Exercise 1.5 Solution
- Chapter No.1 Introduction to Biology
- Chapter No. 2 Solving a Biological Problem
- Chapter No.3 Biodiversity
- Chapter No.4 Cells and Tissues
- Chapter No.5 Cell Cycle
Q.1- Find the determinant of following matrices;
A = [[- 1, 1], [2, 0]]
B = [[1, 3], [2, – 2]]
C = [[3, 2], [3, 2]]
D = [[3, 2], [1, 4]]
Solution:
To find the determinant of a 2×2 matrix, you can use the following formula:
For a 2×2 matrix A = [[a, b], [c, d]], the determinant (det) is given by:
det(A) = ad – bc
Let’s find the determinants of the given matrices:
A = [[-1, 1], [2, 0]]
det(A) = (-1 * 0) – (1 * 2)
det(A) = 0 – 2
det(A) = -2
B = [[1, 3], [2, -2]]
det(B) = (1 * -2) – (3 * 2)
det(B) = -2 – 6
det(B) = -8
C = [[3, 2], [3, 2]]
det(C) = (3 * 2) – (2 * 3)
det(C) = 6 – 6
det(C) = 0
D = [[3, 2], [1, 4]]
det(D) = (3 * 4) – (2 * 1)
det(D) = 12 – 2
det(D) = 10
To summarize:
det(A) = -2
det(B) = -8
det(C) = 0
det(D) = 10
Q.2- Find which of the following matrices are singular or non-singular?
A = [[3, 6], [2, 4]]
B = [[4, 1], [3, 2]]
C = [[7, – 9], [3, 5]]
D = [[5, – 10], [- 2, 4]]
Solution:
A matrix is singular if its determinant is equal to zero (det(A) = 0). Otherwise, it is non-singular.
Let’s find the determinants of the given matrices:
A = [[3, 6], [2, 4]]
det(A) = (3 * 4) – (6 * 2)
det(A) = 12 – 12
det(A) = 0
B = [[4, 1], [3, 2]]
det(B) = (4 * 2) – (1 * 3)
det(B) = 8 – 3
det(B) = 5
C = [[7, -9], [3, 5]]
det(C) = (7 * 5) – (-9 * 3)
det(C) = 35 + 27
det(C) = 62
D = [[5, -10], [-2, 4]]
det(D) = (5 * 4) – (-10 * -2)
det(D) = 20 – 20
det(D) = 0
To summarize:
Matrix A: det(A) = 0 → Singular
Matrix B: det(B) = 5 → Non-singular
Matrix C: det(C) = 62 → Non-singular
Matrix D: det(D) = 0 → Singular
So, matrices A and D are singular, while matrices B and C are non-singular.
Q.3- Find the multiplicative inverse (if it exists) of each.
A = [[- 1, 3], [2, 0]]
B = [[1, 2], [- 3, – 5]]
C = [[- 2, 6], [3, – 9]]
D = [[1/2, 3/4], [1, 2]]
Solution:
To find the multiplicative inverse (also called the inverse or reciprocal) of a square matrix, we need to check if the matrix is invertible (non-singular) and then compute its inverse.
A square matrix A is invertible if its determinant (det(A)) is non-zero. If the determinant is zero, the matrix is singular, and its multiplicative inverse does not exist.
If A is invertible, its multiplicative inverse (A^(-1)) is given by:
A^(-1) = (1/det(A)) * adj(A)
Where adj(A) is the adjugate (adjoint) of matrix A, and det(A) is the determinant of matrix A.
Let’s find the multiplicative inverses (if they exist) for the given matrices:
A = [[-1, 3], [2, 0]]
det(A) = (-1 * 0) – (3 * 2) = 0 – 6 = -6
Since det(A) ≠ 0, the matrix A is invertible.
The inverse of A is:
A^(-1) = (1/det(A)) * adj(A)
Adj(A) = [[0, -3], [-2, -1]] (interchanging the elements on the main diagonal and changing the sign of the off-diagonal elements)
A^(-1) = (1/(-6)) * [[0, -3], [-2, -1]]
A^(-1) = [[0, 1/2], [1/3, 1/6]]
B = [[1, 2], [-3, -5]]
det(B) = (1 * -5) – (2 * -3) = -5 + 6 = 1
Since det(B) ≠ 0, the matrix B is invertible.
The inverse of B is:
B^(-1) = (1/det(B)) * adj(B)
Adj(B) = [[-5, -2], [3, 1]] (interchanging the elements on the main diagonal and changing the sign of the off-diagonal elements)
B^(-1) = (1/1) * [[-5, -2], [3, 1]]
B^(-1) = [[-5, -2], [3, 1]]
C = [[-2, 6], [3, -9]]
det(C) = (-2 * -9) – (6 * 3) = 18 – 18 = 0
Since det(C) = 0, the matrix C is singular, and its multiplicative inverse does not exist.
D = [[1/2, 3/4], [1, 2]]
det(D) = (1/2 * 2) – (3/4 * 1) = 1 – 3/4 = 1/4
Since det(D) ≠ 0, the matrix D is invertible.
The inverse of D is:
D^(-1) = (1/det(D)) * adj(D)
Adj(D) = [[2, -3/4], [-1, 1/2]] (interchanging the elements on the main diagonal and changing the sign of the off-diagonal elements)
D^(-1) = (1/(1/4)) * [[2, -3/4], [-1, 1/2]]
D^(-1) = 4 * [[2, -3/4], [-1, 1/2]]
D^(-1) = [[8, -3], [-4, 2]]
To summarize:
A^(-1) = [[0, 1/2], [1/3, 1/6]]
B^(-1) = [[-5, -2], [3, 1]]
C is singular, and its multiplicative inverse does not exist.
D^(-1) = [[8, -3], [-4, 2]]
Q.4- If A = [[1, 2], [4, 6]] B = [[3, – 1], [2, – 2]] then A(Adj A) = (Adj A) A = (det A)I (ii) B * B ^ – 1 = I = B ^ – 1 *B
Solution:
Let’s perform the calculations step by step to verify the given properties:
Given matrices:
A = [[1, 2], [4, 6]]
B = [[3, -1], [2, -2]]
(i) A(Adj A) = (Adj A)A = (det A)I
Find the determinant of matrix A:
det(A) = (1 * 6) – (2 * 4) = 6 – 8 = -2
Find the adjugate (adjoint) of matrix A:
Adj(A) = [[6, -2], [-4, 1]] (interchange elements on the main diagonal and change the sign of the off-diagonal elements)
Calculate A(Adj A):
A(Adj A) = [[1, 2], [4, 6]] * [[6, -2], [-4, 1]]
A(Adj A) = [[(16)+(2(-4)), (1(-2))+(21)], [(46)+(6(-4)), (4(-2))+(61)]]
A(Adj A) = [[6-8, -2+2], [24-24, -8+6]]
A(Adj A) = [[-2, 0], [0, -2]]
Calculate (Adj A)A:
(Adj A)A = [[6, -2], [-4, 1]] * [[1, 2], [4, 6]]
(Adj A)A = [[(61)+(-24), (62)+(-26)], [(-41)+(14), (-42)+(16)]]
(Adj A)A = [[6-8, 12-12], [-4+4, -8+6]]
(Adj A)A = [[-2, 0], [0, -2]]
As we can see, A(Adj A) = (Adj A)A = [[-2, 0], [0, -2]] = (-2) * I
(ii) B * B^(-1) = I = B^(-1) * B
Find the determinant of matrix B:
det(B) = (3 * -2) – (-1 * 2) = -6 + 2 = -4
Find the inverse of matrix B (B^(-1)):
B^(-1) = (1/det(B)) * adj(B)
B^(-1) = (1/(-4)) * [[-2, 1], [-2, 3]] (interchange elements on the main diagonal and change the sign of the off-diagonal elements)
B^(-1) = [[1/2, -1/4], [1/2, -3/4]]
Calculate B * B^(-1):
B * B^(-1) = [[3, -1], [2, -2]] * [[1/2, -1/4], [1/2, -3/4]]
B * B^(-1) = [[(31/2)+(-11/2), (3(-1/4))+(-1(-3/4))], [(21/2)+(-21/2), (2(-1/4))+(-2(-3/4))]]
B * B^(-1) = [[3/2 – 1/2, -3/4 + 3/4], [1 – 1, -1/2 + 3/2]]
B * B^(-1) = [[1, 0], [0, 1]]
Calculate B^(-1) * B:
B^(-1) * B = [[1/2, -1/4], [1/2, -3/4]] * [[3, -1], [2, -2]]
B^(-1) * B = [[(1/23)+(-1/42), (1/2(-1))+(-1/4(-2))], [(1/23)+(-3/42), (1/2(-1))+(-3/4(-2))]]
B^(-1) * B = [[3/2 – 1/2, -1/2 + 1/2], [3/2 – 3/2, -1/2 + 3/2]]
B^(-1) * B = [[1, 0], [0, 1]]
As we can see, B * B^(-1) = B^(-1) * B = I, where I is the identity matrix.
Both properties have been verified for the given matrices A and B.