Question

Prove That

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

Answer 1

$det\left(\begin{array}{ccc}y+k& y& y\\ y& y+k& y\\ y& y& y+k\end{array}\right)$

$=(y+k)det\left(\begin{array}{cc}y+k& y\\ y& y+k\end{array}\right)-ydet\left(\begin{array}{cc}y& y\\ y& y+k\end{array}\right)+ydet\left(\begin{array}{cc}y& y+k\\ y& y\end{array}\right)$

$det\left(\begin{array}{cc}y+k& y\\ y& y+k\end{array}\right)=2ky+k^2$

$det\left(\begin{array}{cc}y& y\\ y& y+k\end{array}\right)=ky$

$det\left(\begin{array}{cc}y& y+k\\ y& y\end{array}\right)=-ky$

$=(y+k)(2ky+k^2)-yky+y(-ky)$

$=k^3+3k^{2}y$

$=k^2(3y+k)$