- If \(R\) is an integral domain and \(A\) is an \(n \times n\) matrix over \(R\), prove that if a system of linear equations \(A x=0\) has a nonzero solution then \(\operatorname{det} A=0\). Is the converse true? What if we drop the assumption that \(R\) is an integral domain?
Solution
1. If \(R\) is an integral domain, \(A\in M_n(R)\), and \(Ax=0\) for some \(0\neq x\in R^n\), then \(\det A=0\).
The adjugate identity \(\operatorname{adj}(A)A=(\det A)I_n\) gives \((\det A)x=\operatorname{adj}(A)Ax=0\). Some coordinate \(x_j\) is nonzero, and \((\det A)x_j=0\) in the domain \(R\) forces \(\det A=0\).
2. Over an integral domain the converse holds: if \(\det A=0\), then \(Ax=0\) has a nonzero solution in \(R^n\).
Let \(F=\operatorname{Frac}(R)\). Since \(\det A=0\), \(A\) is singular over the field \(F\), so \(Ay=0\) for some \(0\neq y\in F^n\). Write \(y_i=a_i/b_i\) with \(a_i,b_i\in R\), \(b_i\neq0\), and put \(d=b_1\cdots b_n\neq0\). Then \(x=dy\in R^n\) is nonzero and \(Ax=d\,Ay=0\).
3. Without the domain hypothesis, a nonzero solution of \(Ax=0\) does not force \(\det A=0\).
Over \(R=\ZZ/4\ZZ\) with \(n=1\), \(A=(2)\) and \(x=(2)\) satisfy \(Ax=4=0\) and \(x\neq0\), while \(\det A=2\neq0\).
4. Over an arbitrary nonzero commutative ring \(R\), \(Ax=0\) has a nonzero solution in \(R^n\) if and only if \(r\det A=0\) for some \(0\neq r\in R\) (McCoy’s theorem).In particular, \(\det A=0\) still implies a nonzero solution.
This is McCoy’s theorem for an \(n\times n\) matrix, whose ideal of \(n\times n\) minors is \((\det A)\). If \(\det A=0\), then \(r=1\) annihilates it.