Express $\left|\begin{array}{cc} a_1+b_1 & c_1+d_1\\ a_2+b_2 & c_2+d_2 \end{array}\right|$ as a sum of four determinants who's entries contains no sum

My primitive thoughts on this:

I was thinking that I was able to split the determinants but that was impossible because DET(a+b) is not DET(a)+DET(b)

I really had no clue what to do from here

a different attempt was done as I expanded out the determinant to get the following


I was trying to group the above line so that I was going to create more determinants but that efforts proved to be in vain

and thus I request for some help

  • $\begingroup$ Which definition of determinant do you have? The simplest solution to this problem uses the fact that a determinant is a linear function of each of its rows. $\endgroup$ – Semiclassical May 7 '17 at 4:24
  • $\begingroup$ @Semiclassical im not really familiar with linear algebra but the definition that I'm using is pertaining to matrixes not the calculus one $\endgroup$ – John Rawls May 7 '17 at 4:55

Expanding the determinant and regrouping will get you the correct answer, you just need a little more perseverance! Try taking each positive term of your expansion and grouping it with each negative term of your expansion. Those may be the determinants you want.

Whenever you get stuck at some point like that, clever regrouping or looking at the problem at a different angle is necessary. Take a step back and see which parts may go nicely together. Use your intuition to try to guess the answer, and then back it up with math.

If you are in a class where you are looking at things from a more abstract setting, you may want to use the fact that a determinant is a linear function of its rows, and that for any linear function $T$, $$ T(x + y) = T(x) + T(y) .$$


Here's a way to handle the algebra in an organized way. First, we expand the determinant: $$\begin{vmatrix}a_1+b_1 & c_1+d_1\\ a_2+b_2 & c_2+d_2 \end{vmatrix}=(a_1+b_1)(c_2+d_2)-(a_2+b_2)(c_1+d_1)$$ When we multiply out the first term, we'll end up with four different products depending on whether we're distributing the first or second part of each factor. If we group these contributions from each of the two factors together, we get

$$(a_1 c_2-a_2 c_1)+(b_1 c_2-b_2 c_1)+(a_1d_2-a_2 d_1)+(b_1 d_2-b_2 d_1)$$ which we can recognize as the desired sum of determinants

$$ \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2\end{vmatrix} +\begin{vmatrix} b_1 & c_1 \\ b_2 & c_2\end{vmatrix} +\begin{vmatrix} a_1 & d_1 \\ a_2 & d_2\end{vmatrix} +\begin{vmatrix} b_1 & d_1 \\ b_2 & d_2\end{vmatrix}.$$

  • $\begingroup$ The observant reader will note that this approach would have worked just as well had each matrix element had $n$ terms instead of 2. This reflects the fact that the determinant is a multilinear function of its rows. $\endgroup$ – Semiclassical May 8 '17 at 17:01

$a_1c_2+a_1d_2+b_1c_2+b_1d_2-a_2c_1-a_2d_1-b_2c_1-b_2d_1$ - I think you're on the right track here.

Take, for example, $a_1c_2+a_1d_2$. Look at the definition of the determinant for a $2x2$ matrix and see if there's any connection and resemblance between it and $a_1c_2+a_1d_2$. There is one.

UPD: also remember that $a+b$ is $a -(-b)$.


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