If A be square matrix of order n and k is a scalar, then adj (KA) is:

Engineering

Mathematics

Transpose and Adjoint of a Matrix

Question

K^n–1(adjA)

Kⁿ⁺¹(adjA)

Kⁿ(adjA)

K (adj A)

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Given :- A is a square matrix of order n × n matrix and k is scalar

$(KA)^{- 1} = \frac{adj (KA)}{\det (KA)}$

we know,

$\Rightarrow A^{- 1} = \frac{adj (A)}{| A |}$

⇒ adj(KA) = |KA| (KA)^–1

⇒ adj(KA) = Kⁿ|A| (KA)^–1

$\Rightarrow adj (kA) = k^{n} | A | \frac{A^{- 1}}{A}$

⇒ adj(KA) = K^n–1|A| A^–1

$3 adj (kA) = k^{n - 1} \frac{adj A}{| A |} | A | [∵ A^{- 1} = \frac{adj (A)}{| A |}]$

⇒ adj(KA) = K^n–1adjA