Question Number 60303 by Cheyboy last updated on 19/May/19
Answered by tw000001 last updated on 18/Oct/19
![det determinant ((2,(−1),5),(3,4,0),((−2),1,9))=154 → [(2,(−1),5),(3,4,0),((−2),1,9) ]^(−1) = [(((18)/(77)),(1/(11)),(−((10)/(77)))),((−((27)/(154))),(2/(11)),((15)/(154))),((1/(14)),0,(1/(14))) ] det determinant ((9,7,5),(1,4,3),(7,5,(−2)))=−161 → [(9,7,5),(1,4,3),(7,5,(−2)) ]^(−1) = [((1/7),(−((39)/(161))),(−(1/(161)))),((−(1/7)),((53)/(161)),((22)/(161))),((1/7),(−(4/(161))),(−((29)/(161)))) ]](https://www.tinkutara.com/question/Q71620.png)
$$\mathrm{det}\begin{vmatrix}{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{5}}\\{\mathrm{3}}&{\mathrm{4}}&{\mathrm{0}}\\{−\mathrm{2}}&{\mathrm{1}}&{\mathrm{9}}\end{vmatrix}=\mathrm{154} \\ $$$$\rightarrow\begin{bmatrix}{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{5}}\\{\mathrm{3}}&{\mathrm{4}}&{\mathrm{0}}\\{−\mathrm{2}}&{\mathrm{1}}&{\mathrm{9}}\end{bmatrix}^{−\mathrm{1}} =\begin{bmatrix}{\frac{\mathrm{18}}{\mathrm{77}}}&{\frac{\mathrm{1}}{\mathrm{11}}}&{−\frac{\mathrm{10}}{\mathrm{77}}}\\{−\frac{\mathrm{27}}{\mathrm{154}}}&{\frac{\mathrm{2}}{\mathrm{11}}}&{\frac{\mathrm{15}}{\mathrm{154}}}\\{\frac{\mathrm{1}}{\mathrm{14}}}&{\mathrm{0}}&{\frac{\mathrm{1}}{\mathrm{14}}}\end{bmatrix} \\ $$$$\mathrm{det}\begin{vmatrix}{\mathrm{9}}&{\mathrm{7}}&{\mathrm{5}}\\{\mathrm{1}}&{\mathrm{4}}&{\mathrm{3}}\\{\mathrm{7}}&{\mathrm{5}}&{−\mathrm{2}}\end{vmatrix}=−\mathrm{161} \\ $$$$\rightarrow\begin{bmatrix}{\mathrm{9}}&{\mathrm{7}}&{\mathrm{5}}\\{\mathrm{1}}&{\mathrm{4}}&{\mathrm{3}}\\{\mathrm{7}}&{\mathrm{5}}&{−\mathrm{2}}\end{bmatrix}^{−\mathrm{1}} =\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{7}}}&{−\frac{\mathrm{39}}{\mathrm{161}}}&{−\frac{\mathrm{1}}{\mathrm{161}}}\\{−\frac{\mathrm{1}}{\mathrm{7}}}&{\frac{\mathrm{53}}{\mathrm{161}}}&{\frac{\mathrm{22}}{\mathrm{161}}}\\{\frac{\mathrm{1}}{\mathrm{7}}}&{−\frac{\mathrm{4}}{\mathrm{161}}}&{−\frac{\mathrm{29}}{\mathrm{161}}}\end{bmatrix} \\ $$