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 \(1 + 21 + 147 + 7 + 126 + 36 = 328\).

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TopNewestLet \(R_1, R_2, R_3\) 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 \(1 + 21 + 147 + 7 + 126 + 36 = 328\).

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