Let A be a 2 × 2 real matrix and I be the identity matrix of order 2. If the roots of the equation |A-x I|=0 be -1 and 3, then the sum of the diagonal elements of the matrix A² is

Solution

Maths sir · 4 steps

Expand |A - xI| and read off trace and determinant; the trace of A² then follows from them.

1. Expand the determinant. Take A = (a, b; c, d). Then |A - xI| = (a-x)(d-x) - bc = x² - (a+d)x + (ad - bc).
2. Use the roots. The sum of the roots is a + d = -1 + 3 = 2. The product of the roots is ad - bc = (-1)(3) = -3.
3. Diagonal of A squared. Multiplying out, the diagonal entries of A² are a² + bc and bc + d². Their sum can be written using only a + d and ad - bc.

   a² + 2bc + d² = (a+d)² - 2(ad - bc)

4. So. The sum is 2² - 2(-3) = 4 + 6 = 10. As a check, the roots of A²'s equation are (-1)² = 1 and 3² = 9, which also add to 10.

The common slip: 4. Squaring the trace of A gives 2² = 4, not 10. The trace of A² is the sum of the squares of the roots, not the square of their sum.

The idea to keep. For a 2 × 2 matrix, |A - xI| = x² - (trace)x + det A. Sum of roots gives the trace and product gives the determinant, and the trace of A² is the sum of the squares of the roots.

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