In the previous section,
we completed a discussion on symmetric and skew symmetric matrices. In this
section, we will see elementary operations. Later in this section, we will see invertible matrices also.
Elementary Operation of a Matrix
• There are six operations on a matrix.
♦ They are known as elementary operations.
♦ They are also known as transformations.
Operation I:
This can be explained in 2 steps:
1. First pick two suitable rows. They need not be adjacent rows. Then interchange those two rows.
• For example, suppose that, we pick the first and third rows. After the operation, we will get a new matrix such that:
♦ First row in the original matrix, is the third row in the new matrix.
♦ Third row in the original matrix, is the first row in the new matrix.
2. Symbolically, we write this operation as: $R_i \leftrightarrow R_j$.
• The example operation that we saw above, can be written symbolically as follows:
Applying $R_1 \leftrightarrow R_3$ to $\left[\begin{array}{r}
3 &{ -10 } &{ 8 } \\
11 &{ -4 } &{ -9 } \\
2 &{ 12 } &{ 13 } \\
\end{array}\right]
$, gives $\left[\begin{array}{r}
2 &{ 12 } &{ 13 } \\
11 &{ -4 } &{ -9 } \\
3 &{ -10 } &{ 8 } \\
\end{array}\right]$.
Operation II:
This can be explained in 2 steps:
1. First pick a single suitable row. Then multiply each element of that row by a non zero number k.
• For example, suppose that, we pick the second row. After the operation, we will get a new matrix such that:
Each element in the second row is k times the original element.
2. Symbolically, we write this operation as: $R_i \rightarrow k R_i$.
• The example operation that we saw above, can be written symbolically as follows:
Applying $R_2 \rightarrow 3 R_2$ to $\left[\begin{array}{r}
3 &{ -10 } &{ 8 } \\
11 &{ -4 } &{ -9 } \\
2 &{ 12 } &{ 13 } \\
\end{array}\right]
$, gives $\left[\begin{array}{r}
3 &{ -10 } &{ 8 } \\
33 &{ -12 } &{ -27 } \\
2 &{ 12 } &{ 13 } \\
\end{array}\right]$.
Operation III:
This can be explained in 2 steps:
1. First pick a single suitable row. Then multiply each element of that row by a non zero number k. Finally, add each member of the new row with the corresponding element of another suitable row.
2. Symbolically, we write this operation as: $R_i \rightarrow R_i + k R_j$.
• Let us see an example:
Applying $R_3 \rightarrow R_3 - 2 R_1$ to $\left[\begin{array}{r}
3 &{ -10 } &{ 8 } \\
11 &{ -4 } &{ -9 } \\
2 &{ 12 } &{ 13 } \\
\end{array}\right]
$, gives $\left[\begin{array}{r}
3 &{ -10 } &{ 8 } \\
11 &{ -4 } &{ -9 } \\
-4 &{ 32 } &{ -3 } \\
\end{array}\right]$.
Operation IV:
• This is similar to the operation I. Instead of rows, we deal with columns.
• An example:
Applying $C_2 \leftrightarrow C_3$ to $\left[\begin{array}{r}
3 &{ -10 } &{ 8 } \\
11 &{ -4 } &{ -9 } \\
2 &{ 12 } &{ 13 } \\
\end{array}\right]
$, gives $\left[\begin{array}{r}
3 &{ 8 } &{ -10 } \\
11 &{ -9 } &{ -4 } \\
2 &{ 13 } &{ 12 } \\
\end{array}\right]
$.
Operation V:
• This is similar to the operation II. Instead of rows, we deal with columns.
• An example:
Applying $C_3 \rightarrow 2C_3$ to $\left[\begin{array}{r}
3 &{ -10 } &{ 8 } \\
11 &{ -4 } &{ -9 } \\
2 &{ 12 } &{ 13 } \\
\end{array}\right]
$, gives $\left[\begin{array}{r}
3 &{ -10 } &{ 16 } \\
11 &{ -4 } &{ -18 } \\
2 &{ 12 } &{ 26 } \\
\end{array}\right]
$.
Operation VI:
• This is similar to the operation III. Instead of rows, we deal with columns.
• An example:
Applying $C_1 \rightarrow C_1 + 1.5 C_2$ to $\left[\begin{array}{r}
3 &{ -10 } &{ 8 } \\
11 &{ -4 } &{ -9 } \\
2 &{ 12 } &{ 13 } \\
\end{array}\right]
$, gives $\left[\begin{array}{r}
-12 &{ -10 } &{ 8 } \\
5 &{ -4 } &{ -9 } \\
20 &{ 12 } &{ 13 } \\
\end{array}\right]
$.
Invertible Matrices
This can be explained in 5 steps:
1. Consider two matrices A and B.
2. Suppose that, A and B satisfy three conditions:
(i) Both A and B are square matrices.
(ii) Both A and B are of the same order m.
(iii) AB = BA = I
3. If the three conditions are satisfied, then:
• B is called the inverse matrix of A.
♦ The inverse matrix of A is denoted as A-1.
♦ So B = A-1.
• A is called the inverse matrix of B.
♦ The inverse matrix of B is denoted as B-1.
♦ So A = B-1.
4. If it is possible to find A-1, then we say that:
A is an invertible matrix.
5. An example is shown below:
• In the above example, we can write:
♦ AB = BA = I
♦ B is the inverse of A. In other words, B = A-1.
♦ A is the inverse of B. In other words, A = B-1.
♦ A is an invertible matrix.
♦ B is an invertible matrix.
Note:
A rectangular matrix will not have an inverse. This can be demonstrated using an example. It can be written in 6 steps:
1. Suppose that, A is a rectangular matrix, of order 2 × 3
2. Let B be the inverse of A
Then AB = BA
3. If A is to be multiplied by B, then a possible order of B is: 3 × 2
• In that case, the order of AB will be 2 × 2.
4. The order of BA will be 3 × 3.
5. So we get:
♦ Order of AB = 2 × 2
♦ Order of BA = 3 × 3
• In such a situation, AB cannot be equal to BA.
• So B cannot be the inverse of A.
6. We can write:
Both A and B must be square matrices and both should be of the same order.
In the next section, we will see theorems related to invertible matrices.
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