Let \[ A(z)=\begin{pmatrix} 4z^2&1&-1\\ -1&2z^2&0\\ 3&0&1 \end{pmatrix}. \] How many distinct values \(z\) satisfy \[ |z|<1 \] and make \(A(z)\) noninvertible?
Solution
1. The determinant of \(A(z)\) is\[ \det A(z) = 8z^4+6z^2+1. \]
Expanding along the first row gives \[ \begin{aligned} \det A(z) &= 4z^2 \begin{vmatrix} 2z^2&0\\ 0&1 \end{vmatrix} - \begin{vmatrix} -1&0\\ 3&1 \end{vmatrix} - \begin{vmatrix} -1&2z^2\\ 3&0 \end{vmatrix}\\ &= 8z^4+1+6z^2. \end{aligned} \]
2. The determinant factors as\[ \det A(z) = (4z^2+1)(2z^2+1). \]
Direct multiplication gives \[ (4z^2+1)(2z^2+1) = 8z^4+6z^2+1. \]
3. The matrix \(A(z)\) is noninvertible exactly for\[ z \in \left\{ \frac{i}{2}, -\frac{i}{2}, \frac{i}{\sqrt2}, -\frac{i}{\sqrt2} \right\}. \]
A square matrix over \(\CC\) is noninvertible exactly when its determinant is zero. By step 2, \[ \det A(z)=0 \] if and only if \[ z^2=-\frac14 \qquad\text{or}\qquad z^2=-\frac12, \] which gives the four displayed values.
4. All four values from step 3 lie in the open unit disk and aredistinct.
Their absolute values are respectively \[ \frac12 \qquad\text{and}\qquad \frac1{\sqrt2}, \] both strictly less than \(1\). The two absolute values are different, and within each pair the two roots are negatives of one another and nonzero, so all four roots are distinct.
5. Therefore the requested number of values is\[ \boxed{4}. \]
Steps 3 and 4 identify exactly four distinct points in \(\abs{z}<1\) at which \(A(z)\) is noninvertible.
6. Q.E.D.
Step 5 is the required count.