# Inverse Of Matrix

In linear algebra, matrix inverse holds a special place because there is not division in matrix algebra. You cannot divide two matrices. Fortunately, the division is possible when a matrix is multiplied with its inverse which is unique.

The inverse is not possible with just any kind of matrix, a matrix must be square and invertible and the reasons are explained in this article along with several identities and examples involving inverse matrices.

An inverse matrix can also be used for finding the solution for system of linear equations, that is, $Ax = b$ where $A$ is augmented matrix, $x$ is the solution vector and $b$ is the constant vector.

### What Is The Need For Inverse?

Inverse means opposite of some operation performed and the result obtained is identity of that operation.

For example,

Additive identity
------------------
If $a + 0 = a$ then $0$ is additive identity.

$3 + 0 = 3$

but

$3 - 3 = 0$ (additive identity)

Therefore, subtraction is inverse operation of addition.

Multiplicative identity
-----------------------

If $a \times 1 = a$ then $1$ is multiplicative identity because it gives $a$ as result.

$4 \times 1 = 4$

but

$4 \times \frac{1}{4} = 1$ ( multiplicative identity)

Therefore, multiplying with reciprocal or division is inverse operation of multiplication.

The same idea can be extended to matrix since we are unable to divide two matrices directly. If $A$ is a square matrix and invertible , then find an inverse matrix $A^{-1}$ such that multiplying it with $A$ will give an identity matrix $I$ of same order.

$A.A^{-1} = A^{-1}.A = I$

For example,

Let $A$ be a square matrix of order 2 x 2. The inverse of matrix $A$ is $A^{-1}$.

$A = \begin{bmatrix} 2 & 1\\ 2 & 6\end{bmatrix}$

Compute the determinant

$Det(A) = 2\times 6 - 1 \times 2 = 12 - 2 = 10$

Add negative to following elements in matrix $A$

$A = \begin{bmatrix} 2 & 1\\ 2 & 6\end{bmatrix} = \begin{bmatrix} + & -\\ - & +\end{bmatrix} = \begin{bmatrix} 2 & -1\\ -2 & 6\end{bmatrix}$

Swap the positive entries

$\begin{bmatrix} 2 & -1\\ -2 & 6\end{bmatrix} = \begin{bmatrix} 6 & -1\\ -2 & 2\end{bmatrix}$

Multiply the above result with $\frac{1}{det(A)}= \frac{1}{10}$

$A^{-1} = \frac{1}{10}\begin{bmatrix} 6 & -1\\ -2 & 2\end{bmatrix}$
$= \begin{bmatrix} \frac{3}{5} & \frac{-1}{10}\\ \frac{-1}{5} & \frac{-1}{5}\end{bmatrix}$

Let us now verify whether $A.A^{-1} = A^{-1}.A = I$

$\begin{bmatrix} 2 & 1\\ 2 & 6\end{bmatrix} \times \begin{bmatrix} \frac{3}{5} & \frac{-1}{10}\\ \frac{-1}{5} & \frac{-1}{5}\end{bmatrix}$

$= \begin{bmatrix} \frac{6}{5}- \frac{1}{5} & \frac{-1}{5}+ \frac{1}{5}\\ \frac{6}{5}- \frac{6}{5} & \frac{-1}{5}+ \frac{6}{5}\end{bmatrix}$

$= \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$

### Why Square Matrix ?

The inverse deals with negative power such as $A^{-1}$, a non-square matrix is cannot be used because it is undefined( cannot multiply).

The second reason for using square matrix is the identity matrix. An identity matrix is a square matrix only. A product of non-square matrix with its inverse will not result in an identity matrix.

If a square matrix $A$ has inverse matrix $B$ such that

$AB = BA = I$

Then the matrix $A$ is called invertible matrix and matrix $B$ is its inverse. If there is no $B$ for matrix $A$, then it is called Singular matrix.

A matrix is singular and has no inverse if its determinant is 0. You will learn about determinants in future lessons.

Suppose $A$ is a singular matrix of order 2 x 2.

$A = \begin{bmatrix}a & b\\c & d\end{bmatrix}$

$ad - bc = 0$

In the same manner, determinants of higher order matrices is found.

Therefore, only square matrix is used to find inverse which is also a square matrix of size $n \times n$.

### Uniqueness Of Inverse Matrix

If a square matrix $A$ is invertible, then it has exactly one inverse.

Proof :

Suppose that there are two inverse $B$ and $C$ for matrix $A$. We get

$AB = BA = I$  - (1)

$AC = CA = I$  - (2)

We know that any matrix multiplied by Identity matrix will result itself. Therefore, the following is true.

$B = B.I$

$B = B(AC)$ \\by (2)

$B = (BA)C$  \\ by associativity property

$B = I.C$  \\ by (1)

$B = C$

Therefore, inverse is unique.

### Use Of Inverse Matrix

The purpose of using matrix is to solve for $A.x = b$ where $A$ represents the augmented matrix obtained from the system of linear equations, $x$ is the vector of unknowns or solution vector and $b$ is the constant vector.

$A.x = b => \begin{bmatrix}a & b\\c & d\end{bmatrix}.\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}b_1 \\ b_2\end{bmatrix}$

Multiply both sides by $A^{-1}$. Note that the order of operation is important.

$(A^{-1}A)x = A^{-1}b$

By $A^{-1}A = AA^{-1} = I$, we get

$Ix = A^{-1}b => x = A^{-1}b$

Let us try to solve a system of equation using above result where $A$ matrix is invertible and square. Suppose the system of equations have following equation.

$2x + 4y = 10$
$3x + 2y = 7$
Let $A$ be a square and invertible augmented matrix of order 2 x 2 derived from the system of equations above. Therefore, $A.x = b$ is as follows.

$A.x = b$

$\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix} . \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}10 \\ 7\end{bmatrix}$

Let us find the inverse of matrix $A$. But, first we must find the determinant of matrix $A$.

$Det(A) = ad - bc = 4 - 12 = -8$

Change the sign and swap the positive entries. Then multiply it with $\frac{1}{Det(A)}$ to get the inverse of matrix $A$.

$\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix} = \begin{bmatrix}+ & -\\- & +\end{bmatrix} = \begin{bmatrix}2 & -4\\-3 & 2\end{bmatrix}$

Swap the positive entries.

$\begin{bmatrix}2 & -4\\-3 & 2\end{bmatrix} = \begin{bmatrix}2 & -4\\-3 & 2\end{bmatrix}$

Multiply with $\frac{\strut 1}{\strut -8}$

$= \frac{\strut 1}{\strut -8} \times \begin{bmatrix}2 & -4\\-3 & 2\end{bmatrix}$

$A^{-1} = \begin{bmatrix}\frac{\strut -1}{\strut 4} & \frac{\strut 1}{\strut 2}\\\frac{\strut 3}{\strut 8} & \frac{\strut -1}{\strut 4}\end{bmatrix}$

We need to verify if this is correct inverse .

$A^{-1}. A = \begin{bmatrix}\frac{\strut -1}{\strut 4} & \frac{\strut 1}{\strut 2}\\\frac{\strut 3}{\strut 8} & \frac{\strut -1}{\strut 4}\end{bmatrix} . \begin{bmatrix}2 & 4\\3 & 2\end{bmatrix}= I$

$= \begin{bmatrix}\frac{\strut -1}{\strut 2}+ \frac{\strut 3}{\strut 2} & -1 + 1\\ \frac{\strut 3}{\strut 4}+ \frac{\strut -3}{\strut 4} & \frac{\strut 3}{\strut 2}+ \frac{\strut -1}{\strut 2}\end{bmatrix}$

$= \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$

The inverse is correct and compute the value of solution vector $x$ using $x = A^{-1}.b$ in the same order.

$x = A^{-1}. b$ where $A^{-1} = \begin{bmatrix}\frac{\strut -1}{\strut 4} & \frac{\strut 1}{\strut 2}\\ \frac{\strut 3}{\strut 8} & \frac{\strut -1}{\strut 4}\end{bmatrix}$ and $b = \begin{bmatrix}10 \\ 7\end{bmatrix}$

$x = \begin{bmatrix}\frac{\strut -10}{\strut 4}+\frac{\strut 7}{\strut 2}\\ \frac{\strut 30}{\strut 8}+\frac{\strut -7}{\strut 4} \end{bmatrix}$

$x = \begin{bmatrix}\frac{\strut -5}{\strut 2}+\frac{\strut 7}{\strut 2}\\\frac{\strut 15}{\strut 4}+\frac{\strut -7}{\strut 4} \end{bmatrix}$

$x = \begin{bmatrix}1 \\ 2\end{bmatrix}$

We must verify the solution by substitution in the system of linear equations.

$2(1) + 4(2) = 10$
$3(1) + 2(2) = 7$

Similarly, we can verify some other interesting results in the following section.

### Other Interesting Results

In this section, we will verify some other interesting results concerning inverse matrices.

(a) Product of two or more invertible matrices are invertible matrix.

$(AB)^{-1} = B^-1A^{-1}$ //order is important

Proof:

Let $A$ and $B$ be two invertible matrices of order $n \times n$. Then $(AB) = P$. If matrix $P$ is invertible, then its inverse is $P^{-1} = B^{-1}A^{-1}$.

Therefore,

$PP^{-1} = P^{-1}P = I_n$

$(AB)(B^{-1}A^{-1})$

$= A(BB^{-1})A^{-1}$ //because $AA^{-1}=I_n$

$= I_nAA^{-1}$

$= I_n$

Example #1

Let $A$ and $B$ be $2 \times 2$ invertible matrix.

$A = \begin{bmatrix} 1 & 5\\0 & 9\end{bmatrix}$ $B = \begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}$

Let $AB = C$ then

$C = \begin{bmatrix} 1 & 5\\0 & 9\end{bmatrix}\times \begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}$

$C = \begin{bmatrix}2 + 15 & 1+20 \\0 + 27 & 0 + 36\end{bmatrix}$

$C = \begin{bmatrix}17 & 21\\ 27 & 36\end{bmatrix}$

We will now find the inverse of the product matrix $C$, that is, $C^{-1}$. First compute the determinant of $C$ matrix.

$Det(C) = 17\times36 - 21\times27 = 612 - 567 = 45$

Now change value of element $a_{12}$ and $a_{21}$ to negative in matrix $C$. Then swap the remaining positive values. Multiply the resultant matrix with $\frac{1}{45}$.

$C^{-1} = \frac{\strut 1}{\strut 45}\times \begin{bmatrix}36 & -21\\-27&17\end{bmatrix}$

$C^{-1}= \begin{bmatrix}\frac{\strut 36}{\strut 45} & \frac{\strut -21}{\strut 45}\\\frac{\strut -27}{\strut 45}&\frac{\strut 17}{\strut 45}\end{bmatrix}$

$C^{-1} = \begin{bmatrix}\frac{\strut 4}{\strut 5} & \frac{\strut -7}{\strut 15}\\\frac{\strut -3}{\strut 5}&\frac{\strut 17}{\strut 45}\end{bmatrix}$

We must find the product matrix $B^{-1}A^{-1}$.

$B^{-1} = \begin{bmatrix}\frac{\strut 4}{\strut 5} & \frac{\strut -1}{\strut 5}\\ \frac{\strut -3}{\strut 5}& \frac{\strut 2}{\strut 5}\end{bmatrix}$

$A^{-1} = \begin{bmatrix}1 & \frac{\strut -5}{\strut 9}\\ 0 & \frac{\strut 1}{\strut 9}\end{bmatrix}$

$B^{-1}.A^{-1} = \begin{bmatrix}\frac{\strut 4}{\strut 5} & \frac{\strut -1}{\strut 5}\\ \frac{\strut -3}{\strut 5}& \frac{\strut 2}{\strut 5}\end{bmatrix} \times \begin{bmatrix}1 & \frac{\strut -5}{\strut 9}\\ 0 & \frac{\strut 1}{\strut 9}\end{bmatrix}$

$B^{-1}.A^{-1} = \begin{bmatrix}\frac{\strut 4}{\strut 5} + 0 & \frac{\strut -20}{\strut 45} + \frac{\strut -1}{\strut 45}\\ \frac{\strut -3}{\strut 5}+0 & \frac{\strut 15}{\strut 45}+\frac{\strut 2}{\strut 45}\end{bmatrix}$

$B^{-1}.A^{-1} = \begin{bmatrix}\frac{\strut 4}{\strut 5}&\frac{\strut -7}{\strut 15}\\ \frac{\strut -3}{\strut 5}& \frac{\strut 17}{\strut 45}\end{bmatrix}$

Therefore, $(AB)^{-1} = B^{-1}.A^{-1}$

(b) Inverse of inverse matrix is the original matrix.

Let $A^{-1}$ be a $1 \times 1$ invertible matrix. Let $a \in A$. Therefore, inverse of matrix $A$ is the matrix $A^{-1}$ where $\frac{\strut 1}{\strut a} \in A^{-1}$.

We know that $A.A^{-1} = A^{-1}.A = I_n$.

$\begin{bmatrix}a\end{bmatrix} \times \begin{bmatrix}\frac{\strut 1}{\strut a}\end{bmatrix} = \begin{bmatrix}1\end{bmatrix}$

Therefore,

$(A^{-1})^{-1} = (\begin{bmatrix}\frac{\strut 1}{\strut a}\end{bmatrix})^{-1}$

$= (\begin{bmatrix}\frac{\strut 1}{\strut a}\end{bmatrix})^{-1} = \begin{bmatrix}a\end{bmatrix}$

$= \begin{bmatrix}a\end{bmatrix}= A$

$(A^{-1})^{-1} = A$

Example #2

Let $A$ be a $2 \times 2$ invertible matrix.

$A = \begin{bmatrix}1 & 2\\ 3 & 9\end{bmatrix}$

The inverse of the matrix $A$ is

$A^{-1} = \begin{bmatrix}3 & \frac{\strut -2}{\strut 3}\\-1 & \frac{\strut 1}{\strut 3}\end{bmatrix}$

Let us take inverse of inverse matrix $A^{-1}$

$Det(A^{-1}) = 1 - \frac{\strut 2}{\strut 3} = \frac{\strut 1}{\strut 3}$

Change signs and swap positive values in $A^{-1}$

$= \begin{bmatrix}\frac{\strut 1}{\strut 3} & \frac{\strut 2}{\strut 3}\\1 & 3\end{bmatrix}$

Multily above result with $\frac{\strut 1}{\strut Det(A^{-1})}$.

$(A^{-1})^{-1} = 3 \times \begin{bmatrix}\frac{\strut 1}{\strut 3} & \frac{\strut 2}{\strut 3}\\1 & 3\end{bmatrix}$

$(A^{-1})^{-1} = \begin{bmatrix}1 & 2 \\3 & 9\end{bmatrix}= A$


Therefore, $(A^{-1})^{-1} = A$.

(c) If non-negative power of a invertible square matrix $A$ is $A^n = A.A.A.A(n-times)$, then negative power of invertible square matrix is

$A^{-n} = A^{-1}.A^{-1}.A^{-1}.A^{-1}(n-times)$

Example #3

Let $A$ be a invertible square matrix of order $2 \times 2$. Let $n = 3$ be a positive integer.

$A^3 = \begin{bmatrix}2 & 3 \\ 1 & 5\end{bmatrix} \times \begin{bmatrix}2 & 3 \\ 1 & 5\end{bmatrix} \times \begin{bmatrix}2 & 3 \\ 1 & 5\end{bmatrix}$

$A^3 = \begin{bmatrix}35 & 126 \\ 42 & 161\end{bmatrix}$

Let $A^{-1}$ be the inverse matrix for $A$.

$A^{-1} = \begin{bmatrix}\frac{\strut 5}{\strut 7}& \frac{\strut -3}{\strut 7}\\ \frac{\strut -1}{\strut 7}& \frac{\strut 2}{\strut 7}\end{bmatrix}$

$A^{-3} = \begin{bmatrix}\frac{\strut 23}{\strut 49}& \frac{\strut -18}{\strut 49}\\ \frac{\strut -6}{\strut 49}& \frac{\strut 5}{\strut 49}\end{bmatrix}$

But we know that $A^nA^{-n} = A^{n-n} = A^0 = I$

$A^3A^{-3} = A^{3-3} = A^0 = I$

$A^3A^{-3} = \begin{bmatrix}35 & 126 \\ 42 & 161\end{bmatrix} \times \begin{bmatrix}\frac{\strut 23}{\strut 49}& \frac{\strut -18}{\strut 49}\\ \frac{\strut -6}{\strut 49}& \frac{\strut 5}{\strut 49}\end{bmatrix}$

$A^3A^{-3} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = I$

Therefore, $A^{-n} = A^{-1}.A^{-1} (n-times)$.

(d) If $k$ is a non-zero scalar and $kA$ is invertible square matrix, then

$kA = \frac{\strut 1}{\strut k}A^{-1}$

Proof:

We know that $A^{-1}A = AA^{-1} = I$ and also following algebraic identities applies in the case of matrix multiplication with scalars.

$a(bP) = abP$ (1)

$aP(Q) = P(aQ)$ (2)

where $P$ and $Q$ are defined matrices.

Using equation $AA^{-1}$ we get

$(kA)\left(\frac{\strut 1}{\strut k} \right). A^{-1}= I$

Using equation (2)

$\frac{\strut 1}{\strut k}.(kA)(A^{-1})= I$

Using equation (1)

$\frac{\strut 1}{\strut k} . k(A)(A^{-1})= I$

$\frac{\strut 1}{\strut k} . k . I = I$

$(1) . I = I$


Therefore, $(kA)^{-1} = \frac{\strut 1}{\strut k}A^{-1}$ is true.

Example #4

Let $k = 2$ and matrix $A$ is invertible and order 2 x 2.

$A = \begin{bmatrix}3 & 1 \\4 & 2\end{bmatrix}$ $A^{-1} = \begin{bmatrix}1 & \frac{\strut -1}{\strut 2} \\-2 & \frac{\strut 3}{\strut 2}\end{bmatrix}$

Multiply $k$ with matrix $A$ and take inverse.

$kA = 2 \times \begin{bmatrix}3 & 1 \\4 & 2\end{bmatrix} = \begin{bmatrix}6 & 2 \\8 & 4\end{bmatrix}$

Take determinant of the matrix $kA$

$Det(A) = 6 \times 4 - 8 \times 2 = 24 - 16 = 8$

Take negative of $a_{12}$ and $a_{21}$ and swap positive values. Multiply with $\frac{\strut 1}{\strut Det(A)}$

$(kA)^{-1} = \frac{\strut 1}{\strut 8} \times \begin{bmatrix}4 & -2 \\-8 & 6\end{bmatrix} = \begin{bmatrix}\frac{\strut 1}{\strut 2} & \frac{\strut -1}{\strut 4} \\-1 & \frac{\strut 3}{\strut 4}\end{bmatrix}$ (3)

We must compute the value of $\frac{\strut 1}{\strut k}A^{-1}$

$\frac{\strut 1}{\strut k}A^{-1}= \frac{\strut 1}{\strut 2} \times\begin{bmatrix}1 & \frac{\strut -1}{\strut 2} \\-2 & \frac{\strut 3}{\strut 2}\end{bmatrix} = \begin{bmatrix}\frac{\strut 1}{\strut 2} & \frac{\strut -1}{\strut 4} \\-1 & \frac{\strut 3}{\strut 4}\end{bmatrix}$ (4)

$(kA)^{-1} = \frac{\strut 1}{\strut k}A^{-1}$ (5)

(e) If $A$ is an invertible matrix of order $n \times n$ then the transpose $A^T$ is also invertible and equal to transpose of inverse matrix $A^T$.

$\left(A^T\right)^{-1}= \left(A^{-1}\right)^{-1}$

Example #5

Let $A$ matrix of order $2 \times 2$.

$A = \begin{bmatrix}1 & 2\\3 & 8\end{bmatrix}$ $A^T = \begin{bmatrix}1 & 3\\2 & 8\end{bmatrix}$

Inverse of Transpose $A^T$

$\left(A^T\right)^{-1} = \frac{\strut 1}{\strut 2} \times \begin{bmatrix}8 & -3\\-2 & 1 \end{bmatrix}= \begin{bmatrix}4 & \frac{\strut -3}{\strut 2}\\-1 & \frac{\strut 1}{\strut 2} \end{bmatrix}$

Transpose of Inverse $A^{-1}$

$A^{-1} = \frac{\strut 1}{\strut 2} \times \begin{bmatrix}8 & -2\\-3 & 1\end{bmatrix}$

$\left(A^{-1}\right)^{T}= \begin{bmatrix}4 & -1\\\frac{\strut -3}{\strut 2} & \frac{\strut 1}{\strut 2}\end{bmatrix}^T$

$\left(A^{-1}\right)^{T}= \begin{bmatrix}4 & \frac{\strut -3}{\strut 2}\\-1 & \frac{\strut 1}{\strut 2}\end{bmatrix}$

Therefore, $(A^T)^{-1} = (A^{-1})^{T}$.

In this article, we explained why and what are inverse of matrix. Next, we discuss how to obtain inverse of small to large invertible matrices using different available methods.