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Question Number 178870 by ARUNG_Brandon_MBU last updated on 22/Oct/22

Answered by Rasheed.Sindhi last updated on 22/Oct/22

Rule: determinant (( determinant ((a)),( ), determinant ((b))),(( ), determinant ((((ad+bc)/2))),( )),( determinant ((c)),( ), determinant ((d)))) Applying rule: determinant (( determinant (((x+2))),( ), determinant (((x+1)))),(( ), determinant ((((((x+2)(3)+(x)(x+1))/2)=1))),( )),( determinant ((( x ))),( ), determinant ((( 3 ))))) (((x+2)(3)+(x)(x+1))/2)=1 3x+6+x^2 +x=2 x^2 +4x+4=0 (x+2)^2 =0 x=−2

$$\mathrm{Rule}: \\ $$$$\begin{array}{|c|c|c|}{\begin{array}{|c|}{{a}}\\\hline\end{array}}&\hline{\:\:\:\:\:}&\hline{\begin{array}{|c|}{{b}}\\\hline\end{array}}\\{\:\:\:}&\hline{\begin{array}{|c|}{\frac{{ad}+{bc}}{\mathrm{2}}}\\\hline\end{array}}&\hline{\:\:}\\{\begin{array}{|c|}{{c}}\\\hline\end{array}}&\hline{\:\:\:}&\hline{\begin{array}{|c|}{{d}}\\\hline\end{array}}\\\hline\end{array}\:\:\: \\ $$$$\mathrm{Applying}\:\mathrm{rule}: \\ $$$$\begin{array}{|c|c|c|}{\begin{array}{|c|}{{x}+\mathrm{2}}\\\hline\end{array}}&\hline{\:\:\:\:\:}&\hline{\begin{array}{|c|}{{x}+\mathrm{1}}\\\hline\end{array}}\\{\:\:\:}&\hline{\begin{array}{|c|}{\frac{\left({x}+\mathrm{2}\right)\left(\mathrm{3}\right)+\left({x}\right)\left({x}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{1}}\\\hline\end{array}}&\hline{\:\:}\\{\begin{array}{|c|}{\:\:\:\:{x}\:\:\:}\\\hline\end{array}}&\hline{\:\:\:}&\hline{\begin{array}{|c|}{\:\:\:\:\mathrm{3}\:\:}\\\hline\end{array}}\\\hline\end{array} \\ $$$$\:\:\:\frac{\left({x}+\mathrm{2}\right)\left(\mathrm{3}\right)+\left({x}\right)\left({x}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{1} \\ $$$$\:\:\:\mathrm{3}{x}+\mathrm{6}+{x}^{\mathrm{2}} +{x}=\mathrm{2} \\ $$$$\:\:\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\left({x}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:{x}=−\mathrm{2} \\ $$$$\:\:\: \\ $$

Commented by ARUNG_Brandon_MBU last updated on 22/Oct/22

Great! Thanks Sir

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