Let A=[aᵢⱼ] be a 3 × 3 matrix such that A[0; 1; 0]=[0; 0; 1], A[4; 1; 3]=[0; 1; 0] and A[2; 1; 2]=[1; 0; 0], then a₂₃ equals :

Solution

Maths sir · 4 steps

You only need row 2 of A, so read off the second entry of each output and get three equations.

1. Row times column. The second entry of Ax is row 2 of A dotted with x. Call row 2 (a₂₁, a₂₂, a₂₃). The second entries of the three outputs are 0, 1 and 0.
2. Three equations. From (0,1,0): a₂₂ = 0. From (4,1,3): 4a₂₁ + a₂₂ + 3a₂₃ = 1. From (2,1,2): 2a₂₁ + a₂₂ + 2a₂₃ = 0.
3. Use the third one. With a₂₂ = 0, the third equation gives a₂₁ = -a₂₃. Put this into the second equation.

   4(-a₂₃) + 3a₂₃ = 1 ⇒ -a₂₃ = 1

4. So. a₂₃ = -1, which is option (b).

The idea to keep. When you need just one entry of A, do not find all of A or any inverse. Fix the row it sits in and take that row's entry from each given output.

Answer from the NTA final answer key. Solution by PrepWiser.