Question

Show that $\left|\begin{array}{ccc}a+b+2c&a&b\\c&b+c+2a&b\\c&a&c+a+2b\end{array}\right|$ = 2(a + b + c)³

Answer 1

Answer:

See the attachment

.

Step-by-step explanation:

In the attachment I have answered this problem.

See the attachment for detailed solution.

I hope this answer helps you

Answer 2

$\left|\begin{array}{ccc}a+b+2c&a&b\\c&b+c+2a&b\\c&a&c+a+2b\end{array}\right|$

R3-> R3-R2

$\left|\begin{array}{ccc}a+b+2c&a&b\\c&b+c+2a&b\\c-c&a-b-c-2a&c+a+2b-b\end{array}\right|$

$\left|\begin{array}{ccc}a+b+2c&a&b\\c&b+c+2a&b\\0&-(a+b+c)&(a+b+c)\end{array}\right|$

taking (a+b+c) common from R3

$(a+b+c)\left|\begin{array}{ccc}a+b+2c&a&b\\c&b+c+2a&b\\0&-1&1\end{array}\right|$

C3->C3+C2

$(a+b+c)\left|\begin{array}{ccc}a+b+2c&a&b+a\\c&b+c+2a&2a+2b+c\\0&-1&0\end{array}\right|$

R1-> R1-R2

$(a+b+c)\left|\begin{array}{ccc}a+b+c&-a-b-c&-a-b-c\\c&b+c+2a&2a+2b+c\\0&-1&0\end{array}\right|$

taking (a+b+c) common from R1

$(a+b+c)^{2}\left|\begin{array}{ccc}1&-1&-1\\c&b+c+2a&2a+2b+c\\0&-1&0\end{array}\right|$

now expand the determinant along R3

$= 1(2a + 2b + 2c)( {a + b + c})^{2} \\ \\ = 2(a + b + c)( {a + b + c})^{2} \\ \\ = 2 {(a + b + c)}^{3} \\ \\ =R.H.S.$

Hence proved