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The following is written in the solution of my textbook.

$$|A|= \left| \begin{array} {cccc} 1 & 2& -1& 4 \\ 0& 5& -1& 6 \\ 0& -3& 3& -6 \\ 0& 2& 2& -1\\ \end{array} \right| = \left| \begin{array} {ccc} 5& -1& 6 \\ -3& 3& -6 \\ 2& 2& -1\\ \end{array} \right|$$

where $A=\left[ \begin{array} {cccc} 1 & 2& -1& 4 \\ 0& 5& -1& 6 \\ 0& -3& 3& -6 \\ 0& 2& 2& -1\\ \end{array} \right]$

I can see that we are ignoring column 1 and row 1 of $A$ to compute the determinant.

But I don't understand why this is a valid operation. Can someone please show me what are the properties of determinants/matrices that are being used to justify this?

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2 Answers 2

up vote 7 down vote accepted

The rule for a determinant is a recursive rule, namely the Laplace expansion. Thus, you can show that the determinant is equal to: $$A_{11}\cdot|M_{11}|-A_{21}\cdot|M_{21}|+A_{31}\cdot|M_{31}|-A_{41}\cdot|M_{41}|$$ Where $|M_{ij}|$ is the determinant of the minor - the $(n–1) \times (n–1)$ matrix that results from deleting the i-th row and the j-th column of $A$.

Since $A_{21},A_{31},A_{41}$ are all zero, and $A_{11}=1$, the determinant of $A$ happens to be equal to the determinant of $M_{11}$.

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Please add the alternating sign (checkers board pattern) to your formula. As it is (all plus operations), recursive application would compute the permanent, not the determinant of the matrix. –  LutzL Mar 3 at 8:27
@LutzL - Thanks for the fix, just a copy paste mistake. –  nbubis Mar 3 at 8:35

I am going to try to sharpen your intuition a bit on this. The other answer is completely fine.

Hopefully you are familiar with the "signed sum of permutations" definition of the determinant known as the Leibniz Formula:

$$\det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i = 1}^n a_{\sigma(i), i}$$

Where $\operatorname{sgn}(\sigma)$ is the sign of a permutation (+1 for even, -1 for odd) and $S_n$ is the permutation group on $n$ elements.

Think about what happens when you try to construct a permutation that doesn't include the 1 in the upper left hand corner. Try it a few times to convince yourself.

Any time you select a different column from that first row, it forces you to select a 0 from one of the other rows in order to complete the permutation. This means that these permutations will not affect the sum (multiplying by zero gives zero) and they can be safely ignored.

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Unfortunately, this is the first time I am hearing about the "signed sum of permutations" (I am a first year Maths & Econs undergrad). Nevertheless, thank you for your answer and I will study it during my semester break. –  mauna Mar 3 at 18:53
@mauna okay, for some background then, there are two equivalent ways (actually more than that) of defining the determinant. One of them is the Leibniz formula I have used here. The other main one I can think of is the Laplace expansion in the other answer. –  Tim Seguine Mar 3 at 18:57

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