Determinant without computing

Question

How can I show without computing the determinant that the equation is true?

$$\det \begin{pmatrix} b1 + c1 & c1 + a1 & a1 + b1\\ b2 + c2 & c2 + a2 & a2 + b2\\ b3 + c3 & c3 + a3 & a3 + b3 \end{pmatrix} = 2 \det \begin{pmatrix} a1 & b1 & c1\\ a2 & b2 & c2\\ a3 & b3 & c3 \end{pmatrix}$$

I added latex to your question. Please check that I "translated" it properly. — xavierm02, Dec 06 '13 at 22:33

score 5 · Accepted Answer · answered Dec 06 '13 at 22:36

Hints:

(1) Determinant is a multilinear function. For example

$$\det\begin{pmatrix}a+b&c&d\\a'+b'&c'&d'\\a''+b''&c''&d''\end{pmatrix}=\det\begin{pmatrix}a&c&d\\a'&c'&d'\\a''&c''&d''\end{pmatrix}+\det\begin{pmatrix}b&c&d\\b'&c'&d'\\b''&c''&d''\end{pmatrix}$$

and likewise for each column/row.

(2) The determinant is an alternating function, i.e.: if a matrix has two equal columns/rows then its determinant is zero and if we interchange two rows/columns the determinant is multiplied by $\;-1\;$ .

score 0 · Answer 2 · answered Dec 06 '13 at 22:35

0

Try expanding by multilinearity. The determinant is linear in each column (or in each row, if you prefer).

answered Dec 06 '13 at 22:35

Eric Auld

28,127

@arbautjc I don't mean expanding by cofactors...see Don's answer. – Eric Auld Dec 06 '13 at 22:38
Ah, ok, didn't understand. I'll remove the comment, and upvote in few hours after vote blocking ;-) – Jean-Claude Arbaut Dec 06 '13 at 22:39
@arbautjc Don't worry about it. :) – Eric Auld Dec 06 '13 at 22:39

Determinant without computing

2 Answers2