# Prove that the determinant is $0$ by expressing as a product

I need to prove that the determinant $$\begin{vmatrix} my+nz & mq-nr & mb+nc \\ kz-mx & kr-mp & kb-ma \\ nx+ky & np+kq & na+kb \end{vmatrix}=0$$

In my book it is given as hint that the determinant can be expressed as a product of two other determinants whose value will evaluate to $0$.But I'm not being able to express the given determinant as a product of two other determinants.How should I do it?Please guide me through the procedure.

• Can you tell me how to use that rule to decompose the determinant?I read that wiki...but can't figure it out :/ @DietrichBurde – user220382 Jan 5 '16 at 16:02
• I suspect that the $-$ in the first row, second column should be $+$. And the $kb-ma$ in the second row, third column should be $kc-ma$. If that's the case, it can be written as a product of two determinants. – KittyL Jan 5 '16 at 16:15
• I nearly manage to write it as the product of two matrices, but some of the signs are wrong, and I think that for the given matrix is just won’t work. Mathematica also seems to agree with @DietrichBurde that the determinant is not necessarily zero, so I suspect that there is some sign error. – Jendrik Stelzner Jan 5 '16 at 16:16
• @KittyL I guess there is a printing mistake in my book then...!BTW for that case you stated can you please add how to decompose the determinant? – user220382 Jan 5 '16 at 16:18
• @SanchayanDutta: I posted my way of getting this matrix, until the point where the signes don’t work. – Jendrik Stelzner Jan 5 '16 at 16:23

Assuming there is the typo in your post pointed out by KittyL in the comment section, notice that: $$\begin{pmatrix}my-nz&mq+nr&mb+nc\\kz-mx&kr-mp&kc-ma\\nx+ky&np+kq&na+kb\end{pmatrix}=\begin{pmatrix}m&n&0\\0&k&-m\\k&0&n\end{pmatrix}\times\begin{pmatrix}y&q&b\\z&r&c\\x&p&a\end{pmatrix}.$$

• Great!Alright,now how to do it manually?I can't figure out! :-( – user220382 Jan 5 '16 at 16:21
• @SanchayanDutta: Now use the fact that $\det(AB)=\det(A)\det(B)$. – KittyL Jan 5 '16 at 16:22
• I will add that you may want to expand the determinant of the first matrix with respect of its first line. – C. Falcon Jan 5 '16 at 16:25

As already pointed out there is probably some sign error in the given matrix. Otherwise one could decompose this matrix as follows:

Notice that the first column contains $x,y,z$, the second $p,q,r$ and the third $a,b,c$. So it makes sense for the second matrix to be $$B = \begin{pmatrix} x & p & a \\ y & q & b \\ z & r & c \end{pmatrix},$$ where we may have to change the signs of the entries. (We can actually assume w.l.o.g. that the signs in the first column of $B$ are already right.) Using the first column of your matrix we find that the first matrix must be $$A = \begin{pmatrix} 0 & m & n\\ -m & 0 & k\\ n & k & 0 \end{pmatrix}.$$ The problem is now that we still need to fix the signs in the second and third column of $B$, i.e. we need $$\begin{pmatrix} my + nz & mq - nr & mb + nc \\ kz - mx & kr - mp & kb - ma \\ nx + ky & np + kq & na + kb \end{pmatrix} = \begin{pmatrix} 0 & m & n\\ -m & 0 & k\\ n & k & 0 \end{pmatrix} \begin{pmatrix} x & ?p & ?a \\ y & ?q & ?b \\ z & ?r & ?c \end{pmatrix}.$$ The problem is that currently the sign of $r$ must be both $+$ and $-$.

Notice also that $\det(A) = 0$, which is what we want.

PS: Using KittyL’s suggestion we actually want to work with $$\begin{pmatrix} my + nz & mq \color{red}{+} nr & mb + nc \\ kz - mx & kr - mp & k\color{red}{c} - ma \\ nx + ky & np + kq & na + kb \end{pmatrix}.$$ Then the matrix $A$ stays the same, but we can now solve the sign problem with $$\begin{pmatrix} my + nz & mq + nr & mb + nc \\ kz - mx & kr - mp & kc - ma \\ nx + ky & np + kq & na + kb \end{pmatrix} = \begin{pmatrix} 0 & m & n\\ -m & 0 & k\\ n & k & 0 \end{pmatrix} \begin{pmatrix} x & p & a \\ y & q & b \\ z & r & c \end{pmatrix}.$$

Using the rule of Sarrus I get the result that $$\det(A)=kn(2amry + 2anrz - bkqz - bkry + 2bkrz - bmpy + bmqx - 2bmrx - b npz - bnrx + ckqz - ckry + cmpy - cmqx + cnpz - cnrx)$$ which is not identically zero. So, if I am not mistaken, something has to be changed in the question.

As Dietrich Burde mentions (I didn't know this had a name, thank you!), use the Rule of Sarrus (which works ONLY for $3 \times 3$ matrices).

Here's the method:

1. Replicate the first two columns.

$$\begin{bmatrix} my + nz & mq - nr & mb + nc & my + nz & mq - nr \\ kz - mx & kr - mp & kb - ma & kz - mx & kr - mp\\ nx + ky & np + kq & na + kb & nx + ky & np + kq \end{bmatrix}$$

1. Find the product of each individual cell in all length-3 diagonals starting from the top row to the bottom row, and add them.

These length 3-diagonals going from the top row to the bottom row are given by the following:

$$\begin{bmatrix} \color{blue}{my + nz} & mq - nr & mb + nc & my + nz & mq - nr \\ kz - mx & \color{blue}{kr - mp} & kb - ma & kz - mx & kr - mp\\ nx + ky & np + kq & \color{blue}{na + kb} & nx + ky & np + kq \end{bmatrix}$$

$$\begin{bmatrix} my + nz & \color{blue}{mq - nr} & mb + nc & my + nz & mq - nr \\ kz - mx & kr - mp & \color{blue}{kb - ma} & kz - mx & kr - mp\\ nx + ky & np + kq & na + kb & \color{blue}{nx + ky} & np + kq \end{bmatrix}$$

$$\begin{bmatrix} my + nz & mq - nr & \color{blue}{mb + nc} & my + nz & mq - nr \\ kz - mx & kr - mp & kb - ma & \color{blue}{kz - mx} & kr - mp\\ nx + ky & np + kq & na + kb & nx + ky & \color{blue}{np + kq} \end{bmatrix}$$ which gives $(my+nz)(kr-mp)(na+kb)+(mq-nr)(kb-ma)(nx+ky)+(mb+nc)(kz-mx)(np+kq)$.

1. Find the product of each individual cell in all length-3 diagonals starting from the bottom row to the top row, and add them.

$$\begin{bmatrix} my + nz & mq - nr & \color{red}{mb + nc} & my + nz & mq - nr \\ kz - mx & \color{red}{kr - mp} & kb - ma & kz - mx & kr - mp\\ \color{red}{nx + ky} & np + kq & na + kb & nx + ky & np + kq \end{bmatrix}$$

$$\begin{bmatrix} my + nz & mq - nr & mb + nc & \color{red}{my + nz} & mq - nr \\ kz - mx & kr - mp & \color{red}{kb - ma} & kz - mx & kr - mp\\ nx + ky & \color{red}{np + kq} & na + kb & nx + ky & np + kq \end{bmatrix}$$

$$\begin{bmatrix} my + nz & mq - nr & mb + nc & my + nz & \color{red}{mq - nr} \\ kz - mx & kr - mp & kb - ma & \color{red}{kz - mx} & kr - mp\\ nx + ky & np + kq & \color{red}{na + kb} & nx + ky & np + kq \end{bmatrix}$$ which gives $(nx+ky)(kr-mp)(mb+nc)+(np+kq)(kb-ma)(my+nz)+(na+kb)(kz-mx)(mq-nr)$.

1. Take what you get in step 2, and subtract what you get in step 3.

So, the determinant is given by $$(my+nz)(kr-mp)(na+kb)+(mq-nr)(kb-ma)(nx+ky)+(mb+nc)(kz-mx)(np+kq)-\left[(nx+ky)(kr-mp)(mb+nc)+(np+kq)(kb-ma)(my+nz)+(na+kb)(kz-mx)(mq-nr)\right]\text{.}$$

• The OP says he has to prove that this is identically zero. But this is not the case, isn't it ? – Dietrich Burde Jan 5 '16 at 16:18
• @DietrichBurde It sure doesn't look like it. If I had a CAS on me right now, I'd check – Clarinetist Jan 5 '16 at 16:19
• Just enter a few numbers, say $n:=1; m:=1;k:=-1; r:=0; b:=1; p:=1; y:=1; c:=0;x:=0; z:=0;$. then $\det(A)=1$. – Dietrich Burde Jan 5 '16 at 16:21