Math Homework: Solutions for Matrix Algebra and Decay Problems

This document presents solutions to a set of math homework problems. The solutions cover a range of topics including exponential decay calculations and matrix operations. Specifically, the first problem calculates the time it takes for a substance to decay to half its original amount using an exponential decay formula. The subsequent problems involve solving systems of linear equations using matrix forms and augmented matrices. These solutions demonstrate the process of reducing matrices into triangular forms and using back substitution to find the values of x and y. Finally, the document includes the subtraction of two 2x2 matrices. All steps are clearly presented, providing a comprehensive guide to solving these types of mathematical problems. This resource is designed to assist students in understanding and solving similar problems.

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