Noon Academy
Using properties of determinants prove that:
contexto.140
2 years ago
Solution:
![\begin{array}{l} \left|\begin{array}{ccc} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array}\right| \\ =\left|\begin{array}{ccc} a+b & a+b & -(a+b) \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array}\right|\left[R_{1}^{\prime}=R_{1}+R_{2}\right] \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4c90e2aa0169fb26bd3c7fe388275c74_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} =(a+b)\left|\begin{array}{ccc} 1 & 1 & -1 \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array}\right|\left[R_{1}^{\prime}=R_{1} /(a+b)\right] \\ =(a+b)\left|\begin{array}{ccc} 1 & 1 & -1 \\ -c-b & b+c & b+c \\ -b & -a & a+b+c \end{array}\right|\left[R_{2}^{\prime}=R_{2}+R_{3}\right] \\ =(a+b)\left|\begin{array}{ccc} 1 & 1 & -1 \\ -(b+c) & b+c & b+c \\ -b & -a & a+b+c \end{array}\right| \\ =(a+b)(b+c)\left|\begin{array}{ccc} 1 & 1 & -1 \\ -1 & 1 \\ -b & -a & a+b+c \end{array}\right|\left[R_{2}^{\prime}=R_{1} /(b+c)\right] \\ =(a+b)(b+c)\left|\begin{array}{ccc} 0 & 2 & 0 \\ -1 & 1 & 1 \\ -b & -a & a+b+c \end{array}\right|\left[R_{1}^{\prime}=R_{1}+R_{2}\right] \\ =(a+b)(b+c)\{0+2(-b+a+b+c)+0\}[e x p a n s i o n b y \text { first row }] \\ =\mathbf{2}(a+b)(b+c)(c+a) \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a2f34cccc1cf90f64d010f9ebae6f89d_l3.png "Rendered by QuickLaTeX.com")
