I am sure we all might have multiplied a number with it’s reciprocal to find the value of a particular equation avoiding the division operation for simplicity. This is exactly what inversion means. As in the case of matrices, there is no division operator, we multiply by the inverse matrix for dividing purpose.

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For example, if you had to solve (5/8)Â *x* = 10, you could still divide both sides by 5/8, but it would probably be easier to multiply both sides by 8/5. The reciprocal fraction 8/5 is the inverse of 5/8 when you multiply the two fractions, you get 1, which is, here, called as “the (multiplicative) identity”. 1 is called the identity because multiplying something by 1 doesn’t change its value.

This terminology and these facts are very important for matrices too. If you are given a matrix equation like ABÂ = C, where AÂ and C is given and you are to find out*Â *B, you would like to “divide off” the matrix A. But we just learned that division is not possible with matrices.

Well, what if you could find the inverse of A, something similar to finding the reciprocal fraction above? The inverse of A, written as A^{-1}Â and pronounced as A inverse, would allow you to cancel off the A from the matrix equation and then solve for B.

AB = C

A^{â€“1}ABÂ = A^{â€“1}C

IBÂ = A^{â€“1}C

B = A^{â€“1}C

**Browse more Topics under Matrices**

- Matrix
- Types of Matrices
- Addition of Matrices
- Scalar Multiplication of Matrices
- Symmetric and Skew-Symmetric Matrices
- Multiplication of Matrices
- Elementary Operation of a Matrix
- Transpose of a Matrix

## Invertible Matrices

If A is a square matrix 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^{â€“1}. For example, letÂ $$A=\begin{bmatrix} 2 & 3\\ 1 & 2\end{bmatrix} and \: B=\begin{bmatrix} 2 & -3\\ -1 & 2\end{bmatrix}$$ be two matrices.

$$A B =\begin{bmatrix} 2 & 3\\ 1 & 2\end{bmatrix}\begin{bmatrix} 2 & -3\\ -1 & 2\end{bmatrix}$$

$$ A B =\begin{bmatrix} 4 Â – Â 3 & -6 + 6\\ 2 Â – 2 & -3 + 4\end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$$

$$ B A=\begin{bmatrix} 1 Â – Â 0\\ 0 + 1\end{bmatrix} = I$$

Thus B is the inverse of A. In other words B = A^{-1 Â }and A is the inverse of B.Â ie. A = B^{-1}

### Theorem 1

(Uniqueness of inverse) The Inverse of a square matrix, if it exists, is unique. Proof Let A = [a_{ij}] be a square matrix of order m. If possible, let B and C be two inverses of A. We shall show that B = C. Since B is the inverse of A

AB = BA = I ……….(1)

Since C is also the inverse of A

AC = CA = I ……….(2)

Thus B = BI = B (AC) = (BA) C = IC = C

### Theorem 2

If A and B are invertible matrices of the same order, then (AB)^{â€“1} = B^{â€“1} A^{â€“1}.

**Proof:** From the definition of the inverseÂ matrix, we have

(AB) (AB)^{â€“1}Â = 1

A^{â€“1} (AB) (AB)^{â€“1}Â = A^{â€“1} I

(A^{â€“1}A) B (AB)^{â€“1}Â = A^{â€“1
}IB (AB)^{â€“1}Â = A^{â€“1
}B (AB)^{â€“1}Â = A^{â€“1
}B^{â€“1} B (AB)^{â€“1}Â = B^{â€“1} A^{â€“1
}I (AB)^{â€“1}Â Â =Â B^{â€“1} A^{â€“1
}Hence (AB)^{â€“1} = B^{â€“1} A^{â€“1}

## Inverse of a Matrix by Elementary Operations

If A is a matrix such that inverse of a matrix (A^{â€“1}) exists, then to find an inverse of a matrix using elementary row or column operations, write Â A = IA and apply a sequence of row or column operation on A = IA till we get, I = BA. The matrix B will be the inverse matrix of A. By using elementary operations, find the inverse matrix

In order to use elementary row operations, we may write A = IA.

Similarly, we can also obtain the inverse of matrix A by applying elementary column operations.

- Obtain the inverse of the matrix for 3Â Ã— 3 matrix given below

$$A =\begin{bmatrix} 1 & 2 & 0\\ 2 Â & 0 & -1\\ 2 & 3 & -1\end{bmatrix}$$

- Now we write A = IA

$$A =\begin{bmatrix} 1 & 0 & 0\\ 0 Â & 1 & 0\\ 0 & 0 & 1\end{bmatrix} Â \begin{bmatrix} 1 & 2 & 0\\ 2 Â & 0 & -1\\ 2 & 3 & -1\end{bmatrix}$$

- By applying R
_{2}Â â†’Â âˆ’R_{2}Â + R_{3}, we get,

$$A =\begin{bmatrix} 1 & 0 & 0\\ 0 Â & -1 & 0\\ 0 & 0 & 1\end{bmatrix} Â \begin{bmatrix} 1 & 2 & 0\\ 0 Â & 3 & 0\\ 2 & 3 & -1\end{bmatrix}$$

- By applying R
_{2}Â â†’Â R_{2}/ 3 and R_{3}Â â†’ – R_{3}+ 2R_{1}, we get,

$$A =\begin{bmatrix} 1 & 0 & 0\\ 0 Â & -1/3 & 1/3\\ 2 & 0 & -1 \end{bmatrix} Â \begin{bmatrix} 1 & 2 & 0\\ 0 Â & 1 & 0\\ 0 & 1 & 1\end{bmatrix}$$

- By applying R
_{1}Â â†’Â R_{1}– 2R_{2}and R_{3}Â â†’ R_{3}+ R2, we get,

$$A =\begin{bmatrix} 1 & 2/3 & -2/3\\ 0 Â & -1/3 & 1/3\\ 2 & 1/3 & -4/3 \end{bmatrix} Â \begin{bmatrix} 1 & 0 & 0\\ 0 Â & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$$

- Therefore,

$$A =\begin{bmatrix} 1 & 2/3 & -2/3 \\ 0 Â & -1/3 & 1/3\\ 2 & 1/3 & -4/3 \end{bmatrix}Â $$

## Properties of Inverse of a Square Matrix

- Square matrix A is invertible if and only if |A|â‰ 0
- (A
^{-1})^{-1}=A - (Aâ€™)
^{-1}= (A^{-1})â€™ - (AB)
^{-1}= B^{-1}A^{-1}Â In general (A_{1}A_{1}A_{1}… A_{n})^{-1}= A_{n}^{-1}A_{n}â€“ 1^{-1}… A_{3}^{-1}A_{2}^{-1}A_{1}^{-1} - If a non-singular square matrix A is symmetric, then A-1 is also symmetric.
- |A
^{-1}| = |A|^{-1} - AA
^{-1Â }= A^{-1}A = I - (A
^{k})^{-1}= (A^{-1})A^{k}k âˆˆ N - If $$A = \begin{bmatrix}a & 0 & 0\\ 0 & b & 0 \\ 0 & 0 & c\end{bmatrix}$$ and abcÂ â‰ 0, then A
^{-1Â }= \begin{bmatrix} 1/a & 0 & 0\\ 0 & 1/b & 0 \\ 0 & 0 & 1/c\end{bmatrix}

## Solved Examples For You

**Question 1: If AÂ is a square matrix of order 3Â and det A Â = 5Â then what is detÂ [(2A) ^{âˆ’1}]Â equal to?**

**1/10****2/5****8/5****1/40**

**Answer :** If A is of order 3 then, A^{-1} is also of order 3.Â Now, det (cA) = c^{n} (det A) where n is the order of the matrix. And, det A = 1 / det A^{-1.Â }Thus det [(2A)^{-1}] = 2^{3} det[A] and det[A-1] = 1 / 8 . 5 = 1 / 40

**Question 2: How can we find the inverse of a 3 x 3 matrix?**

**Answer:** For finding out the inverse of a 3 x 3 matrix, first of all, we have to evaluate the determinant of that matrix. If the determinant will be zero, the matrix will not be having any inverse. Then, move the matrix by writing the first row again as the first column, the middle row as the central column, and the 3rd row as the 3rd column.

**Question 3: How do we calculate the inverse?**

**Answer:** We can do the calculation of the Inverse of a Matrix by the following steps:

1st Step: Calculate the matrix of minors.

2nd Step: Then convert that into the cofactorâ€™s matrix.

3rd Step:Â Then, Adjugate.

4th Step: Lastly, multiply it with 1/Determinant.

**Question 4: What is the inverse of a 4 x 4 matrix?**

**Answer:** According to the definition; the inverse of an â€˜nâ€™ x â€˜nâ€™ matrix â€˜Aâ€™ is the â€˜nâ€™ x â€˜nâ€™ matrix â€˜Bâ€™. Just like â€˜ABâ€™ = â€˜BAâ€™ = â€˜Iâ€™. Hence, if we will get the inverse of a 4 x 4 matrix â€˜Aâ€™ to be â€˜Bâ€™, then for checking our work, we will have to simply multiply â€˜ABâ€™ and â€˜BAâ€™.

**Question 5: Define the rank of a matrix.**

**Answer:** The rank of a matrix is the extreme number of linearly self-determining column vectors in the matrix.

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