### JEE Main On 8 April 2017 Question 3

##### Question: Let A be any $ 3\times 3 $ invertible matrix. Then which one of the following is not Always true? [JEE Online 08-04-2017]

#### Options:

A) $ adj(adj(1))=|A{{|}^{2}}.{{(adj(1))}^{-1}} $

B) $ adj(adj(1))=|A|.{{(adj(1))}^{-1}} $

C) $ adj(adj(1))=|A|.A $

D) $ adj(1)=|A|.{A^{-1}} $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

Given that the matrix A is 3x3 invertible matrix then, (adj(A))A=|A|⋅I….(1)

Replacing matrix A by adj(A) in equation (1), we get

(adj(adj(A)))adj(A)=|adj(A)|⋅I….(2)

Since the matrix is 3x3 ⇒ | adj(A) | = $|A|^2$ Equation (2), becomes

(adj(adj(A)))adj(A)=$|A|^2$⋅I….(3)

Since A is invertible matrix implies that adj(A) is a invertible matrix.

Apply $(adj(A))^{−1}$ on both sides of equation (3), we get

(adj(adj(A)))adj(A)$(adj(A))^{−1}$=$|A|^2$ $(adj(A))^{−1}$ Since adj(A)$(adj(A))^{−1}$=I (adj(adj(A)))I=$|A|^2$ $(adj(A))^{−1}$ adj(adj(A))=$|A|^2$ $(adj(A))^{−1}$

From property (1) of adjoint of matrix we get

A(adj(A))=|A|⋅I….(3)

Replacing matrix A by adj(A) in equation (3), we get adj(A)(adj(adj(A)))=|adj(A)|⋅I….(4)

Multiplying equation (4) matrix A from left side, we get A⋅adj(A)(adj(adj(A)))=A{|adj(A)|⋅I} (A⋅adj(A))(adj(adj(A)))=|adj(A)|{A⋅I}

From property (1), we get (|A|⋅I)(adj(adj(A)))=|adj(A)|{A⋅I}

|A|(adj(adj(A)))=|adj(A)|⋅A For 3x3 matrix property (2) is | adj(A) | = $|A|^2$ |A|(adj(adj(A)))=$|A|^2$⋅A By cancelling |A| form both sides, we get adj(adj(A))=|A|⋅A

Since A is invertible implies $A^{-1}$ exist. Applying $A^{-1}$ on both sides of equation (1), we get (adj(A))A⋅$A^{-1}$=|A|⋅$A^{-1}$ Since A⋅$A^{-1}$=I ⇒(adj(A))I=|A|⋅$A^{-1}$ ⇒adj(A)=|A|⋅$A^{-1}$

If the option b is always true implies that option a is always not true

Note: proof of property (2) is given below for nxn matrix.

Let A be any nxn matrix. By property (1), we have A(adj(A))=|A|⋅I

Taking determinants on both sides, we get | A(adj(A)) | = | |A|⋅I | Since det(AB) = det(A)det(B) | A | | adj(A) | = | |A|⋅I | For nxn matrix, det(kA) = kndet(A), where k is constant number | A | | adj(A) | =|A|n | I |

Since the determinant of the identity matrix is 1. | A | | adj(A) | =$A^{n-1}$ | A | | adj(A) | =|A|n By cancelling |A| on both sides, we get | adj(A) | =$A^{n-1}$