Matrices of non trivial solution

Question

Let set $A$ contain all the (3 × 3) matrices whose entries are either 0, 1 or –1. Two of these entries are 1, two are –1 and five are 0. How many matrices $B \in A$ are there, for which the equation,

$$ B \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] $$ has non-trivial solution?

I calculated the no. of matrices in set A as 9!/(5!2!2!) = 756. Then i assumed B to be matrix having terms $$ \left[ \begin{array}{ccc} a_1 &a_2 &a_3 \\ b_1 &b_2 &b_3\\ c_1 &c_2 &c_3\\ \end{array} \right] $$ Now for a trivial solution determinant should not be zero. Therefore if we find

total matrices - matrices having non zero determinant

we get the required no. of matrices having non trivial solution. I got stuck here trying to find how many matrices would exist. I tried using permutation/combination for the terms but I do not understand what should I be finding. Please help me out

The solution provided which doesnt make sense to me

Answer 1

This questions seems complicated, but the condition on entries has made it much easier.

Entries are either 0,1 or –1. Two of these entries are 1, two are –1 and five are 0

You have to find the number of singular matrices. Total matrices in set $A$ , as you've calculated, are $$\dfrac{9!}{5!2!2!}$$

Since, you have to use 5 zeroes, out of 9 places, Where each row/column having 3 places each. Our possible cases of arranging zeroes reduce significantly.

Also, while calculating the determinant of such matrices, you can always take the negative sign out from a row or column , or multiply a row or column with it, determinant won't change (since there's just zero, one and negative one)

It's easier to calculate the number of non-singular matrices and then subtract them from total. Give it a try, before further reading the answer.

It may seem tedious and impossible to list out all the possible arrangments of zeroes, and even then there might be possiblity of rest of terms (1 and -1) cancelling out on further calculation.

But there are few key points you need to know before starting.

1. A row/column cannot have 3 zeroes. So our only possible arrangments is of kind $(2,2,1)$

2. Since there are just $1$ and $-1$ other zeroes, Either one can be transformed into another by simply multiplying it by $-1$. This restricts the arrangements of elements a lot. That is, even if you have arrangment like $$ \begin{vmatrix} 0 & 0 & -1\\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$$

It's going to be zero. Two rows/columns shouldn't be identical.

3. There will always be atleast, one zero being included while expanding the determinant. So you don't have to worry about $1$ and $-1$

4. Arrange zeroes first.

5. Diagonal, any row or column can't have 3 zeroes.

6. Don't just start randomly arranging zeroes. Start sequentially and systemically.

You don't have to see all possible ways of rows and columns separately, since it's a square matrix and all the arrangments will be included.

As mentioned above, only possible arrangments are of kind $(2,2,1)$ which means we have to list out 3 different cases, each having one row having only 1 zero and rest having exactly two, and they can't be identical.

I started by fixing two zeroes in $a_{11}$ and $a_{12}$ . Now for the next two zeroes, you can't choose $a_{21}, a_{22}$ or $a_{31},a_{32}$

First let's fix the next two zeroes side by side (consecutive columns). So, for second row, only possible way would be \begin{vmatrix} 0 & 0 & \\ & 0 & 0 \\ & & \end{vmatrix}

Note that the last zero can't be in any place except $a_{31}$

That's the only possible arrangment for having zero in $a_{11},a_{12}$ and $a_{22} ,a_{23}$

I'm gonna list out all the possible cases, you should do that by yourself because I may miss a case.

• Two zeroes fixed in first row in first two columns ,two consecutive zeroes in second row

$\begin{vmatrix} 0 & 0 & \\ & 0 & 0 \\ 0 & & \end{vmatrix}$

• Two zeroes fixed in first row in first two columns ,two separated zeroes in second row

$ \begin{vmatrix} 0 & 0 & \\ 0 & & 0 \\ & 0 & \end{vmatrix} $ and $ \begin{vmatrix} 0 & 0 & \\ 0 & & 0 \\ & & 0 \end{vmatrix} $

• Two zeroes fixed in first row in first two columns ,two consecutive zeroes in third row

$\begin{vmatrix} 0 & 0 & \\ & & 0 \\ & 0 & 0 \end{vmatrix}$ and $\begin{vmatrix} 0 & 0 & \\ 0 & & \\ & 0 & 0 \end{vmatrix}$

• Two zeroes fixed in first row in first two columns ,two separated zeroes in third row

$\begin{vmatrix} 0 & 0 & \\ & & 0 \\ 0 & & 0 \end{vmatrix}$

• Two zeroes fixed in first row, in last two columns ,two consecutive zeroes in second row

$\begin{vmatrix} & 0 & 0\\ 0& 0& \\ & & 0 \end{vmatrix}$ and $\begin{vmatrix} & 0 & 0\\ 0& 0& \\ 0& & \end{vmatrix}$

• Two zeroes fixed in first row in last two columns ,two separated zeroes in second row

$\begin{vmatrix} & 0 & 0\\ 0& & 0 \\ & 0 & \end{vmatrix}$ and $\begin{vmatrix} & 0 & 0\\ 0& & 0 \\ 0& & \end{vmatrix}$

• Two zeroes fixed in first row in last two columns ,two consecutive zeroes in third row

$\begin{vmatrix} & 0 & 0\\ & & 0 \\ 0 & 0 & \end{vmatrix}$ and $\begin{vmatrix} & 0 & 0\\ 0& & \\ 0 & 0 & \end{vmatrix}$

• Two zeroes fixed in first row in last two columns ,two separated zeroes in third row

$\begin{vmatrix} & 0 & 0\\ 0& & \\ 0 & & 0 \end{vmatrix}$ and $\begin{vmatrix} & 0 & 0\\ & 0 & \\ 0 & & 0 \end{vmatrix}$

• Two separated zeroes in first row, two consecutive zeroes in second row

$\begin{vmatrix} 0 & & 0\\ 0& 0& \\ & 0 & \end{vmatrix}$ , $\begin{vmatrix} 0 & & 0\\ & 0& 0\\ & 0 & \end{vmatrix}$ and $\begin{vmatrix} 0 & & 0\\ 0& 0& \\ 0 & & \end{vmatrix}$

• Two separated zeroes in first row, two consecutive zeroes in third row

$\begin{vmatrix} 0 & & 0\\ & 0& \\ 0 & 0 & \end{vmatrix}$ , $\begin{vmatrix} 0 & & 0\\ & & 0 \\ 0 & 0 & \end{vmatrix}$ and $\begin{vmatrix} 0 & & 0\\ 0& & \\ & 0 & 0 \end{vmatrix}$

Now only the case in which first row had 1 zero are left.

$\begin{vmatrix} 0 & & \\ 0& & 0 \\ & 0 & 0 \end{vmatrix}$ , $\begin{vmatrix} 0 & & \\ &0 & 0 \\ 0 & 0 & \end{vmatrix}$

$\begin{vmatrix} & 0 & \\ 0& & 0 \\ & 0 & 0 \end{vmatrix}$, $\begin{vmatrix} & 0 & \\ 0& & 0 \\ 0 & 0 & \end{vmatrix}$ , $\begin{vmatrix} & 0 & \\ 0& 0 & \\ 0 & & 0 \end{vmatrix}$ and $\begin{vmatrix} & 0 & \\ &0 & 0 \\ 0 & &0 \end{vmatrix}$

$\begin{vmatrix} & & 0 \\ 0& & 0 \\ 0 & 0 & \end{vmatrix}$ , $\begin{vmatrix} & & 0\\ 0&0 & \\ 0 & & 0 \end{vmatrix}$ , $\begin{vmatrix} & & 0\\ 0&0 & \\ & 0& 0 \end{vmatrix}$ and $\begin{vmatrix} & & 0\\ &0 & 0\\ 0 &0 & \end{vmatrix}$

Sigh so.. these are 30 cases, do count by yourself and see if I've missed any case, or did a counting mistake.

Rest places cn be filled in $$\dfrac{4!}{2!2!}$$

So total singular matrices are $756-6\times30=576$

I know that's a very tiring way, has a risk of missing some cases, double counting, but I avoided using permutations, because it will include duplicate cases.