Let A be a 3 × 3 real matrix such that
A(1; 0; 1)=2(1; 0; 1), A(-1; 0; 1)=4(-1; 0; 1), A(0; 1; 0)=2(0; 1; 0) .
Then, the system (A-3 I)(x; y; z)=(1; 2; 3) has :

Solution

Maths sir · 3 steps

Each given vector is an eigenvector, so just check whether 3 is an eigenvalue of A. It is not.

1. Read the eigenvalues. Each condition has the form Av = λ v. So A has eigenvalues 2, 4, 2 with eigenvectors (1,0,1), (-1,0,1), (0,1,0). These three vectors are independent, since their determinant is -2 ≠ 0, so these are all the eigenvalues of A.
2. Shift by 3I. If Av = λ v, then (A - 3I)v = (λ - 3)v. So A - 3I has eigenvalues 2-3, 4-3, 2-3, that is -1, 1, -1.

   det(A - 3I) = (-1)(1)(-1) = 1 ≠ 0

3. So. A - 3I is invertible, so (x, y, z)ᵀ = (A - 3I)⁻¹(1, 2, 3)ᵀ is the only solution. The system has a unique solution, option (c).

The idea to keep. For (A - kI)X = B, the determinant of A - kI is the product of (λ - k) over the eigenvalues. If k is not an eigenvalue, the matrix is invertible and the solution is unique.

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