Determinants(Ex. 4.5)

Exercise 4.5

Find adjoint of each of the matrices in Exercises 1 and 2

Solun:- We know that the adjoint of the matrix is the transpose of the cofactor matrix.

⇒ Cofactor of 1 = (-1)1+1M11 = 4

⇒ Cofactor of 2 = (-1)1+2M12 = -3

⇒ Cofactor of 3= (-1)2+1M21 = -2

⇒ Cofactor of 4 = (-1)2+2M22 = 1

⇒ Adjoint matrix = Transpose of Cofactor of matrix

Solun:- We know that the adjoint of the matrix is the transpose of the cofactor matrix.

Verify A(adj A) = (adj A)A = |A|I in Exercise 3 and 4

Solun:- We know that the adjoint of the matrix is the transpose of the cofactor matrix.

⇒ A11 = (-1)1+1M11 = -6

⇒ A12 = (-1)1+2M12 = 4

⇒ A21 = (-1)2+1M21 = -3

⇒ A22 = (-1)2+2M22 = 2

⇒ Adjoint matrix = Transpose of Cofactor of matrix

Taking L.H.S.

Taking M.H.S.

Taking R.H.S.

⇒ A(adj A) = (adj A)A = |A|I   (Hence Proved……)

Solun:- We know that the adjoint of the matrix is the transpose of the cofactor matrix.

Taking L.H.S.

Taking M.H.S.

Taking R.H.S.

⇒ A(adj A) = (adj A)A = |A|I   (Hence Proved……)

Find the inverse of each of the matrices (if it exists) given in Exercise 5 to 11.

Solun:- We know that inverse exists if the matrix is nonsingular, i.e. |A|≠0.

So, matrix A is nonsingular then the inverse exists.

We know A-1 = 1/|A|(adj A)

Cofactor matrix = (-1)i+jMij

⇒ A11 = (-1)1+1M11 = 3

⇒ A12 = (-1)1+2M12 = -4

⇒ A21 = (-1)2+1M21 = 2

⇒ A22 = (-1)2+2M22 = 2

⇒ Adjoint matrix = Transpose of Cofactor of matrix

Solun:- We know that inverse exists if the matrix is nonsingular, i.e. |A|≠0.

So, matrix A is nonsingular then the inverse exists.

We know A-1 = 1/|A|(adj A)

Cofactor matrix = (-1)i+jMij

⇒ A11 = (-1)1+1M11 = 2

⇒ A12 = (-1)1+2M12 = 3

⇒ A21 = (-1)2+1M21 = -5

⇒ A22 = (-1)2+2M22 = -1

⇒ Adjoint matrix = Transpose of Cofactor of matrix

Solun:- We know that inverse exists if the matrix is nonsingular, i.e. |A|≠0.

So, matrix A is nonsingular then the inverse exists.

We know A-1 = 1/|A|(adj A)

Cofactor matrix = (-1)i+jMij

Solun:- We know that inverse exists if the matrix is nonsingular, i.e. |A|≠0.

So, matrix A is nonsingular then the inverse exists.

We know A-1 = 1/|A|(adj A)

Cofactor matrix = (-1)i+jMij

Solun:- We know that inverse exists if the matrix is nonsingular, i.e. |A|≠0.

So, matrix A is nonsingular then the inverse exists.

We know A-1 = 1/|A|(adj A)

Cofactor matrix = (-1)i+jMij

Solun:- We know that inverse exists if the matrix is nonsingular, i.e. |A|≠0.

So, matrix A is nonsingular then the inverse exists.

We know A-1 = 1/|A|(adj A)

Cofactor matrix = (-1)i+jMij

Solun:- We know that inverse exists if the matrix is nonsingular, i.e. |A|≠0.

So, matrix A is nonsingular then the inverse exists.

We know A-1 = 1/|A|(adj A)

Cofactor matrix = (-1)i+jMij

 We know that inverse exists if the matrix is nonsingular, i.e. |AB|≠0.

 We know that inverse exists if the matrix is nonsingular, i.e. |A|≠0 and |B|≠0.

⇒ |A| = 1 and |B| = -2

Given Eq. is A2-5A+7I = 0

Taking L.H.S.

L.H.S. = R.H.S.   (Hence proved……)

⇒ A2-5A+7I = 0

Premultiply by A-1

⇒ A-1(A2-5A+7I)=O

⇒ IA-5I+7A-1 = O    (We know AA-1 = I and IA = A)

⇒ A-5I = -7A-1

⇒ A-1 = -1/7(A-5I)

Compare corresponding elements

⇒ 4+a = 0

⇒ a=-4

⇒ 3+a+b=0

⇒ 3-4+b=0

⇒ b=1

⇒ a = - 4 and b=1

Show that A3-6A2+5A+11I=0. Hence find A-1.

Given Eq. is A3-6A2+5A+11I = 0

Taking L.H.S.

L.H.S. = R.H.S.  (Hence Proved……)

⇒ A3-6A2+5A+11I = 0

Premultiply by A-1

⇒ IA2-6AI+5I+11IA-1 = 0   (We know A2I = A2)

⇒ A2-6A+5I+11A-1 = 0

⇒ -11A-1 = A2-6A+5I

⇒ A-1 = -1/11(A2-6A+5I)

Verify that A3-6A2+9A-4I=0 and  Hence find A-1.

Given Eq. is A3-6A2+9A-4I = O

Taking L.H.S.

L.H.S. = R.H.S.  (Hence Proved……)

⇒ A3-6A2+9A-4I = 0

Premultiply by A-1

⇒ IA2-6AI+9I-4IA-1 = 0   (We know A2I = A2)

⇒ A2-6A+9I-4A-1 = 0

⇒ 4A-1 = A2-6A+9I

⇒ A-1 = 1/4(A2-6A+9I)

17. Let A be a nonsingular matrix of order 3x3. Then |adj A| is equal to

(A) |A|      (B) |A|2      (C) |A|3       (D) 3|A|

Solun:- Given A be a nonsingular matrix i.e. |A| is not equal to 0.

We know that A(adj A) = |A|I

Answer is ………….B.

18. If A is an invertible matrix of order 2, then det(A-1) is equal to

(A) det(A)     (B) 1/det(A)     (C) 1    (D) 0

Solun:- We know that AA-1 = I

⇒ |A| |A-1| = |I|

⇒ det(A-1) = 1/det(A)

Answer is…………B

Download PDF of Exercise 4.5

See Also:-

Notes of Determinants

Exercise 4.1

Exercise 4.3

Exercise 4.4

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