Question

solve the problem.....
PROVE through properties of determinant ​

Answer 1

Answer:

Hope the attachment helps you

Answer 2

To Prove:

$\left|\begin{array}{ccc}x+y+2z&x&y\\z&y+z+2x&y\\z&x&z+x+2y\end{array}\right|=2(x+y+z)^{3}$

Solution:

$\sf{L.H.S.}=\left|\begin{array}{ccc}x+y+2z&x&y\\z&y+z+2x&y\\z&x&z+x+2y\end{array}\right|$

$\mathrm{C_{1}\longrightarrow C_{1}+C_{2}+C_{3}}$

$\left|\begin{array}{ccc}2(x+y+z)&x&y\\2(x+y+z)&y+z+2x&y\\2(x+y+z)&x&z+x+2y\end{array}\right|$

On taking 2(x+y+z) common from C₁, we get

$2(x+y+z)\left|\begin{array}{ccc}1&x&y\\1&y+z+2x&y\\1&x&z+x+2y\end{array}\right|$

On transposing the above determinant, we get

$2(x+y+z)\left|\begin{array}{ccc}1&1&1\\x&y+z+2x&x\\y&y&z+x+2y\end{array}\right|$

$\mathrm{C_{2}\longrightarrow C_{2}-C_{1}\:and\:C_{3}\longrightarrow C_{3}-C_{1}}$

$2(x+y+z)\left|\begin{array}{ccc}1&0&0\\x&x+y+z&0\\y&0&x+y+z\end{array}\right|$

On expanding above determinant, we get

$\mathrm{2(x+y+z)\Big(1(x+y+z)^{2}+0+0\Big)}$

$\mathrm{2(x+y+z)(x+y+z)^{2}}$

$\mathrm{2(x+y+z)^{3}}$

$\sf{=R.H.S.}$

Hence Proved

Note: Value of determinant does not change on transposing.