Question
By using properties of determinants, show that:
$\begin{vmatrix}y+k&y&y\\y&y+k&y\\y&y&y+k\end{vmatrix}=k^2(3y+k)$
$\begin{vmatrix}y+k&y&y\\y&y+k&y\\y&y&y+k\end{vmatrix}=k^2(3y+k)$
$=\begin{vmatrix}3y+k&y&y\\3y+k&y+k&y\\3y+k&y&y+k\end{vmatrix}\ \left[\text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3\right]$
$=(3y+k)\begin{vmatrix}1&y&y\\1&y+k&y\\1&y&y+k\end{vmatrix}$
$=(3y+k)\begin{vmatrix}1&y&y\\0&k&0\\0&0&k\end{vmatrix}\ \left[\text{operating}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_1\ \text{and R}_3\rightarrow\text{R}_3-\text{R}_1\right]$
$ =(3y+k).1\begin{vmatrix}k&0\\0&k\end{vmatrix}=(3y+k)k^2=k^2(3y+k)=\text{R.H.S.}$ Proved.
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