Inverse matrix
The transpose of the matrix obtained by replacing the element of the square matrix A by the corresponding cofactor is called adjoint of the matrix of A and it denote by adj(A)
A square matrix A is said to be an invertible matrix if there exist a square matrix B such that AB = BA = I the matrix B is called invers of A matrix and it denoted by A⁻¹.
If A is an inversible then \({{A}^{-1}}=\frac{adj(A)}{\det (A)}\) .
Example: find the adjoint and inverses of the following matrices
1) \(A=\left[ \begin{matrix} 2 & -3 \\ 4 & 6 \\\end{matrix} \right]\).
Solution: Given that \(A=\left[ \begin{matrix} 2 & -3 \\ 4 & 6 \\\end{matrix} \right]\),
\(A=\left[ \begin{matrix} a & b \\ c & d \\\end{matrix} \right]=\left[ \begin{matrix} 2 & -3 \\ 4 & 6 \\\end{matrix} \right]\),
\(adj(A)=\left[ \begin{matrix} d & -b \\ -c & d \\\end{matrix} \right]=\left[ \begin{matrix}6 & 3 \\ -4 & 2 \\\end{matrix} \right]\),
|A| = 12 – (- 12) = 24
\({{A}^{-1}}=\frac{adj(A)}{\det (A)}=\frac{1}{24}\left[ \begin{matrix} 6 & 3 \\ -4 & 2 \\\end{matrix} \right]\).
2) \(A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]\).
Solution: Given that \(A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]\),
\(A=\left[ \begin{matrix} a & b \\ c & d \\\end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]\),
\(adj(A)=\left[ \begin{matrix} d & -b \\ -c & d \\\end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]\),
|A| = 1 – 0 = 1
\({{A}^{-1}}=\frac{adj(A)}{\det (A)}=\frac{1}{1}\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]\),
\({{A}^{-1}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right]\).