JEE Main 2020 · 4 September · evening · Single correct · +4 / −1
Suppose the vectors x₁, x₂ and x₃ are the
solutions of the system of linear equations,
Ax = b when the vector b on the right side is equal to b₁, b₂ and b₃ respectively. if
x₁ = [ 1; 1; 1 ], x₂ = [ 0; 2; 1 ], x₃ = [ 0; 0; 1 ]
b₁ = [ 1; 0; 0 ], b₂ = [ 0; 2; 0 ] and b₃ = [ 0; 0; 2 ],
then the determinant of A is equal to :
- (a) 3/2
- (b) 4
- (c) 2
- (d) 1/2
Answer and solution
Answer (c) 2
Solution
Maths sir · 4 steps
Stack the three solutions side by side and three equations become one matrix equation, AX = B.
- Combine the systems. Put x₁, x₂, x₃ as the columns of a matrix X, and b₁, b₂, b₃ as the columns of B. Since Ax₁ = b₁, Ax₂ = b₂ and Ax₃ = b₃, column by column this is AX = B.
det A · det X = det B
- Find det X. X = (1, 0, 0; 1, 2, 0; 1, 1, 1) is lower triangular, so its determinant is the product of the diagonal: 1 × 2 × 1 = 2.
- Find det B. B = (1, 0, 0; 0, 2, 0; 0, 0, 2) is diagonal, so det B = 1 × 2 × 2 = 4.
- So. det A = (det B)/(det X) = 4/2 = 2, which is option (c).
The idea to keep. When one matrix A acts on several vectors, line them up as columns: AX = B. Then det A = det B / det X, and you never have to find A itself.
Answer from the NTA final answer key. Solution by PrepWiser.