Addition of matrices

Addition of Matrices is an operation done on two or more matrics according to their corresponding entries. Matrix of same order can only be added. If there orders are different then addition operation cannot be applied on them.

Adding Matrices requires addition of corresponding elements of them. For example if two Matrices are added then elements in first row and first column of the two matrices will be added together to get the element in the first row and first column of the resultant matrix.

Two Matrices of order m × n (m is number of rows and n is number of columns) can be added:

Let sum of these = C = A+B then
C11 = a11 + b11, C12 = a12 + b12….,  C21 = a21 + b21, C22 = a22 + b22 and so on.

A + B = C =  ( )
Order of matrix C will also be m × n.
If two matrix are equal but opposite in sign, then there addition will be null matrix (whose all elements are zero). For example if A = ((1@2@5)) and B = ((-1@-2@-5)) then A+B = 0 as A+B = ((1+(-1)@2+(-2)@5+(-5)))= ((0@0@0))

Example 1) Add Matrices A and B if A = ((1&9@5&2))  and B = ((2&4@5&8))
Solution) Let sum of A and B be C. order of C will be same as that of A and B i.e. 2x2.
Elements of matrix C will be: C11, C 12, C 21, C22.
C11 = A11 + B11 = 1+2,
C12 = A12 + B12 = 9+4,
C21 = A21 + B21 = 5+5,
C22 = A22 + B22 = 2+8,
C = ((C11 = A11 + B11 &C12 = A12 + B12 @C21 = A21 + B21&C22 = A22 + B22))
C =   ((1 + 2&9 + 4@5 + 5 &2 + 8))
C = ((3&13@10&10))
Example 2) Perform addition of Matrices P and Q given as:
P =   ((2&0&- 3@5&3&7@3&- 7&1)), Q =  ((- 4&6&2@5&2&5@21&- 15&- 4))
Solution) let P+Q = R
R11 = P11 + Q11 = 2 +(-4) =-2,
R12 = P12 + Q12 = 0+6 = 6,
R13 = P13 + Q12 = -3+2 =-1
R21 = P21 + Q21 = 5+5 =10,
R22 = P22 + Q22 = 3+2 =5,
R23 = P23 + Q23 = 7+5 = 12
R31 = P31 + Q31 = 3+21=24
R32 = P32 + Q32 = -7+(-15) =-22
R33 = P33 + Q33 = 1+(-4)= -3
R = ((R_11&R_12&R_13@R_21&R_22&R_23@R_31&R_32&R_33 ))
R = ((-2&6&-1@10&5&12@24&-22&-3))

Gaussjordan elimination method

With the concept of matrices one can solve the system of linear equations apart from usual algebraic methods. Even in matrices topic, a system there are different methods to solve and one such method is called as Gauss Jordan elimination method named after the two mathematicians who introduced this elimination method. It is also called in short as Gauss elimination method or as Jordan elimination.
Let us first study the concept behind the Gauss elimination method. A system of equations is expressed in the matrix form as, AX = B, where A is the matrix of the coefficients, X is the matrix of the variables and B is the matrix of the constants on the right side of the system of equations. If the augmented matrix [A¦B] can be expressed in reduced row echelon form, by the method of Gauss-Jordan, that is, in the form [I¦C] (I being the identity matrix), then X = C, which gives us the solution to the system.
Now let us do Gauss-Jordan elimination step by step, with an example for clear understanding. Let a system of equations is,


-3x + 4y + z = -5
x +3y -2z = -6
2x –y + 3z = 9

   Then,A is,

-3	4	1
1	3	-2
2	-1	3

x

y

z

X is, 

-5

-6

9

B is

The following are the steps of elimination.

-3	4	1	-5
1	3	-2	-6
2	-1	3	9

[A│B]

1	-4/3	-1/3	5/3
1	3	-2	-6
2	-1	3	9

[R₁│-3]

1	-4/3	-1/3	5/3
0	13/3	-5/3	-23/3
2	-1	3	9

[R₂│-R₁]

1	-4/3	-1/3	5/3
0	13/3	-5/3	-23/3
0	5/3	11/3	17/3

[R₃-2R₁]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	5/3	11/3	17/3

[R₂/(13/3]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	0	168/39	336/39

[R₃│-(5/3)R₂]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	0	1	2

[R₃/(168/39)]

1	-4/3	0	7/3
0	1	-5/13	-23/13
0	0	1	2

[R₁+(R₃/3)]

1	-4/3	0	7/3
0	1	0	-1
0	0	1	2

[R₂+(5R₃/13)]

1	0	0	1
0	1	0	-1
0	0	1	2

[R₁+(4R₂/3)]

Now the finally reduced matrix is in the form [I│C], where,

1

-1

2

           C and therefore X = 

Hence, the solution to the given system of equations is, x = 1, y = -1 and z = 2. 

Thus, we had seen the used of the elimination method introduced by Gauss and Jordon.

Learning to find the determinant value of a 3x3 matrix

Consider a matrix A. If A is a square matrix then there is some value that can be assigned to A. This is called the determinant of matrix A. It is denoted by the symbol |A| or det(A). The entries of the matrices are used and an arithmetic expression is formed out of them to come to a value of the matrix. Such square matrices are most commonly used in solving system of linear equations. If the number of variables in the system is 2, then the square matrix would be a 2x2 matrix. Similarly if the number of variables in the system is n, then the square matrix would be an nXn matrix.
Determinant of a 3x3 matrix:
The determinant of a 3x3 matrix is also called a third order determinant. If we have to solve a system of three simultaneous linear equations in three variables, we shall have to deal with third order determinants or 3x3 determinant. For real numbers a1, a2, a3, b1, b2, b3, c1, c2, c3, the symbol
|a1 a2 a3|
|b1 b2 b3|
|c1 c2 c3|

Represents the determinant of 3x3 matrix.

The numbers a1, a2, a3, b1, b2, b3, c1, c2, c3 are the entries or elements of the det.  The first, the second and the third rows of the det are respectively: a1, a2, a3; b1, b2, b3 and c1, c2, c3. The first, the second and the third columns are respectively;
a1 a2 a3
b1; b2; b3;
c1 c2 c3

The value of this determinant of 3x3 matrix is defined as:

|a1 a2 a3|
|b1 b2 b3|
|c1 c2 c3|

= a1*|b2 b3| - a2 * |b1 b3| + a3 * |b1 b2|
          |c2 c3|            |c1 c3|             |c1 c2|

= a1 * (b2c3 – b3c2) – a2 * (b1c3 – b3c2) + a3 * (b1c2 – b2c1)

The above formula can be used to find the value of any 3x3 determinant. This is not the only method used to evaluate a 3x3 determinant. We can also use the Sarrus’ expansion method. It is as follows:
Suppose D
= |a b c|
   |d e f|
   |g h i|
We write five columns like this:

a b c a b
d e f d e
g h I g h
Then we multiply the terms long the red and blue lines as shown in the picture below:

The red diagonals are aei, bfg and cdh. The blue diagonals are gec, hfa and idb.
Then D = aei + bfg + cdh – gec – hfa - idb

Solution of a quadratic equation by completing the square

One method of obtaining the roots of a quadratic equation is the factoring method. Another method that is used is completing the square method.

Consider the following situation:
The product of Sara’s age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age?
To answer this, let her present age (in years) be = x. Then the product of her ages two years ago and four years from now is (x-2)*(x+4).
Therefore, (x-2)*(x+4) = 2x+1
I.e, x^2 + 2x – 8 = 2x + 1
i.e, x^2 – 9 = 0

So Sara’s present age satisfies the quadratic equation x^2 – 9 = 0
We can write this as x^2 = 9. Taking square roots, we get x = 3 or x = -3. Since age is always a positive number, x = 3 would be our answer.

Now consider the quadratic equation (x+2)^2 – 9 = 0. To solve this we can write it as (x+2)^2 = 9. Taking roots we have x+2 = 3 and x+2 = -3. Therefore x = 1 or x = -5. These are the roots.

In both the examples above, the term containing x is completely inside a square, and we found the roots easily by taking square roots. But what if we are given the equation x^2 + 4x – 5 = 0 to solve? We would probably apply factorization unless we (somehow!) realize that x^+4x-5 = (x+2)^2 – 9.

So solving x^2+4x-5 = 0 is equivalent to solving (x+2)^2-9 = 0, which we have seen is faster. In fact, we can convert any quadratic equation to the form (x+a)^2 – b^2 = 0 and then we can easily find its roots. This process is called completing a square.

Suppose we were to complete the square for x^2+4x. The process is as follows:
X^2+4x
=(x^2 + (4/2)x) + (4/2)x
= x^2+2x+2x
= (x+2)x + 2*x
= (x+2)x + 2*x + 2*2 – 2*2
= (x+2)x + (x+2)2 – 2*2
= (x+2)(x+2) – 2^2
= (x+2)^2 – 4
So, x^2+4x-5 = (x+2)^2 -4 -5 = (x+2)^2 – 9.
So turning x^2+4x-5 to (x+2)^2-9 makes a complete square of the expression.

In brief, this can be shown as follows:
X^2+4x = (x+(4/2))^2 – (4/2)^2 = (x+(4/2))^2 – 4
So, x^2+4x-5 = 0 can be rewritten as
(x+(4/2))^2 -4-5 = 0
i.e., (x+2)^2 – 9 = 0. This was the completing square method in brief.

Addition of like fractions

We know that whole numbers can be added, subtracted, divided or multiplied. Similarly fractions can also be added, subtracted, divided or multiplied. A problem of addition of fractions can be of either of the two types:
(a) Adding fractions like denominators
(b) Adding fractions – unlike denominators

Adding like denominators fractions:

Adding fractions with like denominators word problems may be worded using different terms such as: “How to add fractions with common denominators?’ or ‘adding fractions with same denominators’.  These words, “common denominators” or “ same denominators” mean the same thing as “like denominators”. By definition, if two fractions have the same or equal denominators they are called, like fractions.

For being able to add fractions, the denominators have to be equal or same. In other words, for adding fractions it is important that all the addends are like fractions and have the same common denominator. For adding fractions with unlike denominators, we first make the denominators equal by taking the lowest common multiple of the denominators and converting each of the fractions to its corresponding like fraction that has denominator equal to the lowest common multiple found.

For adding like fraction, the following steps are to be followed:

Step 1: Make sure that the fractions are like fractions. That means, ensure that the denominators of all the fractions to be added are equal.
Step 2: Add up all the numbers on the numerators of all the fractions.
Step 3: Put the result found in step 2 over the common denominator that we confirmed in step 1.
Step 4: Simplify the fraction if needed. That is, reduce the fraction to its lowest terms if needed.

Let us illustrate the above process with an example:

Example 1: Add 1/10, 3/10, 1/10

Solution:
Step 1: The fractions to be added are 1/10, 3/10 and 1/10. Note that the denominators of all the three fractions are same and equal to 10. So we can move on to step 2.

Step 2: The numerators of the fractions are 1,3 and 1. Adding them we have,
1 + 3 + 1 = 5.

Step 3: The result found in step 2 was the number 5. We put this number over the common denominator that we confirmed in step 1 which was 10. So we have,
= 5/10

Step 4: Now we reduce the fraction thus found to its lowest terms. So,
5/10 = (5*1)/(5*2), the common factor 5 cancels out and we are left with,
= ½ <- answer.="answer." p="p">

Word Problems on Addition

Word problems in mathematics mean a text representation of a mathematical expression. Word problems make it easy for one to solve a problem. Words problems include all the four mathematical operations – addition, subtraction, multiplication and division. Let’s have a look at word problems on addition in this post.
Word problems on addition are nothing but a text representation of addition operation. For example: Maria bought nine bed rails from Bed Guard India

collection online. Shiva gave three Bed Guard India bed rails to Maria. How many number of bed rails is Maria left with? This problem is the word problem of addition operation (9+3=?)

Solving word problems on addition:

Solving word problems on addition include three steps. Firstly, identify the numbers to be added. Secondly, convert the word problem into a mathematical additional operation and finally add the numbers. For example: Patrick bought two Angry Bird toys from children online shopping store. Joe bought five Chota Bheem toys from the same children online shopping
store. He gave these five toys to Patrick. How many toys do Patrick has now? Identifying the numbers, Patrick bought 2 toys and Joe bought 5 toys. Converting the word problem into mathematical operation, (2+5); adding the numbers (2+5=7). Thus, the answer is 7 toys.

Examples of Word Problems on Addition

1. Mary has one baby bottle from Bottle Sterilizer India collection. Joe bought 2 baby bottles from Bottle Sterilizer India
from online store and Thomas bought 5 baby sterilizers from local store. Three of them gave these to Philip. How many baby bottles and sterilizer are there in total with Philip?Answer: 1 baby bottle + 2 baby bottles + 5 baby sterilizers
(1+2+5) = 8 baby products from Bottle Sterilizer India
collection.2. There are five play schools in the city. One play school is in the process of construction and Thomas is planning to complete 2 play schools in two different corners of the city by the end of next year. How many play schools will be there by the end of next year?
Answer: 5 play schools + 1 constructing play school + 2 planned play school
(5+1+2) = 8 play schools in the city by the end of next year.
These are examples of word problems on addition.