Question Number 55841 by gunawan last updated on 05/Mar/19
![If A= [(( 4),(x+2)),((2x−3),(x+1)) ] is symmetric, then x=](https://www.tinkutara.com/question/Q55841.png)
$$\mathrm{If}\:{A}=\begin{bmatrix}{\:\:\:\:\mathrm{4}}&{{x}+\mathrm{2}}\\{\mathrm{2}{x}−\mathrm{3}}&{{x}+\mathrm{1}}\end{bmatrix}\:\mathrm{is}\:\mathrm{symmetric},\:\mathrm{then}\:{x}= \\ $$
Answered by gunawan last updated on 05/Mar/19
![A=A^t [(4,(x+2)),((2x−3),(x+1)) ]= [(4,(2x−3)),((x+2),(x+1)) ] x+2=2x−3 x=5](https://www.tinkutara.com/question/Q55848.png)
$${A}={A}^{{t}} \\ $$$$\begin{bmatrix}{\mathrm{4}}&{{x}+\mathrm{2}}\\{\mathrm{2}{x}−\mathrm{3}}&{{x}+\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{4}}&{\mathrm{2}{x}−\mathrm{3}}\\{{x}+\mathrm{2}}&{{x}+\mathrm{1}}\end{bmatrix} \\ $$$${x}+\mathrm{2}=\mathrm{2}{x}−\mathrm{3} \\ $$$${x}=\mathrm{5} \\ $$