The positive value of the determinant of the matrix A, whose
Adj(Adj(A)) = ( 14, 28, - 14; - 14, 14, 28; 28, - 14, 14 ), is _____________.

Solution

Maths sir · 4 steps

Do not hunt for A itself. Take the determinant of the given matrix and use the power rule for adjoints.

1. Power rule. For an n × n matrix, |adj A| = |A|ⁿ⁻¹. Apply it twice: |adj(adj A)| = (|A|ⁿ⁻¹)ⁿ⁻¹ = |A|^((n-1)²). Here n = 3, so the power is 4.
2. Take out 14. Every entry is a multiple of 14, so pull 14 out of each of the three rows. The determinant becomes 14³ times a small determinant.

   14³|1, 2, -1; -1, 1, 2; 2, -1, 1| = 14³[1(1+2) - 2(-1-4) - 1(1-2)] = 14³ × 14 = 14⁴

3. Match the powers. So |A|⁴ = 14⁴, which gives |A| = ± 14.
4. So. The positive value of |A| is 14.

The common slip: 196. Using |A|ⁿ⁻¹ = |A|² for the double adjoint, as if only one adjoint were taken, gives |A|² = 14⁴ and so 196, not 14.

The idea to keep. Each adjoint raises the determinant to the power n-1. Two adjoints give |A|^((n-1)²), so for a 3 × 3 matrix that is |A|⁴.

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