Suppose that in the \(3 \times 3\) version of Lights Out,
only the top-left, top-right, and bottom-left squares have the lights on.
Then pressing square 1, square 2, and square \(i\) will solve the puzzle.
What is \(i\)?
(According to our convention,
the three squares in the top row are labelled 1,2,3 starting from the left,
the three squares in the middle row are labelled 4,5,6 starting from the left,
and the three squares in the bottomw row are labelled 7,8,9 starting from the left.)
The answer is 4.
As shown in the notes,
the solution to the \(3\times 3\) version of Lights Out is given by
\(\begin{bmatrix}
x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\\x_8\\x_9
\end{bmatrix}
=
\begin{bmatrix}
b_1+b_3+b_6+b_7+b_8 \\
b_5+b_7+b_8+b_9 \\
b_1+b_4+b_3+b_8+b_9 \\
b_3+b_5+b_6+b_9 \\
b_2+b_4+b_5+b_6+b_8 \\
b_1+b_4+b_5+b_7 \\
b_1+b_2+b_6+b_7+b_9 \\
b_1+b_2+b_3+b_5 \\
b_2+b_3+b_4+b_7+b_9
\end{bmatrix}.\)
Having only the top-left, top-right, and bottom-left squares with the lights on
mean \(b_1=b_3=b_7=1\) and the rest of the \(b\) values zero.
This gives \(x_1 = x_2 = x_4 = 1\).