USING PROPERTIES OF INVERSE AND ADJOINT OF A MATRIX

Adjoint of matrix :

Let A be a square matrix of order n. Then the matrix of cofactors of A is defined as the matrix obtained by replacing each element aij of A with the corresponding cofactor Aij. The adjoint matrix of A is defined as the transpose of the matrix of cofactors of A. It is denoted by adj A.

For every square matrix A of order n,

A(adj A) = (adj A) A = |A| In

Properties of adjoint of matrix A :

(adj A)-1=adj A-1=1|A|A|adj A| = |A|n-1adj (adj A)=|A|n-2Aadj (𝜆A)=𝜆n-1 adj (A)|adj (adj A)| = |A|(n-1)2(adj A)T = adj ATA-1=±1|adj A|adj A

Inverse of matrix A :

Let A be a square matrix of order n. If there exists a square matrix B of order n such that

AB = BA = In

then the matrix B is called an inverse of A.

  • If a square matrix has an inverse then it is unique.
  • Let A be square matrix of order n. Then A-1 exists if and only is A is non singular.

Properties of inverses of matrices :

If A is non singular, then 

Problem 1 :

If A = 3545x35 and AT = A-1

Then the value of x is

(1)  -4/5  (2)  -3/5  (3)  3 

Given, A = 3545x35 AT = A-1AT = 35x4535A-1 = adj A|A|adj A = 35-45-x35|A| = 352- x× 45= 925 - 4x5|A| = 9 - 20x25A-1 = 35-45-x359 - 20x25A-1= 259 - 20x × 35-45-x35 AT= A-135x4535 =259 - 20x × 35-45-x35 1 = 259 - 20x9 - 20x = 25-20x = 25 - 9-20x = 16x = -1620x = -45

So, the value of x is -4/5.

Hence, option (1) is correct. 

Problem 2 :

If A = 1tan 𝜃2-tan 𝜃21 and AB = I2, then B =
(1) cos2𝜃2 A

                  (3)  (cos2θ)I
(2) cos2𝜃2 AT
(4) sin2𝜃2 A

Solution :

Given, A = 1tan 𝜃2-tan 𝜃21AB = IMultiplying both side by A-1.A-1AB = IA-1IB = IA-1B = A-1A-1 = adj A|A||A| = 1tan 𝜃2-tan 𝜃21|A| = 1 + tan2𝜃2|A|= sec2𝜃2adj (A) = 1-tan 𝜃2tan 𝜃21A-1 = 1sec2𝜃2 1-tan 𝜃2tan 𝜃21= cos2𝜃2 AT
So, the value of B is cos2𝜃2 AT.

Hence, option (2) is correct. 

Problem 3 :

If A = cos 𝜃 sin 𝜃 -sin 𝜃cos 𝜃 and A(adj A) = k 0 0k, then k =

(1)  0  (2)  sin θ  (3)  cos θ  (4)  1

Solution :

Given, A = cos 𝜃 sin 𝜃 -sin 𝜃cos 𝜃A(adj A) = k 0 0k A(adj A) = k1 0 01 = kI [k = |A|]A(adj A) = |A|I A = cos 𝜃 sin 𝜃 -sin 𝜃cos 𝜃|A| = cos2𝜃 + sin2𝜃|A| = 1k = 1

So, the value of k is 1.

Hence, option (4) is correct.

Problem 4 :

If A = 2 3 5-2 be such that 𝜆A-1 =A, then 𝜆 is

(1)  17  (2)  14  (3)  19  (4)  21

Solution :

Given, A = 2 3 5-2𝜆A-1 =AA-1 = adj A|A||A| = 235-2= (-4 - 15)|A|= -19adj A = -2 -3 -52A-1 = 1-19-2 -3 -52= 1192 3 5-2A-1 = 119A19A-1 = A𝜆A-1 =A𝜆 = 19

So, the value of λ is 19.

Hence, option (3) is correct.

Problem 5 :

If adj A = 2 3 4-1 and adj B = 1 -2 -31 then adj(AB) is
(1) -7 -1 7-9
(3) -7 7 -1-9
(2) -6 5 -2-10
(4) -6 -2 5-10

Solution :

Given, adj A = 2 3 4-1 adj B = 1 -2 -31adj(AB) = adj B · adj A= 1 -2 -31 · 2 3 4-1= 2 - 8 3 + 2-6 + 4 -9 - 1adj(AB) = -6 5 -2-10

Hence, option (2) is correct.

Problem 6 :

The rank of the matrix 1 2 342468-1-2-3-4 is

(1)  1  (2)  2  (3)  4  (4)  3

Solution :

Let A = 1 2 342468-1-2-3-4R2 →R2 - 2R1 2 4 6 8 -2 -4 - 6 -8 0 0 0 0 A ∼ 1 2 340000-1-2-3-4 R3 → R3 + R1 -1 -2 -3 -4 1 2 3 4 0 0 0 0 A ∼ 1 2 3400000000

The echelon matrix of A has only one non zero row.

Therefore rank of A = 1.

So, option (1) is correct.

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