Question

Without expanding the determinants, show that... (we should use the properties of determinant)​

Answer 1

Step-by-step explanation:

$\left|\begin{array}{ccc}1&1&1\\a&b&c\\b+c&c{+a&a+b\end{array}\right|\\ \\R3----> R3+R2\\\left|\begin{array}{ccc}1&1&1\\a&b&c\\a+b+c&c{+a+b&a+b+c\end{array}\right|\\ \\\\Take (a+b+c) common from R3\\\left|\begin{array}{ccc}1&1&1\\a&b&c\\1&1&1\end{array}\right|\\ \\\\Since R1=R3,\\\left|\begin{array}{ccc}1&1&1\\a&b&c\\b+c&c{+a&a+b\end{array}\right|\\ \\ =0$

Answer 2

Answer:

this is your last question's answer

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