Question Number 175193 by Best1 last updated on 22/Aug/22
Answered by Rasheed.Sindhi last updated on 22/Aug/22
![A= [(2,1,0),(1,2,1),(3,3,2) ], A^T = [(2,1,3),(1,2,3),(0,1,2) ] AA^T = [((4+1+0),(2+2+0),(6+3+0)),((2+2+0),(1+4+1),(3+6+2)),((6+3+0),(3+6+2),(9+9+4)) ] AA^T = [(5,4,9),(4,6,(11)),(9,(11),(22)) ] 2AA^T = [((10),8,(18)),(8,(12),(22)),((18),(22),(44)) ] determinant (((2AA^T )))=2∙2 determinant ((5,8,9),(4,(12),(11)),(9,(22),(22))) =4 determinant ((5,8,9),(4,(12),(11)),(9,(22),(22))) =4 determinant ((5,8,( 1)),(4,(12),(−1)),(9,(22),( 0))) C3−C2 =4 determinant ((5,8,( 1)),(9,(20),( 0)),(9,(22),( 0))) R2+R1 =4{1(9∙22−9∙20)} = 4∙9(22−20)=4∙9∙2=72](https://www.tinkutara.com/question/Q175195.png)
$$ \\ $$$${A}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{3}}&{\mathrm{2}}\end{bmatrix},\:{A}^{{T}} =\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{2}}\end{bmatrix} \\ $$$${AA}^{{T}} =\begin{bmatrix}{\mathrm{4}+\mathrm{1}+\mathrm{0}}&{\mathrm{2}+\mathrm{2}+\mathrm{0}}&{\mathrm{6}+\mathrm{3}+\mathrm{0}}\\{\mathrm{2}+\mathrm{2}+\mathrm{0}}&{\mathrm{1}+\mathrm{4}+\mathrm{1}}&{\mathrm{3}+\mathrm{6}+\mathrm{2}}\\{\mathrm{6}+\mathrm{3}+\mathrm{0}}&{\mathrm{3}+\mathrm{6}+\mathrm{2}}&{\mathrm{9}+\mathrm{9}+\mathrm{4}}\end{bmatrix}\: \\ $$$${AA}^{{T}} =\begin{bmatrix}{\mathrm{5}}&{\mathrm{4}}&{\mathrm{9}}\\{\mathrm{4}}&{\mathrm{6}}&{\mathrm{11}}\\{\mathrm{9}}&{\mathrm{11}}&{\mathrm{22}}\end{bmatrix}\: \\ $$$$\mathrm{2}{AA}^{{T}} =\begin{bmatrix}{\mathrm{10}}&{\mathrm{8}}&{\mathrm{18}}\\{\mathrm{8}}&{\mathrm{12}}&{\mathrm{22}}\\{\mathrm{18}}&{\mathrm{22}}&{\mathrm{44}}\end{bmatrix}\: \\ $$$$\begin{vmatrix}{\mathrm{2}{AA}^{{T}} }\end{vmatrix}=\mathrm{2}\centerdot\mathrm{2}\begin{vmatrix}{\mathrm{5}}&{\mathrm{8}}&{\mathrm{9}}\\{\mathrm{4}}&{\mathrm{12}}&{\mathrm{11}}\\{\mathrm{9}}&{\mathrm{22}}&{\mathrm{22}}\end{vmatrix}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\begin{vmatrix}{\mathrm{5}}&{\mathrm{8}}&{\mathrm{9}}\\{\mathrm{4}}&{\mathrm{12}}&{\mathrm{11}}\\{\mathrm{9}}&{\mathrm{22}}&{\mathrm{22}}\end{vmatrix}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\begin{vmatrix}{\mathrm{5}}&{\mathrm{8}}&{\:\:\mathrm{1}}\\{\mathrm{4}}&{\mathrm{12}}&{−\mathrm{1}}\\{\mathrm{9}}&{\mathrm{22}}&{\:\:\mathrm{0}}\end{vmatrix}\:{C}\mathrm{3}−{C}\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\begin{vmatrix}{\mathrm{5}}&{\mathrm{8}}&{\:\:\mathrm{1}}\\{\mathrm{9}}&{\mathrm{20}}&{\:\:\mathrm{0}}\\{\mathrm{9}}&{\mathrm{22}}&{\:\:\mathrm{0}}\end{vmatrix}\:{R}\mathrm{2}+{R}\mathrm{1} \\ $$$$=\mathrm{4}\left\{\mathrm{1}\left(\mathrm{9}\centerdot\mathrm{22}−\mathrm{9}\centerdot\mathrm{20}\right)\right\} \\ $$$$=\:\mathrm{4}\centerdot\mathrm{9}\left(\mathrm{22}−\mathrm{20}\right)=\mathrm{4}\centerdot\mathrm{9}\centerdot\mathrm{2}=\mathrm{72} \\ $$
Answered by kapoorshah last updated on 23/Aug/22
$$ \\ $$$$\mid{A}\mid\:=\:\mathrm{3} \\ $$$$\mid\mathrm{2}{AA}^{{T}} \mid\:=\:\mathrm{2}^{\mathrm{3}} \mid{A}\mid\mid{A}^{{T}} \mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{8}\mid{A}\mid\mid{A}\mid\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{8}.\mathrm{3}.\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{72} \\ $$