# 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

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

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

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

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

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

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