Step 1

An \(n \times n\) square matrix A is called Invertible Matrices, if there exists an n x n square matrix B such that AB = BA = I.

\(AB=BA=I_n\)

\(A_{n \times n} , B_{n \times n} \text{ and } I_n \text{ is identity matrix of } n \times n\)

Step 2

In other words, determinant of A is non - zero, then matric A is invertible. \(det(A)=|A| \neq 0\)

Step 3

For example, let us assume

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

Step 4

So, the determinant of A is

\(|A|=\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)

\(=1 \cdot (3) - 2 \cdot (-1)\)

\(=3+2\)

\(=5\)

\(|A| \neq 0 \rightarrow A \text{ is invertible.}\)

An \(n \times n\) square matrix A is called Invertible Matrices, if there exists an n x n square matrix B such that AB = BA = I.

\(AB=BA=I_n\)

\(A_{n \times n} , B_{n \times n} \text{ and } I_n \text{ is identity matrix of } n \times n\)

Step 2

In other words, determinant of A is non - zero, then matric A is invertible. \(det(A)=|A| \neq 0\)

Step 3

For example, let us assume

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

Step 4

So, the determinant of A is

\(|A|=\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)

\(=1 \cdot (3) - 2 \cdot (-1)\)

\(=3+2\)

\(=5\)

\(|A| \neq 0 \rightarrow A \text{ is invertible.}\)