Solving Exponential Decay, Matrix Operations and Equations

This document explains how to solve problems related to exponential decay, matrix operations and equations with step by step examples. It also mentions the use of matrices in solving equations. The document is suitable for students studying mathematics and related subjects. It does not mention any specific course code, course name or college/university.

5)
Answer :
1

A(t) = 400e−0.045t
At t = 0
A(0) = 400
Therefore,the originalamount is 400.Half of it is 200.Therefore,at half
the original amount:
200 = 400e−0.045t
0.5 = e−0.045t
[Taking log on both sides]
ln 0.5 = −0.045t
−0.6931 = −0.045t
=⇒ t = 15.40
Thus, the substance decays to half its original amount in 15.4 years.
6)
x − 3y = 9
−2x − 2y = 6
Answer:
Matrix form:A · x = B
1 −3
−2 −2
| {z }
A
x
y
|{z}
x
= 9
6
|{z}
B
The boxed element (1) is the pivot.Augmented matrix [A|B] is:
[A|B] = 1 −3 | 9
−2 −2 | 6
[ Multiply Row 1 (R1) by 2 and add to Row 2 (R2) : R2 = R2 + 2 × R1]
= 1 −3 | 9
0 −8 | 24
2

Matrix A is reduced into a triangular matrix and the solution is obtained
by back substitution.
The second row of [A|B] implies:
0 · x − 8 · y = 24 =⇒ y = −3
The first row is:
1 · x − 3y = 9 =⇒ x − 3(−3) = 9 =⇒ x = 0
Thus the solution to the set of equations is:
x
y = 0
−3
7)
x − 6y = 23
2x − 6y = 28
Answer:
Matrix form:A · x = B
1 −6
2 −6
| {z }
A
x
y
|{z}
x
= 23
28
| {z }
B
The boxed element (1) is the pivot.Augmented matrix [A|B] is:
[A|B] = 1 −6 | 23
2 −6 | 28
[ Multiply Row 1 (R1) by 2 and subtract from Row 2 (R2) : R2 = R2 − 2 × R1]
= 1 −6 | 23
0 6 | −18
Matrix A is reduced into a triangular matrix and the solution is obtained
by back substitution.
The second row of [A|B] implies:
0 · x + 6 · y = −18 =⇒ y = −3
3

The first row is:
1 · x − 6y = 23 =⇒ x − 6(−3) = 23 =⇒ x = 5
Thus the solution to the set of equations is:
x
y = 5
−3
8)
Answer :
A = −9 10
−8 −7 B = 9 2
−8 6
Two matrices are compatible for subtraction only when they have same
order.Since A and B both are of order 2, they can be subtracted from each
other.
The subtraction operation is represented as:
[cij ] = [aij ] − [bij ]
Where,
A = [aij ]
represents the element of A in ith column and jth row. Thus,
A − B = −9 10
−8 −7 − 9 2
−8 6
= −9 − 9 10 − 2
−8 − (−8) −7 − 6
= −18 8
0 −13
4