A Matrices Problem - Cayley-Hamilton or Bash?

Here's a cool problem I came across sometime back, and I haven't been able to solve it yet (let's hope that people at Math SE come up with interesting solutions for it!)

$A$ is a square matrix of order 2, with $|A| ≠ 0$ such that $|A + |A|adj(A)| = 0$, where $|A|$ and $adj(A)$ denote the determinant and adjoint/adjugate of matrix A, respectively. Find $|A - |A|adj(A)| = ?$

Is there any way to avoid a bash, by assuming elements of the matrix and then trying to compute the desired determinant? Would it be possible to proceed using Cayley Hamilton theorem?

Also, I think this result could possibly be generalised to a nxn square matrix. So, please help me find the value of the desired determinant, and also let's collectively investigate this problem for square matrices of higher order - could probably be an interesting generalisation, who knows?

Update: I just solved the problem for a 2x2 square matrix - the required determinant is 4, and the determinant of matrix A is 1. Now, the question that remains is, how do I solve it for an nxn square matrix?

• By adjoint do you mean adjugate? Sep 30, 2018 at 3:03
• Yes! That's what it means. Sep 30, 2018 at 3:14
• By "order 2", do you mean that $A^2 = 1$, or that $A$ is a $2 \times 2$ matrix? And just checking, does $\operatorname{adj}(A) = |A| A^{-1}$? Oct 1, 2018 at 6:05
• 1. 2x2 Matrix 2. Yes Oct 1, 2018 at 9:03
• Are you considering these to be matrices over the real numbers? I can find only two $2 \times 2$ matrices over the reals that satisfy the condition in your question. If you include complex numbers, there are infinite families of matrices, all with differing determinant, satisfying the conditions. Oct 1, 2018 at 13:34

You are reading too much into the question. The result does not generalise well to higher dimensions. Let us consider the $$2\times2$$ case first. Presumably $$A$$ is real. Let $$d=|\det(A)|$$, the absolute value of $$\det(A)$$. The given condition can be rewritten as $$\det(A + d^2A^{-1}) = 0$$, which implies that $$\det(A^2 + d^2 I) = 0$$. This simply means that $$-d^2$$ is a negative eigenvalue of $$A^2$$.

Since $$\det(A^2)=d^2$$, when $$A$$ is $$2\times2$$, the other eigenvalue of $$A^2$$ has to be $$-1$$. Thus $$A^2$$ has two negative real eigenvalues $$-d^2$$ and $$-1$$. Therefore $$A$$ must have a conjugate pair of eigenvalues. Hence $$d=1$$ and also $$\det(A)=|\det(A)|$$. It follows that the characteristic polynomial of $$A^2$$ is $$p(x) = (x+d^2)(x+1) = (x+1)^2$$ and $$\det\left(A - \det(A)\operatorname{adj}(A)\right) =\det(A^{-1})\det(A^2 - d^2I) =\frac1d p(d^2) = 4.$$ This concludes the $$2\times2$$ case. When $$A$$ is at least $$3\times3$$, only what we've said in the first paragraph remains valid. The condition $$\det\left(A + \det(A)\operatorname{adj}(A)\right)$$ is true if and only if $$-d^2$$ is an eigenvalue of $$A^2$$, meaning that $$A$$ is similar to $$\left(\begin{array}{c|c}\pmatrix{0&-d\\ d&0}&\ast\\ \hline0&P^{-1}\end{array}\right)$$ for some real matrix $$P$$ with determinant $$\pm d$$ (unlike the $$2\times2$$ case, $$\det(A)$$ is not necessarily equal to $$|\det(A)|$$ here and its sign is controlled by $$P$$). Therefore \begin{align} \det\left(A - \det(A)\operatorname{adj}(A)\right) &=\det(A-d^2A^{-1})\\ &=4d^2\det\left(P^{-1} - d^2P\right)\\ &=4\det(P)\det\left(I - d^2P^2\right), \end{align} whose value can assume any positive, zero or negative real value. The answer by loup blanc is obtained by using a block-diagonal matrix $$A$$ with $$d=P=\frac1z$$.

Firstly, notice that the condition $$\operatorname{det}(A + \operatorname{det}(A) \operatorname{adj}(A)) = 0$$ is invariant under conjugation $$A \mapsto P A P^{-1}$$, and so it suffices to check one element for each similarity class. Each $$2 \times 2$$ real matrix is similar to either an upper triangular matrix, or a "complex number" matrix.

Upper triangular case

$$A = \begin{pmatrix} \lambda_1 & t \\ 0 & \lambda_2 \end{pmatrix} \in \operatorname{Mat}_{2 \times 2}(\mathbb{R})$$ The condition $$\operatorname{det}(A) \neq 0$$ gives that $$\lambda_1$$ and $$\lambda_2$$ are nonzero. We compute $$A + (\operatorname{det} A)(\operatorname{adj} A) = \begin{pmatrix} \lambda_1 + \lambda_1 \lambda_2^2 & -t\lambda_1 \lambda_2 \\ 0 & \lambda_2 + \lambda_1^2 \lambda_2 \end{pmatrix}$$ which has determinant $$\lambda_1 \lambda_2 (1 + \lambda_1^2) (1 + \lambda_2^2)$$. This determinant being zero forces one of $$\lambda_1, \lambda_2$$ to be zero, which is not allowed. So there are no upper-triangular real matrices $$A$$ satisfying the conditions.

Complex number case

$$A = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \in \operatorname{Mat}_{2 \times 2}(\mathbb{R})$$ In this case the matrix $$A$$ works like the complex number $$z = a + ib$$, and we have $$\operatorname{det}(A) = z \overline{z}$$ and $$\operatorname{adj}(A) = \overline{z}$$, where $$\overline{z} = a - ib$$ is the complex conjugate. So $$\operatorname{det}(A + (\operatorname{det} A)(\operatorname{adj} A)) = |z + z \overline{zz}|^2 = |z|^2 |1 + \overline{z}^2|^2 = 0$$ gives that $$z$$ must be a square root of $$-1$$, i.e. one of the two matrices $$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$

Conclusion

So the equation only has solutions when $$A$$ is an invertible $$2 \times 2$$ matrix with eigenvalues $$\pm i$$. In this case, it is easy to check (using either of the two matrices above) that $$\operatorname{det}(A - (\operatorname{det} A)(\operatorname{adj} A)) = 4$$.

I'm not going to do any of the cases with more dimensions, but perhaps this gives you a way to think about what might happen. For example, every $$3 \times 3$$ real matrix is similar either to an upper triangular matrix, or a "block upper-triangular" matrix where along the diagonal we have a real entry, and then a $$2 \times 2$$ "complex number" entry.

For $$n=3$$, the required result depends on the matrix $$A$$.

For example $$A=\begin{pmatrix}z&0&0\\0&0&1/z\\0&-1/z&0\end{pmatrix}$$ where $$z\in\mathbb{R}\setminus\{0\}$$.