# Get the determinant of a block matrix given the submatrices.

I have a matrix $$M$$ that is equal to: $$\begin{bmatrix} 1 & 1 & 0\\ 1 & 0 & 1 \\ 0 & 1 & 1\\ \end{bmatrix}$$

It's easy to compute $$|M| = -2$$, but then given matrix:

$$N = \begin{bmatrix} M & 0 \\ 0 & M \\ \end{bmatrix}$$

What is $$|N|$$? My textbook and multiple sources say it's $$|N| = -1$$. But I couldn't find any explanation as to why. Wikipedia's Determinants page says it's $$det(N) = det(M)det(M)$$ which is 4. That makes sense to me. Why it says -1 in the solutions manual does not. I've even tried expanding the matrix out into a 6x6 matrix and getting the determinant of that (used calculator to make sure), but it's still 4.

$$det(N) = |\begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1\\ \end{bmatrix}| = 4?$$

• You are completely right, $\det(N)=(-2)^2=4$. Did you check the matrix $M$ for typos? – Dietrich Burde Oct 16 '18 at 21:32
• Yes. Very carefully. Thank you for reaffirming my sanity – QuantumHoneybees Oct 16 '18 at 21:53
• Moreover, $|N|=-1$ can’t be correct since that would mean that $|M|$ is imaginary. – amd Oct 16 '18 at 22:47