- Books Name
- Mathmatics Book Based on NCERT

- Publication
- KRISHNA PUBLICATIONS

- Course
- CBSE Class 12

- Subject
- Mathmatics

**Invertible Matrices and Inverse of a matrix by elementary operations**

Elementary Operation (Transformation) of a Matrix

There are six operations (transformations) on a matrix, three of which are due to rows, and three are due to columns, known as elementary operations or transformations.

**[1] **The interchange of any two rows or two columns.

denoted by Ri ↔ Rj and interchange of i th and j th column is denoted by Ci ↔ Cj .

**[2] **The multiplication of the elements of any row or column by a non zero number.

denoted by Ri → kRi . The corresponding column operation is denoted by Ci → kCi

**[3] **The addition to the elements of any row or column, the corresponding elements of any other row or column are multiplied by any non zero number.

denoted by Ri → Ri + kRj . The corresponding column operation is denoted by Ci → Ci + kCj .

Elementary transformations are **operations done on the rows and columns of matrices to change their shape so that the computations become easier**. It is also used to discover the inverse of a matrix, the determinants of a matrix, and to solve a system of linear equations. A square matrix is always an elementary matrix.

**Invertible Matrices **

Suppose a square matrix A of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A, and it is denoted by A^{-1}. Also, matrix A is said to be an invertible matrix here.

Note:-

**[1]** A rectangular matrix does not possess inverse matrix, since for products BA and AB to be defined and to be equal, it is necessary that matrices A and B should be square matrices of the same order.

**[2]** If B is the inverse of A, then A is also the inverse of B.

Theorem 4: If A and B are invertible matrices of the same order,

then (AB)^{–1} = B^{–1 }A^{–1}

Proof : From the definition of inverse of a matrix,

we have (AB) (AB)^{–1} = I

or A^{–1} (AB) (AB)^{–1} =A^{–1}I (Pre multiplying both sides by A^{–1})

or (A^{–1}A) B (AB)^{–1} =A^{–1} (Since A^{–1} I = A^{–1})

or IB (AB)^{ –1} =A^{–1} or B (AB)^{ –1} =A^{–1}

or B^{–1} B (AB)^{ –1} =B^{–1} A^{–1}

or I (AB)^{ –1} =B^{–1} A^{–1}

Hence (AB)^{ –1} =B^{–1} A^{–1}

**Example 1: **

, then find the value of a, b, c, x, y, and z.

**Solution:**

It is given that, the two matrices are equal. Therefore, the corresponding elements present in matrices should be equal to each other. By comparing the corresponding elements in the matrices, we get:

x+3 = 0. **⇒ x = -3**

z +4 = 6 ⇒ z = 6-4

**⇒ z = 2**

2y-7 = 3y-2 ⇒3y-2y =-7+2

** ⇒y = -5 **

a-1 = -3

⇒a = -3+1

** ⇒a=-2**

2c+2 = 0

⇒2c = -2

**⇒ c = -1**

b-3 = 2b+4

⇒2b-b = -3-4

**⇒ b = -7**

Therefore, the values of the variables are:

a = -2

b = -7

c = -1

x = -3

y = -5

z = 2

**Example 2:**

Now, we need to calculate the transpose of AB.

Hence verified.