Matrices and Determinants
6.0 Inverse of a Matrix
6.0 Inverse of a Matrix
A non-singular square matrix of order $n$ is invertible if there exists a squre matrix $B$ of the same order such that $AB$=${I_n}$ =$BA$.
In such a case ,we say that the inverse of $A$ is $B$ and we write , ${A^{ - 1}}$=$B$.
If $A$ is invertible, then the inverse of $A$ is given by ${A^{ - 1}} = \frac{{adjA}}{{\left| A \right|}}$.
Question 13.
Find Inverse of the matrix $A = \left[ {\begin{array}{c} 1&2&3 \\ 0&5&0 \\ 2&4&3 \end{array}} \right]$
Solution:
Let $C$ be the matrix of cofactors of the element in $\left| A \right|$
then $C = \left( {\begin{array}{c} {{C_{11}}}&{{C_{12}}}&{{C_{13}}} \\ {{C_{21}}}&{{C_{22}}}&{{C_{23}}} \\ {{C_{31}}}&{{C_{32}}}&{{C_{33}}} \end{array}} \right)$
Here ,
\[{C_{11}} = \left| {\begin{array}{*{20}\[{C}} 5&0 \\ 4&3 \end{array}} \right| = 15 \]
\[{C_{12}} = - \left| {\begin{array}{*{20}\[{C}} 0&0 \\ 2&3 \end{array}} \right| = 0 \]
\[{C_{13}} = \left| {\begin{array}{*{20}\[{C}} 0&5 \\ 2&4 \end{array}} \right| = - 10 \]
\[{C_{21}} = - \left| {\begin{array}{*{20}\[{C}} 2&3 \\ 4&3 \end{array}} \right| = 6 \]
\[{C_{22}} = \left| {\begin{array}{*{20}\[{C}} 1&3 \\ 2&3 \end{array}} \right| = - 3 \]
\[{C_{23}} = \left| {\begin{array}{*{20}\[{C}} 1&2 \\ 2&4 \end{array}} \right| = 0 \]
\[{C_{31}} = \left| {\begin{array}{*{20}\[{C}} 2&3 \\ 5&0 \end{array}} \right| = - 15 \]
\[{C_{32}} = - \left| {\begin{array}{*{20}\[{C}} 1&3 \\ 0&0 \end{array}} \right| = 0 \]
\[{C_{33}} = \left| {\begin{array}{*{20}\[{C}} 1&2 \\ 0&5 \end{array}} \right| = 5 \]
\[C = \left( {\begin{array}{c} {15}&0&{ - 10} \\ 6&{ - 3}&0 \\ { - 15}&0&5 \end{array}} \right)\therefore AdjA = {C^T} = \left( {\begin{array}{c} {15}&6&{ - 15} \\ 0&{ - 3}&0 \\ { - 10}&0&5 \end{array}} \right)\]
\[\left| A \right| = 1\left| {\begin{array}{c} 5&0 \\ 4&3 \end{array}} \right| - 2\left| {\begin{array}{c} 0&0 \\ 2&3 \end{array}} \right| + 3\left| {\begin{array}{c} 0&5 \\ 2&4 \end{array}} \right| = 1(15 - 0) - 2(0 - 0) + 3(0 - 10) = - 15\]
${A^{ - 1}} = \frac{{adjA}}{{\left| A \right|}}$
${A^{ - 1}} = \frac{{ - 1}}{{15}}\left( {\begin{array}{c} {15}&6&{ - 15} \\ 0&{ - 3}&0 \\ { - 10}&0&5 \end{array}} \right)$
