Find the inverse of the matrix (if it exists) given \(\left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ 4&3 \end{array}} \right]\)
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Find the inverse of the matrix (if it exists) given \(\left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ 4&3 \end{array}} \right]\)
Answer
Let \(A = \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ 4&3 \end{array}} \right]\)
\(\therefore \left| A \right| = \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ 4&3 \end{array}} \right] = 6 - \left( { - 8} \right) = 6 + 8 = 14 \ne 0\)
\(\therefore\) Matrix A is non-singular and hence \({A^{ - 1}}\) exist.
Now adj. A \( = \left[ {\begin{array}{*{20}{c}} 3&2 \\ { - 4}&2 \end{array}} \right]\) And \({A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A = \frac{1}{{14}}\left[ {\begin{array}{*{20}{c}} 3&2 \\ { - 4}&2 \end{array}} \right]\)
\(\therefore \left| A \right| = \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ 4&3 \end{array}} \right] = 6 - \left( { - 8} \right) = 6 + 8 = 14 \ne 0\)
\(\therefore\) Matrix A is non-singular and hence \({A^{ - 1}}\) exist.
Now adj. A \( = \left[ {\begin{array}{*{20}{c}} 3&2 \\ { - 4}&2 \end{array}} \right]\) And \({A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A = \frac{1}{{14}}\left[ {\begin{array}{*{20}{c}} 3&2 \\ { - 4}&2 \end{array}} \right]\)
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