19.13 - Invertible Matrices

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.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com