Inverse Matrix Theorem

Inverse Matrix Theorem

If A is a nonsingular matrix then A is invertible and \({{A}^{-1}}=\frac{adjA}{\det A}\).

Proof: Let A = \(\left( \begin{matrix}   {{a}_{1}} & {{b}_{1}} & {{c}_{1}}  \\   {{a}_{2}} & {{b}_{2}} & {{c}_{2}}  \\   {{a}_{3}} & {{b}_{3}} & {{c}_{3}}  \\\end{matrix} \right)\) be a nonsingular matrix.

Therefore detA not equal to zero

\(adjA=\left( \begin{matrix}   {{A}_{1}} & {{A}_{2}} & {{A}_{3}}  \\   {{B}_{1}} & {{B}_{2}} & {{B}_{3}}  \\   {{C}_{1}} & {{C}_{2}} & {{C}_{3}}  \\\end{matrix} \right) \),

\(A.adjA=\left( \begin{matrix}   {{a}_{1}} & {{b}_{1}} & {{c}_{1}}  \\   {{a}_{2}} & {{b}_{2}} & {{c}_{2}}  \\   {{a}_{3}} & {{b}_{3}} & {{c}_{3}}  \\\end{matrix} \right)\ \left( \begin{matrix}   {{A}_{1}} & {{A}_{2}} & {{A}_{3}}  \\   {{B}_{1}} & {{B}_{2}} & {{B}_{3}}  \\   {{C}_{1}} & {{C}_{2}} & {{C}_{3}}  \\\end{matrix} \right) \),

\(=\left( \begin{matrix}{{a}_{1}}{{A}_{1}}+{{b}_{1}}{{B}_{1}}+{{c}_{1}}{{C}_{1}} &{{a}_{1}}{{A}_{2}}+{{b}_{1}}{{B}_{2}}+{{c}_{1}}{{C}_{2}} &{{a}_{1}}{{A}_{3}}+{{b}_{1}}{{B}_{3}}+{{c}_{1}}{{C}_{3}}  \\  {{a}_{2}}{{A}_{1}}+{{b}_{2}}{{B}_{2}}+{{c}_{2}}{{C}_{1}} &{{a}_{2}}{{A}_{2}}+{{b}_{2}}{{B}_{2}}+{{c}_{2}}{{C}_{2}} &{{a}_{2}}{{A}_{3}}+{{b}_{2}}{{B}_{3}}+{{c}_{2}}{{C}_{3}}  \\   {{a}_{3}}{{A}_{1}}+{{b}_{3}}{{B}_{2}}+{{c}_{3}}{{C}_{1}} &{{a}_{3}}{{A}_{2}}+{{b}_{3}}{{B}_{2}}+{{c}_{3}}{{C}_{2}} &{{a}_{3}}{{A}_{3}}+{{b}_{3}}{{B}_{3}}+{{c}_{3}}{{C}_{3}}  \\\end{matrix} \right)\),

\(=\left( \begin{matrix}   \det A & 0 & 0  \\   0 & \det A & 0  \\   0 & 0 & \det A  \\\end{matrix} \right)=\det A\left( \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\\end{matrix} \right)\).

A.adjA = detA.I

Therefore \(A.\frac{adjA}{\det A}=1\),

Similarly we can prove that,

\(\frac{adjA}{\det A}.A=1\).

\({{A}^{-1}}=\frac{adjA}{\det A}\).

Example 1) find the adjoint and inverses of the following matrix \(A=\left[ \begin{matrix} 1 & -3 \\   4 & 1  \\\end{matrix} \right]\).

Solution: Given that \(A=\left[ \begin{matrix}   1 & -3  \\   4 & 1  \\\end{matrix} \right]\),

\(A=\left[ \begin{matrix}   a & b  \\   c & d  \\\end{matrix} \right]=\left[ \begin{matrix}   1 & -3  \\   4 & 1  \\\end{matrix} \right]\),

\(adj(A)=\left[ \begin{matrix}   d & -b  \\   -c & d  \\\end{matrix} \right]=\left[ \begin{matrix}   1 & 3  \\   -4 & 1  \\\end{matrix} \right]\),

|A| = 1 – (-12) = 13

\({{A}^{-1}}=\frac{adj(A)}{\det (A)}=\frac{1}{13}\left[ \begin{matrix}1 & 3  \\ -4 & 1  \\\end{matrix} \right]\).