Question

If \(\rm A=\begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 1 \\\ 0 & -2 & 4 \end{bmatrix}\) and 6A^-1 = A² + cA + dI. where A^-1 is inverse of A, I is the identity matrix then (c, d) is
1. (-6, 11)
2. (6, -11)
3. (11, -6)
4. (6, 11)

Answer 1

Correct Answer - Option 1 : (-6, 11)

Concept:

Let A be any square matrix.

Characteristic equation of the matrix is given by: det (A - λI) = 0

 

Calculations:

Given matrix is \(\rm A=\begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 1 \\\ 0 & -2 & 4 \end{bmatrix}\)

A - λI = \(=\begin{bmatrix} 1 - λ & 0 & 0 \\\ 0 & 1- λ & 1 \\\ 0 & -2 & 4- λ \end{bmatrix}\)

Characteristic equation is det (A - λI) = 0

⇒ (1 - λ) × [(1 - λ)(4 - λ) + 2] = 0

⇒ (1 - λ) × [4 - 5λ + λ² + 2] = 0

⇒ (1 - λ) × [6 - 5λ + λ2] = 0

⇒ 6 - 5λ + λ² - 6λ + 5λ² - λ³ = 0

⇒ λ³ - 6λ² + 11λ - 6 = 0 

⇒ λ3 - 6λ2 + 11λ = 6

⇒ λ(λ2 - 6λ + 11) = 6

⇒ (λ2 - 6λ + 11) = 6λ^-1

Put λ = A

⇒ (A² - 6A + 11) = 6A^-1

Given: 6A-1 = A2 + cA + dI

So, A2 + cA + dI = A2 - 6A + 11

Therefore c = -6 and d = 11