The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is :-
Step-by-Step Solution
Key Concept: A 3x3 matrix with four 1s and five 0s has a determinant that can be calculated by expanding along rows or columns. A matrix is non-singular if its determinant is non-zero. We need to count how many ways to place four 1s such that the determinant is non-zero.
A 3x3 matrix with four 1s and five 0s has determinant 0 if it contains a row or column of all zeros. Since there are only five 0s, it is impossible to have a row or column of all zeros (which would require at least three 0s in a row/column, and we have five 0s total). However, if two rows are identical, the determinant is 0. If two rows are identical, they must be either (1,1,1) or (0,0,0). Since we only have four 1s, we cannot have two rows of (1,1,1). Thus, we check for rows being identical to (0,0,0) or other combinations. By exhaustive counting or properties, it is found that the number of such non-singular matrices is 0, as any 3x3 matrix with only four 1s will have a determinant of 0. Wait, re-evaluating: actually, for a 3x3 matrix with four 1s, the determinant is always 0. Therefore, the number of non-singular matrices is 0. Since 0 is less than 4, option (A) is correct.
Correct Answer: 4