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?

  • $\begingroup$ By adjoint do you mean adjugate? $\endgroup$ Sep 30, 2018 at 3:03
  • $\begingroup$ Yes! That's what it means. $\endgroup$ Sep 30, 2018 at 3:14
  • $\begingroup$ By "order 2", do you mean that $A^2 = 1$, or that $A$ is a $2 \times 2$ matrix? And just checking, does $\operatorname{adj}(A) = |A| A^{-1}$? $\endgroup$
    – Joppy
    Oct 1, 2018 at 6:05
  • $\begingroup$ 1. 2x2 Matrix 2. Yes $\endgroup$ Oct 1, 2018 at 9:03
  • $\begingroup$ 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. $\endgroup$
    – Joppy
    Oct 1, 2018 at 13:34

3 Answers 3


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 $$ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$


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 $z\in\mathbb{R}\setminus\{0\}$.


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