Question

Show that $\left|\begin{array}{ccc}y+z&x&x\\y&z+x&y\\z&z&x+y\end{array}\right|$ = 4xyz

Answer 1

Answer:

Step-by-step explanation:

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

i am going to split the given determinant into two parts for simplify it

$\left|\begin{array}{ccc}y&x&x\\y&z&y\\z&z&x\end{array}\right| +\left|\begin{array}{ccc}z&x&x\\y&x&y\\z&z&y\end{array}\right| \\\\R_{2} -> R_{2}-R_{1} \\\\\left|\begin{array}{ccc}y&x&x\\0&z-x&y-x\\z&z&x\end{array}\right| +\left|\begin{array}{ccc}z&x&x\\y-z&0&y-x\\z&z&y\end{array}\right| \\\\\\C_{2} -> C_{2}-C_{3}\:\:\:C_{3} -> C_{3}-C_{2}\\\\\left|\begin{array}{ccc}y&0&x\\0&z-y&z-x\\z&z&x\end{array}\right| +\left|\begin{array}{ccc}z&x&0\\y-z&0&y-x\\z&z&y-z\end{array}\right|$

$R_{3} -> R_{3}-R_{1}\:\:\:$

$\left|\begin{array}{ccc}y&0&x\\0&z-y&z-x\\z-y&z&0\end{array}\right| +\left|\begin{array}{ccc}z&x&0\\y-z&0&y-x\\0&z-x&y-z\end{array}\right|$

now expand the determinant

$\left|\begin{array}{ccc}y&0&x\\0&z-y&z-x\\z-y&z&0\end{array}\right| +\left|\begin{array}{ccc}z&x&0\\y-z&0&y-x\\0&z-x&y-z\end{array}\right|\\\\\\=[y(0-z(z-x)+x(-(z-y)^{2} ]+[z(-(z-x)(y-x))-x(y-z)^{2} )]\\\\=-yz^{2} +xyz+x(-z^{2} -y^{2}+2zy)+[z(zy-zx-xy+x^{2})-x(y^{2}+z^{2}-2yz)]\\\\= -yz^{2}+xyz -xz^{2}- xy^{2}+2xyz+yz^{2} +xz^{2}-xyz +xz^{2} +xy^{2} -xz^{2}+2xyz\\\\=4xyz$

hence proved