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240 LearnersLast updated on October 29, 2025

The inverse of a matrix A can be represented as A–1. We use a direct formula to calculate the inverse of a 2 × 2 matrix. For matrices of order 3 or higher, we calculate the determinant and the adjoint to find the inverse. The matrix inversion method is useful for solving linear equations effectively. In this article, we will discuss the inverse of a matrix and its wider applications in real life.
The inverse of a matrix is what results in the identity matrix when multiplied by the original. Note that a matrix can have an inverse only if it has a non-zero determinant, and such matrices are called invertible matrices.
For example:
The inverse of a square matrix A, denoted as A⁻¹, is a matrix that, when multiplied with A, yields the identity matrix. If A is a square matrix, then its inverse satisfies \(A × A⁻¹ = A⁻¹ × A = I\)
\(A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \)
So, the inverse is:
\(A^{-1} = \begin{pmatrix} 4 & 3 \\ -1 & 2 \end{pmatrix} \div (2 \times 4 - 3 \times 1) = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix} \div 5 = \begin{pmatrix} \frac{4}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix} \)
so,
\(A \cdot A^{-1} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \times \begin{pmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \)
and,
\(A^{-1} \cdot A = \begin{pmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{pmatrix} \times \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \)
We represent the inverse of any square matrix A as A–1. Additionally, the product of a matrix and its inverse always results in the identity matrix I.
The inverse of a square matrix A is given by the formula:
\(A–1= 1A× adj (A)\)
Where:
|A| → determinant of matrix A
Adj (A) → adjoint or transpose of cofactors of A
What are the properties of Inverse of a Matrix?
The inverse of the matrix has certain key features, as discussed below:
(A–1)-1= A
\((AB)-1= B–1 A–1\)
\( (A^{T})^{-1} = (A^{-1})^{T} \)
How to Find the Matrix Inverse?
The two common methods to determine the inverse of a matrix are:
Using Matrix Formula
The inverse of a matrix is the adjoint divided by its determinant.
\(
A^{-1} = \frac{\text{Adj } A}{|A|}
\)
To find the matrix inverse, we use two Inverse matrix methods, such as:
Determinant Method: The inverse of a matrix is calculated using the determinant and adjoint. The formula for a square matrix A:
\(
A^{-1} = \frac{1}{\det(A)} \times \text{adj}(A)
\)
Where:
A–1 → inverse of matrix A
det (A)→ determinant of A
adj(A) → adjoint of A
If the determinant of A is zero, the matrix has no inverse.
Elementary Transformation Method: In this method, we find the inverse using a step-by-step process:
Step 1: Express the matrix equation as [A\I]. Here, I is the identity matrix, which is in the same order as the matrix A.
Step 2: To convert matrix A into the identity matrix on the left, use only row operations or only column operations. Apply the same step to the identity matrix on the right side.
The matrix B on the right is the inverse of A once you obtain: \(I = B × A\).
Step 3: Use either row operations or column operations exclusively, not both simultaneously.
Using this method, we can quickly determine the inverse of a 2 × 2 matrix.
2 × 2 matrix Inverse
The inverse of a 2 × 2 matrix can be found using a simple method:
\(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)
Let’s say: Since \( |A| = (ad - bc)\)
\(\text{Adj } A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \)
Using the formula:
\(
A^{-1} = \frac{1}{|A|} \times \text{adj } A
\)
So,
\(A^{-1} = \frac{1}{(ad - bc)} \times \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \)
Hence, the inverse of the 2 × 2 matrix is determined.
Inverse of a 3 × 3 Matrix
In the case of any 3 × 3 matrix:
\(A = \begin{pmatrix} x & y & z \\ l & m & n \\ a & b & c \end{pmatrix} \)
Use the inverse matrix formula:
\(
A^{-1} = \frac{1}{|A|} \times \text{adj}(A)
\)
Determinant of Inverse Matrix
The determinant of an inverse matrix is derived by taking the reciprocal of the original matrix’s determinant:
\(
\det(A^{-1}) = \frac{1}{\det(A)}
\)
Proof (in simple steps):
Step 1: We know, det(A × B) = det(A) × det(B)\(det(A × B) = det(A) × det(B)\)
Step 2: Also, \(A × A⁻¹ = I\) (by inverse matrix property)
Step 3: So, \(det(A × A⁻¹) = det(I)\)
\( \det(A^{-1}) = \frac{1}{\det(A)} \)
\(
\det(A^{-1}) = \frac{1}{\det(A)}
\)
Step 4: Thus, the determinant of the inverse is the reciprocal of the original determinant, as shown.
Understanding the inverse of a matrix is crucial for students in various fields. And here are some useful tips and tricks to master the inverse of a matrix.
The inverse of a matrix is not just a mathematical concept; it has numerous practical uses beyond math. Here are a few real-life instances where it can be applied.
The inverse of a matrix is an important concept in mathematics. However, students often make mistakes while calculating it. Here are a few common mistakes and tips to avoid them.
Find the inverse of a 2×2 matrix. A = [ 3 5 , 2 7]
\(A^{-1} = \begin{pmatrix} \frac{7}{11} & -\frac{5}{11} \\ -\frac{2}{11} & \frac{3}{11} \end{pmatrix} \)
We first find the determinant
\(det (A) = (3) (7) – (5) (2) = 21 – 10 = 11\).
To find the adjoint, we need to swap the main diagonal and change the signs of the off-diagonal elements:
\(\text{Adj}(A) = \begin{bmatrix} 7 & -5 \\ -2 & 3 \end{bmatrix} \)
Now, use the inverse formula:
\(A^{-1} = \frac{1}{11} \begin{pmatrix} 7 & -5 \\ -2 & 3 \end{pmatrix} \)
\(\Rightarrow A^{-1} = \begin{pmatrix} \frac{7}{11} & -\frac{5}{11} \\ -\frac{2}{11} & \frac{3}{11} \end{pmatrix} \)
Find the inverse of a 2×2 matrix. A = [4 7, 2 6]
\(A^{-1} = \frac{1}{\det (A)} \times \text{adj} (A) = \frac{1}{10} \begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix} \)
\(\text{adj}(A) = \begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix} \)
Using the formula for inverse:
\(A^{-1} = \frac{1}{\det (A)} \times \text{adj} (A) = \frac{1}{10} \begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix} \)
Find the inverse of a 3×3 matrix. A = [1 2 3, 0 1 4, 5 6 0]
\(A^{-1} = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & -4 \\ 5 & -4 & 1 \end{pmatrix} \)
The first step is to find the determinant
\(det (A) = 1 (1 × 0 – 4 × 6) – 2 (0 × 0 – 4 × 5) + 3 (0 × 6 – 1 × 5) \)
\(det (A) = 1(0 – 24) – 2(0 – 20) + 3 (0 – 5) = 24 + 40 – 15 = 1\)
Let’s first find the determinant
\(det (A) = (4) (6) – (7) (2) = 24 – 14 = 10\).
We now find the adjoint matrix:
To find the adjoint matrix, we first calculate the cofactor matrix and then take its transpose.
\(\text{Cof}(A) = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & -4 \\ 5 & -4 & 1 \end{pmatrix} \)
Since Adjoint of A = Transpose of the cofactor matrix:
\(\text{adj}(A) = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & -4 \\ 5 & -4 & 1 \end{pmatrix} \)
Here, we use the formula for inverse:
\(
A^{-1} = \frac{1}{\det(A)} \times \text{adj}(A)
\)
so,
\(A^{-1} = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & -4 \\ 5 & -4 & 1 \end{pmatrix} \)
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






