Two non zero distinct numbers a and b are as used elements to make determinants of the third order. The number of determinant whose value is zero for all a,b is:-

my attmept: Determinant of a 3x3 matrix is zero when either any two rows or columns are identical or all the three rows or columns are identical. It also happens when all the elements in any row are 0 but this is ruled out here.

Each element in a row has two choices (a or b) and it implies there are 2∗2∗2 =8 ways to fill a row in a determinant. So, there are a total of 8∗8∗8 =512 different determinants.

Number of determinants such that each row is unique can be formed by selecting 3 combinations out of 8 combinations to fill a row and so, the number of such determinants = 8𝐶3∗3𝑃3 =336 So, the total number of 3x3 determinants with at least two identical rows or columns = 512 - 336 = 176

Therefore, total number of determinants whose value is 0 is 176 but the Answer given is (32)


The given answer is certainly wrong, but you've overlooked some cases too. If one of the rows is $aaa$ and another row is $bbb$ these two rows are linearly dependent and the determinant is $0$. Also, consider the matrix $$\begin{bmatrix}b&b&b\\a&b&a\\b&a&b\end{bmatrix}$$ The determinant is $0$ since the sum of the last two rows is a constant multiple of the first row.

I get $248$ by running the problem through sympy:

from sympy import symbols, Matrix
from itertools import product

a,b = symbols('a,b')
count = 0
for p in product({a,b}, repeat = 9):
    M = Matrix([p[:3],p[3:6],p[6:]])
    if M.det().expand() == 0:
        count += 1

The program prints $248.$ (It also prints out all the singular matrices, but I don't suppose anyone wants to see that.)

I agree with your calculation of $176.$ I haven't tried to verify that the two other cases I've identified account for all $72$ missing singular matrices. I leave that for you.


I have verified that these two cases account for the other $72$ singular matrices ($36$ in each case) but I'll let you fill in the details. They're easy.


To count the case where one row is $aaa$ and another $bbb$ note that the third row must be different from $aaa$ and $bbb$ or we would have counted the matrix already. That gives $6$ choices for the third row, and then there are $3!$ ways to order them, giving $36.$

For the second case, we have one row with two $a$s and a $b$. There are three possible such rows. Once we have chosen this row, the row with two $b$s and one $a$ is determined since the sum of the two rows must be a constant vector. The third row must be $aaa$ or $bbb$ so we have $6$ choices, and $36$ matrices as before.

  • $\begingroup$ will need counting method. $\endgroup$ – maveric Apr 23 '19 at 8:30
  • $\begingroup$ some body please count. $\endgroup$ – maveric Apr 23 '19 at 10:47

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