19.12 - Second Theorem

In the previous section, we saw the theorem 1 related to symmetric and skew symmetric matrices. In this section, we will see theorem 2.

Theorem 2
Any square matrix can be expressed as the sum of a symmetric matrix and a skew symmetric matrix.

Proof:
Let A be a square matrix. First, we will prove that,
$A = \frac{1}{2} \left(A + A' \right)~+~\frac{1}{2} \left(A - A' \right)$

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{{}}    &{{}}    &{\frac{1}{2} \left(A + A' \right)~+~\frac{1}{2} \left(A - A' \right)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} A ~+~\frac{1}{2} A' ~+~\frac{1}{2} A ~-~\frac{1}{2} A'}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} A ~+~\frac{1}{2} A ~+~\frac{1}{2} A' ~-~\frac{1}{2} A'}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} A ~+~\frac{1}{2} A}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{A}    \\
\end{array}$

• Now we can write the proof for theorem 2 in  steps:
1. We just proved that, A can be written as the sum of two terms. $\frac{1}{2} \left(A + A' \right)$ and $\frac{1}{2} \left(A - A' \right)$
2. Consider the first term.
• In this term, we know (from theorem 1) that, (A+A') is a symmetric matrix.
• We need to prove that, $\frac{1}{2} \left(A + A' \right)$ is also a symmetric matrix.
• For that, we need to prove:
$\frac{1}{2} \left(A + A' \right)~=~\left[\frac{1}{2} \left(A + A' \right) \right]'$
• In other words, we need to prove:
$\frac{1}{2} B~=~\left[\frac{1}{2} B \right]'$
   ♦ Where B = A + A'
• This can be proved in 4 steps:
(i) B is symmetric.
   ♦ That means, B = B'
   ♦ That means, $\frac{1}{2} B ~=~ \frac{1}{2} B'$
(ii) For any matrix X, we have: kX' = (kX)'
(iii) So the result in (i) becomes:
$\frac{1}{2} B ~=~ \frac{1}{2} B' ~=~\left[\frac{1}{2} B \right]'$
(iv) By picking the first and last terms in (iii), we get:
$\frac{1}{2} B ~=~\left[\frac{1}{2} B \right]'$.
• This is same as:
$\frac{1}{2} \left(A + A' \right)~=~\left[\frac{1}{2} \left(A + A' \right) \right]'$
• So $\frac{1}{2} \left(A + A' \right)$ is a symmetric matrix.

3. Consider the second term.
• In this term, we know (from theorem 1) that, (A−A') is a skew symmetric matrix.
• We need to prove that, $\frac{1}{2} \left(A - A' \right)$ is also a skew symmetric matrix.
• For that, we need to prove:
$\frac{1}{2} \left(A - A' \right)~=~- \left[\frac{1}{2} \left(A - A' \right) \right]'$
• In other words, we need to prove:
$\frac{1}{2} C~=~- \left[\frac{1}{2} C \right]'$
   ♦ Where C = A − A'
• This can be proved in 4 steps:
(i) C is skew symmetric.
   ♦ That means, C = -C'
   ♦ That means, $\frac{1}{2} C ~=~ -\frac{1}{2} C'$
(ii) For any matrix X, we have: kX' = (kX)'
(iii) So the result in (i) becomes:
$\frac{1}{2} C ~=~ -\frac{1}{2} C' ~=~- \left[\frac{1}{2} C \right]'$
(iv) By picking the first and last terms in (iii), we get:
$\frac{1}{2} C ~=~-\left[\frac{1}{2} C \right]'$.
• This is same as:
$\frac{1}{2} \left(A - A' \right)~=~- \left[\frac{1}{2} \left(A - A' \right) \right]'$
• So $\frac{1}{2} \left(A - A' \right)$ is a skew symmetric matrix.

4. So the first term is a symmetric matrix. Also, the second term is a skew symmetric matrix.
• Thus we effectively wrote matrix A as the sum of a symmetric matrix and a skew symmetric matrix.
• Theorem 2 is proved.

---

Solved example 19.15
Express the matrix B = $\left[\begin{array}{r}                           
3    &{    11    }    &{    2    }    \\
-10    &{    -4    }    &{    12    }    \\
8    &{    -9    }    &{    13    }    \\
\end{array}\right]                           
$ as the sum of a symmetric matrix and a skew symmetric matrix.
Solution:
1. Write the transpose of B:
B' = $\left[\begin{array}{r}                           
3    &{    -10    }    &{    8    }    \\
11    &{    -4    }    &{    -9    }    \\
2    &{    12    }    &{    13    }    \\
\end{array}\right]                           
$

2. Let $P = \frac{1}{2} (B+B')$ and $Q = \frac{1}{2} (B-B')$

3. Finding P and proving that, it is symmetric:

• If we try to write P', we will get the same P. So P is a symmetric matrix.

4. Finding Q and proving that, it is skew symmetric:

 

• We see that, Q' = -Q. So it is a skew symmetric matrix.

5. Checking the sum:

---

The link below gives a few more solved examples:

Exercise 19.3

---

In the next section, we will see elementary operation of a matrix and invertible matrices.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

19.11 - Symmetric and Skew Symmetric Matrices

In the previous section, we saw transpose of a matrix. In this section, we will see symmetric and skew symmetric matrices.

Symmetric Matrix

This can be written in 5 steps:
1. Consider any square matrix A. We know how to write it's transpose A'.
2. If A' is equal to A, then we say that, A is a symmetric matrix.
3. An example is given below:
Let A = $\left[\begin{array}{r}                       
3    &{    \sqrt2    }    &{    11    }\\
\sqrt2    &{    5    }    &{    -8    }\\
11    &{    -8    }    &{    9    }\\
\end{array}\right]                       
$.
• If we write A', we will get A itself. So A is a symmetric matrix.
4. Consider any element say the $\sqrt2$ at second row, first column. Let us interchange the row and column. So we want first row, second column. What is the element at the interchanged position? It is also $\sqrt2$
5. This is applicable to any element of a symmetric matrix.
• We can write:
   ♦ Let a be any element of a symmetric matrix. Let it be at the row i, column j.
   ♦ The element at row j, column i will also be the same element a.

---

Skew Symmetric Matrix

This can be written in 5 steps:
1. Consider any square matrix A. We know how to write it's transpose A'.
2. If A' is equal to −A, then we say that, A is a skew symmetric matrix.
3. An example is given below:
Let A = $\left[\begin{array}{r}                       
0    &{    -3    }    &{    11    }\\
3    &{    0    }    &{    -7    }\\
-11    &{    7    }    &{    0    }\\
\end{array}\right]                       
$.
• If we write A', we will get:
A' = $\left[\begin{array}{r}                       
0    &{    3    }    &{    -11    }\\
-3    &{    0    }    &{    7    }\\
11    &{    -7    }    &{    0    }\\
\end{array}\right]                       
$.

• We see that, A' = −A. So A is a skew symmetric matrix.
4. Consider any non-diagonal element in A. Say the 7 at third row, second column. Let us interchange the row and column. So we want second row, third column. What is the element at the interchanged position? It is -7.
5. This is applicable to any non-diagonal element of a skew symmetric matrix.
• We can write:
   ♦ Let a be any non-diagonal element of a skew symmetric matrix. Let it be at the row i, column j.
   ♦ The element at row j, column i will be the -ve of element a.
6.The property written in (5) is applicable to diagonal elements also.This can be explained in 3 steps:
(i) For any diagonal element, the row number and column number will be the same.
• So any diagonal element, can be represented as a_ii.
(ii) In a skew symmetric matrix, any a_ij must be equal to −a_ji.
• So any a_ii must be equal to −a_ii
• That means:
a_ii = −a_ii
⇒ 2a_ii = 0
⇒ a_ii = 0
(iii) We can write:
If A is a skew symmetric matrix, then all diagonal elements of A will be zero.

---

Now we will see two theorems related to symmetric and skew symmetric matrices.

Theorem 1
For any square matrix A,
(i) (A + A') is a symmetric matrix
(ii) (A – A') is a skew symmetric matrix.

Proof for part (i):
$\begin{array}{ll} {~\color{magenta}    1    }    &{\text{Let}}    &{B}    & {~=~}    &{A+A'}    \\
{~\color{magenta}    2    }    &{\implies}    &{B'}    & {~=~}    &{(A+A')'}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{A' + (A')'}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{A' + A}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{A+A'}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{B}    \\
\end{array}$

◼ Remarks:
(3) (A+B)' = A' + B'
(4) (A')' = A
(5) A+B = B+A

• We see that, B' = B. That means, B is a symmetric matrix. That means, (A+A') is a symmetric matrix.

Proof for part (ii):
$\begin{array}{ll} {~\color{magenta}    1    }    &{\text{Let}}    &{C}    & {~=~}    &{A-A'}    \\
{~\color{magenta}    2    }    &{\implies}    &{C'}    & {~=~}    &{(A-A')'}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{A' - (A')'}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{A' - A}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{-(A-A')}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{-C}    \\
\end{array}$

◼ Remarks:
(3) (A−B)' = [A+(−B)]'= [A+(−1B)]'= [A'+(−1B)']
= [A'+(−1)(B)']  =   A' − B'
(4) (A')' = A
(5) A−B = [A + (−B)] = [(−B) + A] = −[B − A]

• We see that, C' = −C. That means, C is a skew symmetric matrix. That means, (A−A') is a skew symmetric matrix.

---

In the next section, we will see the second theorem.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

 

19.9 - Multiplicative Identity

In the previous section, we saw two properties of multiplication of matrices. In this section, we will see the third property.

Property III: The existence of multiplicative identity
• Consider any square matrix A of the order (m×m).
• We can write an identity matrix I of the same order.
That I will satisfy the equation: AI = IA = A
• Let us see an example. It can be written in 4 steps:

1. Let A = $\left[\begin{array}{r}                           
-8    &{    2    }    &{    5    }    \\
0    &{    -7    }    &{    -3    }    \\
3    &{    2    }    &{    4    }    \\
\end{array}\right]$

• Then I = $\left[\begin{array}{r}                           
1    &{    0    }    &{    0    }    \\
0    &{    1    }    &{    0    }    \\
0    &{    0    }    &{    1    }    \\
\end{array}\right]                           
$               

2. First we find AI:

3. Next we find IA:

4. Based on (2) and (3), we can write:
AI = IA = A

◼ We will see the actual proof in higher classes.

---

Now we will see a solved example.

Solved example 19.14
If A = $\left[\begin{array}{r}                           
1    &{    3    }    &{    2    }    \\
2    &{    0    }    &{    -1    }    \\
1    &{    2    }    &{    3    }    \\
\end{array}\right]                           
$, then show that A³ - 4A² - 3A + 11I = O
Solution:
1. First we will write A²:

2. Next we will write A³:

3. Next we will write 4A²:

3. Next we will write 3A:

4. Finally we will write 11I:

5. Substituting the values, we get:

---

The link below gives a few more solved examples:

Exercise 19.2

---

In the next section, we will see transpose of a matrix.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

19.5 - Solved Examples

In the previous section, we saw negative matrix and difference of two matrices. We also saw multiplication of matrix by a scalar and a solved example. In this section, we will see a few more solved examples.

Solved example 19.10
Find X and Y if:
X + Y = $\left[\begin{array}{r}       
5    &{2}    \\
0    &{9}    \\
\end{array}\right]       
$ and X − Y = $\left[\begin{array}{r}       
3    &{6}    \\
0    &{-1}    \\
\end{array}\right]$.
Solution:

Fig.19.14

Solved example 19.10
Find the values of x and y from the following equation:
$2\left[\begin{array}{r}       
x    &{5}    \\
7    &{y-3}    \\
\end{array}\right] +
\left[\begin{array}{r}       
3    &{-4}    \\
1    &{2}    \\
\end{array}\right] =
\left[\begin{array}{r}       
7    &{6}    \\
15    &{14}    \\
\end{array}\right]       
$
Solution:

Fig.19.15

◼ Remarks:
5,6: Equations are obtained by equating corresponding elements.
7,8: x and y values are obtained by solving the equations.

Solved example 19.11
Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely, Basmati, Permal and Naura. The sale (in rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B.

Fig.19.16

(i) Find the combined sales in September and October for each farmer in each variety.
(ii) Find the decrease in sales from September to October.
(iii) If both farmers receive 2% profit on gross sales, compute the profit for each farmer and for each variety sold in October.
Solution:
Part (i):
The combined sales for the two months can be obtained by adding A and B. We get:

Fig.19.17

• Let us write a sample information that can be obtained from (A+B):
When the two months of September and October are taken together, Gurcharan Singh achieved a sale of Rs.20000/- in the case of Naura rice.

Part (ii):
The decrease in sales can be obtained by subtracting B from A. We get:

Fig.19.18

• Let us write two sample information that can be obtained from (A−B):

1. Compared to September, Ramkishan experienced a decrease of Rs.10000/- in the sale of Permal rice in october. 

2. Compared to September, Gurcharan Singh did not experience any increase or decrease in the sale of Naura rice in october.

Part (iii):

First we will see an example. It can be written in steps:
1. Consider the matrix for October.
2. We see that, Ramkishan achieved a sale of Rs.5000/- in the case of Basmati rice. This amount includes the profit obtained.
3. Given that, profit is 2%.
• So if x is the original cost, then (x  × 1.02) = 5000
4. If p is the amount obtained as profit, then (5000 - x) = p
5. Solving the equations in (3) and (4), we get:
p = 5000 × 0.02
6. It is clear that:
To find the profit amount (or simply profit), we need to multiply the sales amount by 0.02.
7. To find the profit for each product, for each farmer, in just one step, we multiply the October matrix by 0.02. We get:

Fig.19.19

• Let us write a sample information that can be obtained from 0.02B:
In the month of October, Gurcharan Singh obtained a profit of Rs.400/- in the case of Basmati rice.

---

In the next section, we will see multiplication of matrices. 

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

19.3 - Addition of Matrices

In the previous section, we saw equality of matrices. In this section, we will see operations on matrices.

First we will see addition of matrices. It can be explained in 6 steps:
1. An industrialist has two factories. Factory A and Factory B. Both produces bicycles for boys and girls.
2. Bicycles are produced in three price tags: Platinum class, Gold class and Silver class.
◼ For example:
• Factory A produces:
   ♦ 35 Platinum class bicycles for boys and 40 Platinum class bicycles for girls.  
   ♦ 75 Gold class bicycles for boys and 80 Gold class bicycles for girls.  
   ♦ So on . . .
3. This data can be effectively represented using matrices A and B as shown in fig.19.14 below:

Fig.19.14

4. Now, if the industrialist wants to know the total production in each class, he can simply add the two matrices together. The resulting matrix C is shown in the same fig.19.14 above.
• The industrialist now know that, from the two factories,
   ♦ (35+15) = 50 platinum class is produced for boys.
   ♦ (40+25) = 65 platinum class is produced for girls.
   ♦ (75+35) = 110 Gold class is produced for boys.
   ♦ So on . . .
5. We see that, to find the sum, all we need to do, is to add the corresponding terms together.
• But it is important to make sure that, the matrices being added, are of the same order. If they are not of the same order, then the sum is not defined.
6. So we can write:
If A = [a_ij]_m×n and B = [b_ij]_m×n are two matrices of the same order m×n, then the sum of the two matrices A and B is defined as a matrix C = [c_ij]_m×n, where c_ij = a_ij + b_ij for all possible values of i and j.
• Using symbols, this is written as:
A + B = [a_ij] + [b_ij] = [a_ij + b_ij] = C

Solved example 19.7
If A = $\left[\begin{array}{r}           
4    &{\sqrt5}    &{2}    \\
6    &{0}    &{-9}    \\
\end{array}\right]$ and B = $\left[\begin{array}{r}           
\sqrt3    &{7}    &{5}    \\
-2    &{\frac{2}{3}}    &{9}    \\
\end{array}\right]           
$, then find A+B
Solution:
• Both A and B are of the same order. So addition of A and B is defined. We get:
A+B = $\left[\begin{array}{r}           
4+\sqrt3    &{\sqrt5 + 7}    &{2+5}    \\
6 – 2    &{0 + \frac{2}{3}}    &{-9+9}    \\
\end{array}\right]~=~
\left[\begin{array}{c}           
4+\sqrt3    &{\sqrt5 + 7}    &{7}    \\
4    &{\frac{2}{3}}    &{0}    \\
\end{array}\right]$

---

Properties of Matrix addition

Matrix addition satisfy the following four properties:
(i) Commutative law
(ii) Associative law
(iii) Existence of additive identity
(iv) Existence of additive inverse

(i) Commutative law
• If A = [a_ij]_m×n and B = [b_ij]_m×n are two matrices of the same order m×n, then A+B = B+A
• Proof can be written in 3 steps:
1. A + B = [a_ij] + [b_ij] = [a_ij + b_ij]
We already saw this when we discussed matrix addition.
2. a_ij and b_ij are numbers. Addition of numbers is commutative. So (1) can be modified as:
A + B = [b_ij + a_ij]
3. [b_ij + a_ij] means that, every element of B is being added to the corresponding element of A.
• So it is the addition of B and A.
• Thus (2) can be modified as: A+B = B+A

(ii) Associative law
• If A = [a_ij]_m×n , B = [b_ij]_m×n and C = [c_ij]_m×n are three matrices of the same order m×n, then (A+B)+C = A+(B+C)
• Proof can be written in 6 steps:
1. A + B = [a_ij] + [b_ij] = [a_ij + b_ij]
We already saw this when we discussed matrix addition.
2. So (A+B) + C = [a_ij + b_ij]+ [c_ij]
3. (2) can be modified by using the same procedure in (1). We get:
(A+B) + C = [a_ij + b_ij]+ [c_ij] = [(a_ij + b_ij) + c_ij]
4. a_ij , b_ij and c_ij are numbers. Addition of numbers is associative. So (3) can be modified as:
(A+B) + C = [(a_ij + b_ij) + c_ij] = [a_ij + (b_ij + c_ij)]
5. Consider the last term of the result in (4). It indicates that, two matrices are being added. The matrices being added are: A and (B+C).
• That is:
[a_ij + (b_ij + c_ij)] = [a_ij] + [(b_ij + c_ij)] = A + (B+C)
6. based on the results in (4) and (5), we get:
(A+B) + C = A + (B+C)

(iii) Existence of additive identity
• If A = [a_ij]_m×n and O = [0_ij]_m×n are two matrices of the same order m×n, then A+O = O+A = A
• Proof can be written in 4 steps:
1. A + O = [a_ij] + [0_ij] = [a_ij + 0_ij]
We already saw this when we discussed matrix addition.
• But [a_ij + 0_ij] = [a_ij] = A
• So we can write:
A + O = A
2. O + A = [0_ij] + [a_ij] = [0_ij + a_ij]
We already saw this when we discussed matrix addition.
• But [0_ij + a_ij] = [a_ij] = A
• So we can write:
O + A = A
3. Based on (1) and (2), we can write: A+O = O+A = A
4. Whenever we add O to a matrix A, the sum will be the same matrix A. So O is called the additive identity for matrix addition.

(iv) Existence of additive inverse
• If A = [a_ij]_m×n and A = [-a_ij]_m×n are two matrices of the same order m×n, then A+(-A) = (-A)+A = O
(Here, -A is a matrix obtained by multiplying each element of A by -1)
• Proof can be written in 4 steps:
1. A + (-A) = [a_ij] + [-a_ij] = [a_ij + -a_ij] = [a_ij - a_ij]
We already saw this when we discussed matrix addition.
• But [a_ij - a_ij] = [0_ij] = O
• So we can write:
A + (-A) = O
2. (-A) + A = [-a_ij] + [a_ij] = [-a_ij + a_ij]
We already saw this when we discussed matrix addition.
• But [-a_ij + a_ij] = [0_ij] = O
• So we can write:
(-A) + A = O
3. Based on (1) and (2), we can write: A+(-A) = (-A)+A = O
4. Whenever we add -A to a matrix A, the sum will be the zero matrix O. So -A is called the additive inverse of A or negative of A.

---

In the next section, we will see multiplication of a matrix by a scalar. 

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com