Let A be a 3 × 3 real matrix such that
A( 1; 1; 0 ) = ( 1; 1; 0 );A( 1; 0; 1 ) = ( - 1; 0; 1 ) and A( 0; 0; 1 ) = ( 1; 1; 2 ).
If X = (x₁,x₂,x₃)ᵀ and I is an identity matrix of order 3, then the system (A - 2I)X = ( 4; 1; 1 ) has :

Solution

Maths sir · 4 steps

Build A column by column from the given products, then test whether (A-2I)X = b is consistent.

1. Find the columns. Ae₃ = (1, 1, 2)ᵀ is given directly. Since Ae₁ + Ae₃ = (-1, 0, 1)ᵀ, we get Ae₁ = (-2, -1, -1)ᵀ. Since Ae₁ + Ae₂ = (1, 1, 0)ᵀ, we get Ae₂ = (3, 2, 1)ᵀ.

   A = (-2, 3, 1; -1, 2, 1; -1, 1, 2), A - 2I = (-4, 3, 1; -1, 0, 1; -1, 1, 0)

2. Determinant. |A - 2I| = -4(0 - 1) - 3(0 + 1) + 1(-1 - 0) = 4 - 3 - 1 = 0. So there is no unique solution. Either the system has no solution or it has infinitely many.
3. Check consistency. The equations are -4x₁ + 3x₂ + x₃ = 4, -x₁ + x₃ = 1 and -x₁ + x₂ = 1. The last two give x₂ = x₃ = 1 + x₁. Substituting into the first gives -4x₁ + 3 + 3x₁ + 1 + x₁ = 4, that is 4 = 4, which is true for every x₁.
4. So. The first equation is 3 times the third plus the second, and the right side matches (3 × 1 + 1 = 4). With x₁ free, the system has infinitely many solutions.

The idea to keep. When |M| = 0, do not stop there. Check whether the right side satisfies the same dependence as the rows: if it does, there are infinitely many solutions; if not, there is no solution.

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