RD Sharma Solutions for Class 12 Maths Exercise 7.1 Chapter 7 Adjoint and Inverse of a Matrix is provided here. The solutions for the exercise wise answers are prepared by the experts at BYJU’S in the best possible way which are easily understandable by students.
The PDF of RD Sharma Solutions for Class 12, Exercise 7.1 of Chapter 7 Adjoint and Inverse of a Matrix can be downloaded from the given links. This exercise consists of two levels according to the increasing order of difficulties. Let us have a look at the important topics covered in this exercise.
- Definition and meaning of adjoint of a square matrix
- The inverse of a matrix
- Some useful results on invertible matrices
- Determining the adjoint and inverse of a matrix
- Determining the inverse of a matrix when it satisfies the matrix equation
- Finding the inverse of a matrix by using the definition of inverse
- Finding a non – singular matrix when adjoint is given
RD Sharma Solutions For Class 12 Maths Chapter 7 Adjoint and Inverse of a Matrix Exercise 7.1:-Download PDF Here
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1. Find the adjoint of each of the following matrices:
Verify that (adj A) A = |A| I = A (adj A) for the above matrices.
Solution:
(i) Let
A =
Cofactors of A are
C11Â = 4
C12 = – 2
C21 = – 5
C22 = – 3
(ii) Let
A =
Therefore cofactors of A are
C11Â = d
C12 = – c
C21 = – b
C22Â = a
(iii) Let
A =
Therefore cofactors of A are
C11 = cos α
C12 = – sin α
C21 = – sin α
C22 = cos α
(iv) Let
A =
Therefore cofactors of A are
C11Â = 1
C12 = tan α/2
C21 = – tan α/2
C22Â = 1
2. Compute the adjoint of each of the following matrices.
Solution:
(i) Let
A =
Therefore cofactors of A are
C11 = – 3
C21Â = 2
C31Â = 2
C12Â = 2
C22 = – 3
C23Â = 2
C13Â = 2
C23Â = 2
C33 = – 3
(ii) Let
A =
Cofactors of A
C11Â = 2
C21Â = 3
C31 = – 13
C12 = – 3
C22Â = 6
C32Â = 9
C13Â = 5
C23 = – 3
C33 = – 1
(iii) Let
A =
Therefore cofactors of A
C11 = – 22
C21Â = 11
C31 = – 11
C12Â = 4
C22 = – 2
C32Â = 2
C13Â = 16
C23 = – 8
C33Â = 8
(iv) Let
A =
Therefore cofactors of A
C11Â = 3
C21 = – 1
C31Â = 1
C12 = – 15
C22Â = 7
C32 = – 5
C13Â = 4
C23 = – 2
C33Â = 2
Solution:
Given
A =
Therefore cofactors of A
C11Â = 30
C21Â = 12
C31 = – 3
C12 = – 20
C22 = – 8
C32Â = 2
C13 = – 50
C23 = – 20
C33 = 5
Solution:
Given
A =
Cofactors of A
C11 = – 4
C21 = – 3
C31 = – 3
C12Â = 1
C22Â = 0
C32Â = 1
C13Â = 4
C23Â = 4
C33Â = 3
Solution:
Given
A =
Cofactors of A are
C11 = – 3
C21Â = 6
C31Â = 6
C12 = – 6
C22Â = 3
C32 = – 6
C13 = – 6
C23 = – 6
C33Â = 3
Solution:
Given
A =
Cofactors of A are
C11Â = 9
C21Â = 19
C31 = – 4
C12Â = 4
C22Â = 14
C32Â = 1
C13Â = 8
C23Â = 3
C33Â = 2
7. Find the inverse of each of the following matrices:
Solution:
(i) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
Now, |A| = cos θ (cos θ) + sin θ (sin θ)
= 1
Hence, A – 1 exists.
Cofactors of A are
C11 = cos θ
C12 = sin θ
C21 = – sin θ
C22 = cos θ
(ii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
Now, |A| = – 1 ≠0
Hence, A – 1 exists.
Cofactors of A are
C11Â =Â 0
C12 = – 1
C21 = – 1
C22Â = 0
(iii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
(iv) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
Now, |A| = 2 + 15 = 17 ≠0
Hence, A – 1 exists.
Cofactors of A are
C11Â = 1
C12Â = 3
C21 = – 5
C22Â = 2
8. Find the inverse of each of the following matrices.
Solution:
(i) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 1(6 – 1) – 2(4 – 3) + 3(2 – 9)
= 5 – 2 – 21
= – 18≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 5
C21 = – 1
C31 = – 7
C12 = – 1
C22 = – 7
C32Â = 5
C13 = – 7
C23Â = 5
C33 = – 1
(ii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 1 (1 + 3) – 2 (– 1 + 2) + 5 (3 + 2)
= 4 – 2 + 25
= 27≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 4
C21Â = 17
C31Â = 3
C12 = – 1
C22 = – 11
C32Â = 6
C13Â = 5
C23Â = 1
C33 = – 3
(iii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 2(4 – 1) + 1(– 2 + 1) + 1(1 – 2)
= 6 – 2
= – 4≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 3
C21Â = 1
C31 = – 1
C12Â = + 1
C22Â = 3
C32Â = 1
C13 = – 1
C23Â = 1
C33Â = 3
(iv) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 2(3 – 0) – 0 – 1(5)
= 6 – 5
= 1≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 3
C21 = – 1
C31Â = 1
C12 = – 15
C22Â = 6
C32 = – 5
C13Â = 5
C23 = – 2
C33Â = 2
(v) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 0 – 1 (16 – 12) – 1 (– 12 + 9)
= – 4 + 3
= – 1≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 0
C21 = – 1
C31Â = 1
C12 = – 4
C22Â = 3
C32 = – 4
C13 = – 3
C23Â = 3
C33 = – 4
(vi) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 0 – 0 – 1(– 12 + 8)
= 4≠0
Hence, A – 1 exists
Cofactors of A are
C11 = – 8
C21Â = 4
C31Â = 4
C12Â = 11
C22 = – 2
C32 = – 3
C13 = – 4
C23Â = 0
C33Â = 0
(vii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
 – 0 + 0
= – (cos2 α – sin2 α)
= – 1≠0
Hence, A – 1 exists
Cofactors of A are
C11 = – 1
C21Â = 0
C31Â = 0
C12Â = 0
C22 = – cos α
C32 = – sin α
C13Â = 0
C23 = – sin α
C33 = cos α
9. Find the inverse of each of the following matrices and verify that A-1A = I3.
Solution:
(i) We have
|A| =
= 1(16 – 9) – 3(4 – 3) + 3(3 – 4)
= 7 – 3 – 3
= 1≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 7
C21 = – 3
C31 = – 3
C12 = – 1
C22Â = 1
C32Â = 0
C13 = – 1
C23Â = 0
C33Â = 1
(ii) We have
|A| =
= 2(8 – 7) – 3(6 – 3) + 1(21 – 12)
= 2 – 9 + 9
= 2≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 1
C21Â = 1
C31 = – 1
C12 = – 3
C22Â = 1
C32Â = 1
C13Â = 9
C23 = – 5
C33 = – 1
10. For the following pair of matrices verify that (AB)-1 = B-1A-1.
Solution:
(i) Given
Hence, (AB)-1 = B-1A-1
(ii) Given
Hence, (AB)-1 = B-1A-1
Solution:
Given
Solution:
Given
Solution:
Given
Solution:
Solution:
Given
A =
 and B – 1 =
Here, (AB) – 1 = B – 1 A – 1
|A| = – 5 + 4 = – 1
Cofactors of A are
C11 = – 1
C21Â = 8
C31 = – 12
C12Â = 0
C22Â = 1
C32 = – 2
C13Â = 1
C23 = – 10
C33Â = 15
(i) [F (α)]-1 = F (-α)
(ii) [G (β)]-1 = G (-β)
(iii) [F (α) G (β)]-1 = G (-β) F (-α)
Solution:
(i) Given
F (α) =
|F (α)| = cos2 α + sin2 α = 1≠0
Cofactors of A are
C11 = cos α
C21 = sin α
C31Â = 0
C12 = – sin α
C22 = cos α
C32Â = 0
C13Â = 0
C23Â = 0
C33Â = 1
(ii) We have
|G (β)| = cos2 β + sin2 β = 1
Cofactors of A are
C11 = cos β
C21Â = 0
C31 = -sin β
C12Â = 0
C22Â = 1
C32Â = 0
C13 = sin β
C23Â = 0
C33 = cos β
(iii) Now we have to show that
[F (α) G (β)] – 1 = G (– β) F (– α)We have already know that
[G (β)] – 1 = G (– β) [F (α)] – 1 = F (– α)And LHS = [F (α) G (β)] – 1
= [G (β)] – 1 [F (α)] – 1
= G (– β) F (– α)
Hence = RHS
Solution:
Consider,
Solution:
Given
Solution:
Given