ranks and Linear Equations”). I'd tried using a similar Laplace expansion myself but was missing the idea of using the largest dimension at which the minors are not all annihilated by the same non-zero element. I'll try to summarize the argument in somewhat less formal terms, omitting the tangential material included in the book.
Let $A$ be an $m\times n$ matrix over a commutative ring $R$. We want to find a condition for the system of equations $Ax=0$ with $x\in R^n$ to have a non-trivial solution. If $R$ is a field, various definitions of the rank of $A$ coincide, including the column rank (the dimension of the column space), the row rank (the dimension of the row space) and the determinantal rank (the dimension of the lowest non-zero minor). This is not the case for a general commutative ring. It turns out that for our present purposes a useful generalization of rank is the largest integer $k$ such that there is no non-zero element of $R$ that annihilates all minors of dimension $k$, with $k=0$ if there is no such integer.
We want to show that $Ax=0$ has a non-trivial solution if and only if $k\lt n$.
If $k=0$, there is a non-zero element $r\in R$ which annihilates all matrix elements (the minors of dimension $1$), so there is a non-trivial solution
$$A\pmatrix{r\\\vdots\\r}=0\;.$$
Now assume $0\lt k\lt n$. If $m\lt n$, we can add rows of zeros to $A$ without changing $k$ or the solution set, so we can assume $k\lt n\le m$. There is some non-zero element $r\in R$ that annihilates all minors of dimension $k+1$, and there is a minor of dimension $k$ that isn't annihilated by $r$. Without loss of generality, assume that this is the minor of the first $k$ rows and columns. Now consider the matrix formed of the first $k+1$ rows and columns of $A$, and form a solution $x$ from the $(k+1)$-th column of its adjugate by multiplying it by $r$ and padding it with zeros. By construction, the first $k$ entries of $Ax$ are determinants of a matrix with two equal rows, and thus vanish; the remaining entries are each $r$ times a minor of dimension $k+1$, and thus also vanish. But the $(k+1)$-th entry of this solution is non-zero, being $r$ times the minor of the first $k$ rows and columns, which isn't annihilated by $r$. Thus we have constructed a non-trivial solution.
In summary, if $k\lt n$, there is a non-trivial solution to $Ax=0$.
Now assume conversely that there is such a solution $x$. If $n\gt m$, there are no minors of dimension $n$, so $k\lt n$. Thus we can assume $n\le m$. The minors of dimension $n$ are the determinants of matrices $B$ formed by choosing any $n$ rows of $A$. Since each row of $A$ times $x$ is $0$, we have $Bx=0$, and then multiplying by the adjugate of $B$ yields $\det B x=0$. Since there is at least one non-zero entry in the non-trivial solution $x$, there is at least one non-zero element of $R$ that annihilates all minors of size $n$, and thus $k\lt n$.
Specializing to the case $m=n$ of square matrices, we can conclude:
A system of linear equations $Ax=0$ with a square $n\times n$ matrix $A$ over a commutative ring $R$ has a non-trivial solution if and only if its determinant (its only minor of dimension $n$) is annihilated by some non-zero element of $R$, that is, if its determinant is a zero divisor or zero.
The inverse of $A$ is almost certainly not $A$. You have $A^2+2A=-I$, or $A(A+2I)=-I$. Multiplying by $-1$ you get $$A(-A-2I)=I$$
Hence the inverse of $A$ is $-A-2I$.
Best Answer
I would do it using the rank-nullity theorem and properties of linear maps. If $Ker(A) = {0}$ Then the linear map which the matrix represents is injective and has nullity zero. Hence the rank of the matrix is $n$ (if the matrix is $n\times n$,) which means that it is a surjective map. Since the map is linear and a bijection, it is an isomorphism which means it has an inverse. Hence the matrix must be invertible.
(Of course, to say this rigourously you'd have to explicitly state where the map is from and to and all the rest of it, but if you know a little bit about it it's not hard to do.)