Noon Academy
Using properties of determinants prove that:
contexto.140
2 years ago
Solution:
![=\left(\frac{1}{3}\right)\left|\begin{array}{ccc}3 a & 3 a+6 b & 3 a+6 b+9 c \\ 3 a & 4 a+6 b & 5 a+7 b+9 c \\ 6 a & 9 a+12 b & 11 a+15 b+18 c\end{array}\right|\left[R_{1}^{\prime}=3 R_{1}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-22e2143d1dd590a1aa237e0f39eea316_l3.png "Rendered by QuickLaTeX.com")
![=\left(\frac{1}{3}\right)\left|\begin{array}{ccc}0 & -a & -2 a-b \\ 3 a & 4 a+6 b & 5 a+7 b+9 c \\ 6 a & 9 a+12 b & 11 a+15 b+18 c\end{array}\right|\left[R_{1}^{\prime}=R_{1}-R_{2}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6825f886d547d95702fdaf444f5f50ee_l3.png "Rendered by QuickLaTeX.com")
![=\left(\frac{1}{6}\right)\left|\begin{array}{ccc}0 & -\mathrm{a} & -2 \mathrm{a}-\mathrm{b} \\ 6 \mathrm{a} & 8 \mathrm{a}+12 \mathrm{~b} & 10 \mathrm{a}+14 \mathrm{~b}+18 \mathrm{c} \\ 6 \mathrm{a} & 9 \mathrm{a}+12 \mathrm{~b} & 11 \mathrm{a}+15 \mathrm{~b}+18 \mathrm{c}\end{array}\right|\left[\mathrm{R}_{2}^{\prime}=2 \mathrm{R}_{2}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0876db8d4bdd915b5c35e0ba488b3756_l3.png "Rendered by QuickLaTeX.com")
![=\left(\frac{1}{6}\right)\left|\begin{array}{ccc}0 & -a & -2 a-b \\ 0 & -a & -a-b \\ 6 a & 9 a+12 b & 11 a+15 b+18 c\end{array}\right|\left[R_{2}^{\prime}=R_{2}-R_{3}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-48ec79f3e29ebe7f2d2cdcd1e2111333_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} =(1 / 6)[0+0+6 a\{a(a+b)-a(2 a+b)[\text { expansion by first column }] \\ =-a^{3} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-92e43c7fe94bcddc3cbd58a216b456a8_l3.png "Rendered by QuickLaTeX.com")
