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# Delirious Determinant

Consider all 3x3 determinants whose each element can be $$1$$ or $$0$$.Find the number of such determinants whose value is $$0$$ when evaluated.

3 years, 5 months ago

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Let $$R_1, R_2, R_3$$ represent the rows of the matrix. If the determinant is zero, then the rows must be linearly dependent. Therefore there exist integers $$a,b,c$$ such that $$a R_1 + b R_2 + c R_3 = 0$$. So I guess we can look at several mutually exclusive cases:

$$(1)$$ All of the rows are zero. There is just 1 matrix like this.

$$(2)$$ Exactly 2 of the rows are zero. We choose which two ($$\binom{3}{2}$$ ways), and there are 7 possibilities for the remaining nonzero row. (the remaining row can't be zero because we have already counted the case where all 3 rows are zero). Therefore there are 21 ways in this case.

$$(3)$$ Exactly one of the rows is zero. We choose which one ($$\binom{3}{1}$$ ways), and there are $$7*7$$ possibilities for the remaining two nonzero rows. Therefore there are 147 ways in this case.

$$(4)$$ None of the rows are zero, but the rows are all equal to each other. Well, there are 7 possibilities for what the row looks like, so there are 7 ways.

$$(5)$$ Exactly 2 of the rows are equal to each other. There are $$\binom{3}{2}$$ ways to choose which two, 7 ways to decide what the two equal rows look like, and 6 ways remaining to decide how the last one looks (because we want it to be different). Therefore, there are 126 ways in this case.

$$(6)$$ The rows are all different, but two of them add to make the third row.

First we choose which two rows add to make the third one ($$\binom{3}{2}$$ ways). Suppose that $$R_a + R_b = R_c$$ where $$a,b,c$$ are the numbers $$1,2,3$$ in some order.

Now we'll need to split this case into a few subcases:

$$(6a)$$ $$R_a$$ has exactly one '1'. There are 3 possibilities: $$(1,0,0),(0,1,0),(0,0,1)$$. Note that if $$R_{a,k} = 1$$, then we must have $$R_{b,k} = 0$$, because the entries can only be 0 and 1 and we assumed that $$R_a + R_b = R_c$$. For the other two digits in $$R_b$$, there are 3 possibilities (we can't have $$R_b = (0,0,0)$$ since that possibility was covered in the previous cases). Since $$R_a + R_b = R_c$$, there is only 1 choice for what $$R_c$$ is. This gives a total of 9 for this subcase.

$$(6b)$$ $$R_a$$ has exactly two '1's. There are three possibilities: $$(1,1,0),(0,1,1),(1,0,1)$$. Now $$R_b$$ can only be $$(0,0,1),(1,0,0),(0,1,0)$$ respectively (i,e, only one choice for $$R_b$$). Again since $$R_a + R_b = R_c$$, there is only 1 choice for what $$R_c$$ is. This gives a total of 3 for this subcase.

Now $$R_a$$ can't be all zeros since this case was covered earlier. It also can't be all ones, because that would force $$R_b$$ to be all zeros.

Therefore, the total of the subcases is 12, and multiplying that by $$\binom{3}{2}$$ gives 36 for case $$(6)$$.

Hence, the total of all the cases is $$1 + 21 + 147 + 7 + 126 + 36 = 328$$.

- 3 years, 5 months ago