# How many invertible 3x3 matrices?

How many invertible $3 \times 3$ matrices exist over $2$-element field?

Obviously if some field has only $2$ elements, those elements must be $0$ and $1$. A matrix is invertible if and only if its determinant is non zero. I think it will be easier to find matrices which determinants are $0$ and subtract that number from $2^9$ (total number of matrices we can build over that field).

Is there any "clever" way of solving this problem?

$2^3-2^0=7$ choices for the first row as nozero vector. Then $2^3-2^1=6$ choices for the second row vector not in the span of the first. Then $2^3-2^2=4$ choices for the third row not in the span of the first two.

• Wouldn't the second choice be only 5 ? Zero is exclude d, the first row is excluded, and the first row + first row is excluded. – Shailesh Oct 7 '15 at 10:31
• @Shailesh There cannot be first row + first row, since field has only two elements I think – luka5z Oct 7 '15 at 10:32
• @luka5z. Yes.Absolutely. – Shailesh Oct 7 '15 at 10:36

Here we are finding invertible matrices:

None of columns can be entirely zero and the columns must linearly independent. We can construct the first in $2^3$ ways but $(0,0,0)$ is included so we are left with $8-1=7$ ways for the first column. Now again we can construct the second column in $2^3$ ways but $(0,0,0)$ and the arrangement of type of first column should be excluded for linear independence; thus for second column we are left with $8-2=6$ ways; similarly for third column we have $8-1{(0,0,0)}-1$(first column)$-1$ (second column)$-1$ (span of first two columns)$= 4$.

Thus, the total no. of ways = total number of invertible matrices = 7×6×4 = 168.

• Why the 1, for the span of first two columns? Some more detail there would be appreciated, possibly explaining in a way without using 'span' – arya_stark Apr 18 at 18:05