Question
Find the inverse of the matrix (if it exists) given $\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$
$\therefore |A| = \left| {\begin{array}{*{20}{c}} 1&2&3 \\ 0&2&4 \\ 0&0&5 \end{array}} \right| $ = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) $ = 10 \ne 0$
$\therefore {A^{ - 1}}$ exists.
${A_{11}} = + \left[ {\begin{array}{*{20}{c}} 2&4 \\ 0&5 \end{array}} \right] = + \left( {10 - 0} \right) = 10,$ ${A_{12}} = - \left| {\begin{array}{*{20}{c}} 0&4 \\ 0&5 \end{array}} \right| = - \left( {0 - 0} \right) = 0$
${A_{13}} = + \left[ {\begin{array}{*{20}{c}} 0&2 \\ 0&0 \end{array}} \right] = + \left( {0 - 0} \right) = 0,$ ${A_{21}} = - \left| {\begin{array}{*{20}{c}} 2&3 \\ 0&5 \end{array}} \right| = - \left( {10 - 0} \right) = - 10$
${A_{22}} = + \left| {\begin{array}{*{20}{c}} 1&3 \\ 0&5 \end{array}} \right| = + \left( {5 - 0} \right) = 5,$ ${A_{23}} = - \left| {\begin{array}{*{20}{c}} 1&2 \\ 0&0 \end{array}} \right| = - \left( {0 - 0} \right) = 0$
${A_{31}} = + \left| {\begin{array}{*{20}{c}} 2&3 \\ 2&4 \end{array}} \right| = + \left( {8 - 6} \right) = 2,$ ${A_{32}} = - \left| {\begin{array}{*{20}{c}} 1&3 \\ 0&4 \end{array}} \right| = - \left( {4 - 0} \right) = - 4$
${A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&2 \\ 0&2 \end{array}} \right| = + \left( {2 - 0} \right) = 2$
$\therefore adj.A = \left| {\begin{array}{*{20}{c}} {10}&0&0 \\ { - 10}&5&0 \\ 2&{ - 4}&2 \end{array}} \right|'$
$= \left| {\begin{array}{*{20}{c}} {10}&{ - 10}&2 \\ 0&5&{ - 4} \\ 0&0&2 \end{array}} \right|$
$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A=\frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {10}&{ - 10}&2 \\ 0&5&{ - 4} \\ 0&0&2 \end{array}} \right]$
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$\int\limits_{\pi/6}^{\pi/3}\frac{\text{dx}}{1+\sqrt{\tan\text{x}}}$.
| Tablets | Iron | Calcium | Vitamin |
| X | 6 | 3 | 2 |
| Y | 2 | 3 | 4 |