Lesson Objective

Dear learner,

by the end of this lesson you will be able to:

• Define symmetric and skew-symmetric matrices
• Describe elementary operations on matrices
• Determine inverse of a square matrix by using elementary row operations
• Find associated augmented matrix of a linear system of equations

Key Terms

• Symmetric matrix
• Skew-symmetric matrix
• Elementary operation
• Echelon matrix

Brainstorming Activity

Dear learner, try to see the following two matrices and give your description?

1. Find the transpose of the following matrices $A = \begin{pmatrix}
   1 & 2 & 3\\
   4 & 7 & 2 \\
   8 & 1 & 0
   \end{pmatrix}$ and $B= \begin{pmatrix}
   1 & 2 & 3\\
   2 & 7 &5 \\
   3 & 5 & 0
   \end{pmatrix}$ and compare with the original matrices.

Solution:

From your discussion you can observe the following $A^{T}= \begin{pmatrix}
1 & 4&8\\
2 & 7 & 1 \\
3 & 2 & 0
\end{pmatrix}$, while $B^{T} = \begin{pmatrix}
1 & 2 & 3\\
2 & 7 &5 \\
3 & 5 & 0
\end{pmatrix}$. when you compare A with $A^{T}$ they are two different matrices, But when we compare $B$ and $B^{T}$: $B=B^{T}=\begin{pmatrix}
1 & 2 & 3\\
2 & 7 &5 \\
3 & 5 & 0
\end{pmatrix}$

2.1. Symmetric and Skew- symmetric Matrices

I. Symmetric matrix

Definition : Any square matrix A is said to be symmetric if it is equal to its transpose

i.e. $A = A^{T}$

Example 1:

1. Show that $A = \begin{pmatrix}
   1 & 4 & 0\\
   4 & 6 &-2 \\
   0 & -2 & 0
   \end{pmatrix}$ is a symmetric matrix?

Solution:

Assume that A is symmetric implies A and its transpose are the same:

$\implies$ $A^{T} = \begin{pmatrix}
1 & 4 & 0\\
4 & 6 &-2 \\
0 & -2 & 0
\end{pmatrix}$ which is the same as that of matrix A. therefore A is a symmetric matrix.

 ii. skew-symmetric Matrix

Definition: Any square matrix A is said to be skew-symmetric if $A = – A^{T}$ or $A=A^{T}=0$.

Example 2:

1. Which of the following matrices are skew-symmetric?

$A = \begin{pmatrix}0&-1&4\\ 1&0&7\\-4&-7&0\end{pmatrix}$

$B = \begin{pmatrix}1&2&3\\2&5&4\\3&4&6\end{pmatrix}$

Solution:

You have $A = \begin{pmatrix}0&-1&4\\ 1&0&7\\-4&-7&0\end{pmatrix}$ and $A^{t} = \begin{pmatrix}0&1&-4\\ -1&0&4\\4&7&0\end{pmatrix}$

$A + A^{t} = \begin{pmatrix}0+0&-1+1&4+(-4)\\ 1+(-1)&0+0&7+(-7)\\-4+4&-7+7&0+0\end{pmatrix}$ = $\begin{pmatrix}0&0&0\\ 0&0&0\\0&0&0\end{pmatrix}$

Therefore A is skew symmetric matrix.

Let $B = \begin{pmatrix}1&2&3\\2&5&4\\3&4&6\end{pmatrix}$ and $B^{t} =\begin{pmatrix}1&2&3\\2&5&4\\3&4&6\end{pmatrix}$

$B + B^{t}$ = $\begin{pmatrix}1+1 & 2+2 & 3+3\\2+2 & 5+5 & 4+4\\3+3 & 4+4 & 6+6\end{pmatrix}$ = $\begin{pmatrix}2&4&6\\4&10&8\\6&8&12\end{pmatrix}$ is anon- zero matrix therefore B is not skew-symmetric.

2. Let $A = \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}$ be a matrix, then find

a). $B = \frac{A +A^{t}}{2}$ and show B is symmetric

b). $D = \frac{A – A^{t}}{2}$ and show D is skew- symmetric

c). show that B +D = A

Solution:

$A = \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}$ and $A^{t} = \begin{pmatrix}1&4&7\\2&5&8\\3&6&9\end{pmatrix}$

a) $\frac{A +A^{t}}{2}$ = $\frac{ \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} + \begin{pmatrix}1&4&7\\2&5&8\\3&6&9\end{pmatrix}}{2}$

$\implies$ $B = \begin{pmatrix}1&3&5\\3&5&7\\5&7&9\end{pmatrix}$ is symmetric.

b). $D = \frac{A – A^{t}}{2}$ = $\frac{\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} + \begin{pmatrix}1&4&7\\2&5&8\\3&6&9\end{pmatrix}}{2}$

$\implies$ $D = \begin{pmatrix}0&-1&-2\\1&0&-1\\2&1&0\end{pmatrix}$ is a skew- symmetric matrix

c) . B + D = $ \begin{pmatrix}1&3&5\\3&5&7\\5&7&9\end{pmatrix} + \begin{pmatrix}0&-1&-2\\1&0&-1\\2&1&0\end{pmatrix}$ = $\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}$ = A

Note: Any square matrix can be expressed as a sum of symmetric and skew-symmetric matrices

2.2. Elementary Operations on  Matrices

Definition: An elementary operation of matrices is a simple operation made on rows or columns which aims transforming any matrix to row or column equivalent matrix.

I. Elementary row operations:

Methods

 Swapping:-  interchanging rows of a matrix i.e. $R_i \leftrightarrow R_j$

 Re-scaling:- multiplying a row of a matrix by anon zero constant: $R_i \rightarrow rR_i$

 Pivoting :-  adding constant multiple of one row of a matrix on to another row. $R_i \rightarrow R_i + rR_j​$

Example 3:

1. Let $A = \begin{pmatrix}5&2&3\\1&5&6\\7&8&9\end{pmatrix}$ be a matrix then transform A in to a row equivalent matrix using elementary operations?

Solution:

step 1: row interchanging (swapping)

$R_ 1\leftrightarrow R_2$ $\begin{pmatrix}1&5&6\\5&2&3\\7&8&9\end{pmatrix}$

step 2: re-scaling row 1 to eliminate row 2 and row 3

a. to eliminate first entry of row 2 multiply row 1 by -5

$R_1 \rightarrow -5R_1$ $\implies$ $R_{1}$ = (-5 -25 -30)

b. to eliminate first entry of row3 multiply row1 by -7

$R_1 \rightarrow -7R_1$ $\implies$ $R_{1}$ = (-7 -35 -42)

step 3: pivoting (adding the rescaled rows to row 2 and row 3 respectively)

$R_2 \rightarrow R_2 + (-5)R_1​$ $\begin{pmatrix}1&5&6\\5-5 & 2-25 & 3-30 \\7&8&9\end{pmatrix}$ = $\begin{pmatrix}1&5&6\\0&-23&-27\\7&8&9\end{pmatrix}$

$R_3 \rightarrow R_3 + (-7)R_1​$ $\begin{pmatrix}1&5&6\\0&-23&-27\\7-7 & 8-35 & 9-42\end{pmatrix}$ = $\begin{pmatrix}1&5&6\\0&-23&-27\\0&-27&-31\end{pmatrix}$

step 4: again rescale $R_{2}$ to eliminate second entry (-27) of row 3 and pivote third row

$R_2 \rightarrow -\frac{27}{31}R_2$ $\implies$ $R_{2}$ = (0 27 $\frac{729}{23}$)

and $R_3 \rightarrow R_3 + (-27/31)R_2$ $\begin{pmatrix}1 & 5 & 6\\ 0 & -23 &-27\\ 0+0 & -27+27 & -31 + 729/23 \end{pmatrix}$ = $\begin{pmatrix}1 & 5 & 6\\ 0 & -23 & -27\\0 & 0 & 729/23 \end{pmatrix}$

II. Echelon form of a matrix

 Definition:

A matrix is said to be in row echelon form if:

• A zero row( if there is ) comes at the bottom
• The first non-zero entry in each non-zero row is 1
• The number of zero entries down the row is increasing.

Example 4:

1. Identify which of the following matrices are in echelon form?

A= $\begin{pmatrix}1 & 5 & 8\\ 0 & 1 & 3\\0 & 0 & 0 \end{pmatrix}$

B = $\begin{pmatrix}1 & 5 & 6\\ 0 & 0 & 0\\0 & 0 & 1 \end{pmatrix}$

C = $\begin{pmatrix}1 & 5 \\ 0 & 1\\0 & 0 \end{pmatrix}$

D = $\begin{pmatrix}1 & 5 \\ 0 & 0 \end{pmatrix}$

E = $\begin{pmatrix} 0 & 0\\0 & 0 \end{pmatrix}$

Solution:

• matrices A, C, D and E are in row echelon form while matrix B is not because number of zeros in the second row are greater than number of zeros in third row

iii. Row Reduced Echelon form of a Matrix:

Definition: A matrix is said to be row reduced echelon (RREF) if and only if

• it is in echelon form
• the first non-zero entry in each non-zero row is the only non-zero entry in its column.

Example 5:

1. Identify the following matrices are row reduced echelon form or not?

A = $\begin{pmatrix}1 & 0 &0 \\ 0 & 1&0\\0 & 0 &0\end{pmatrix}$

B = $\begin{pmatrix}1 & 2 &0 \\ 0 & 1&0\\0 & 0 &1\end{pmatrix}$

C = $\begin{pmatrix}1 & 2 &0 \\ 0 & 0&1\end{pmatrix}$

Solution:

• Matrices A and C are in RREF
• Matrix B is not in RREF because the first non-zero entry on row 2 is not the only non-zero entry across its column.

2. Reduce the following matrix in to row reduced echelon form?

A = $\begin{pmatrix}1&2&3\\4&5&6\\1&1&7\end{pmatrix}$

Solution:

Since the first entry of the matrix is 1 so you do not neccessarly do swapping

thus $R_{2}\rightarrow R_{2} +(-4) R_{1}$ $\begin{pmatrix}1&2&3\\0&-3&-6\\7&8&9\end{pmatrix}$

$R_{3} \rightarrow R_{3} +(-1) R_{1}$ $\begin{pmatrix}1&2&3\\0&-3&-6\\0&-1&4\end{pmatrix}$

$R_{2} \rightarrow -\frac{R_{2}}{3}$ $\begin{pmatrix}1&2&3\\0&1 &2\\0&-1&4\end{pmatrix}$

$R_{3} \rightarrow R_{3} + R_{2}$ $\begin{pmatrix}1&2&3\\0&1 &2\\0&0 &6\end{pmatrix}$

$R_{3} \rightarrow -\frac{R_{3}}{6}$ $\begin{pmatrix}1&2&3\\0&1 &2\\0&0 &1\end{pmatrix}$

$R_{1} \rightarrow R_{1} + (-2)R_{2}$ $\begin{pmatrix}1&0&-1\\0&1 &2\\0&0 &1\end{pmatrix}$

$\frac{R_{1} \rightarrow R_{1} + R_{3}}{R_{2} \rightarrow R_{1} + (-2)R_{3}}$ $\begin{pmatrix}1&0&0\\0&1 &0\\0&0 &1\end{pmatrix}$ is the RREF