Find the number of A = [a<sub>ij</sub>]<sub>2x2</sub> satisfying a<sub>ij</sub> is 1 or -1 and a<sub>11</sub>a<sub>21</sub> + a<sub>12</sub>a<sub>22</sub> = 0.
Step-by-Step Solution
Key Concept: The condition a11a21 + a12a22 = 0 implies that the dot product of the two columns of the matrix is zero. Since each entry is 1 or -1, the product of two entries is always 1 or -1. For the sum to be zero, one product must be 1 and the other must be -1.
Let A = [[a11, a12], [a21, a22]]. The condition is a11a21 + a12a22 = 0. Since aij \in {1, -1}, a11a21 \in {1, -1} and a12a22 \in {1, -1}. For the sum to be 0, one must be 1 and the other -1. There are 2 cases: (1) a11a21 = 1 and a12a22 = -1, or (2) a11a21 = -1 and a12a22 = 1. For each case, there are 2 choices for each pair (e.g., a11=1, a21=1 or a11=-1, a21=-1). Thus, there are 2*2 = 4 ways for case 1, and 4 ways for case 2. Total = 4 + 4 = 8.
Correct Answer: 8