How to explain the calculation of the determinant of a $4\times4$ matrix

Question

In my linear algebra lecture notes, I am studying an example which concerns the calculation of the determinant of a $4 \times 4$ matrix, by first reducing the matrix to upper triangular form. (See page 40 of the PDF document at this link.

I do not yet understand the 'clever' observation which is referred to in the last sentence of the example.
Why is it true that the determinant of the original matrix is equal to the determinant of this smaller $2 \times 2$ matrix?

Answer 1

This is because at the last but one step, the matrix is block-triangular so its determinant is the product of the determinants of the diagonal blocks: $$\begin{vmatrix}1&2&1&1\\ 0&1&1&2\\ 0&0&4&5\\ 0&0&3&1\end{vmatrix} =\begin{vmatrix}1&2\\ 0&1\end{vmatrix} \cdot\begin{vmatrix} 4&5\\ 3&1\end{vmatrix}=\begin{vmatrix} 4&5\\ 3&1\end{vmatrix}$$

Answer 2

See the step before the final step. Expand ( calculate ) the determinant along the first column. You'll see that only one 3*3 determinant survives. Again expand along first column . You'll see that only one 2*2 determinant survives.

Answer 3

Depending on the approach you take to define determinants, you'll look for different explanations.

In case you use the Leibniz formula to define the determinant of a square matrix $A=(a_{ij})_{i,j=1,\dots,n}$ as $$ \det A = \sum_{\sigma\in S_n} \operatorname{sgn} \sigma \prod_{i=1}^n a_{\sigma(i),i}, $$ you can look at the possible $\sigma\in S_n$ that yield a non-zero summand. Since $a_{k,1}=0$ for $k\neq 1$, the permutation $\sigma$ needs to map $1$ to $1$ to produce a non-zero summand. In the next column, we have $a_{k,2}=0$ for $k\notin \{1,2\}$, so $\sigma$ has to map $2$ to either $1$ or $2$, but since already $\sigma(1)=1$ we conclude $\sigma(2)=2$. Hence, we can drop all other summands and only consider permutations of $\{3,4\}$: $$ \det A = \sum_{\rho\in\operatorname{Sym}\{3,4\}} \operatorname{sgn}\rho \underbrace{a_{11} a_{22}}_1 \prod_{i=3}^4 a_{\rho(i),i} = \det \pmatrix{a_{33} & a_{34} \\ a_{43} & a_{44}}. $$

If you already know more elaborate techniques like column expansion or determinants of block triangular matrices, you can of course apply those, as suggested by Bernard and A Googler.