Browse Questions

# By using the properties of determinants show that $(ii) \begin{vmatrix} y+k&y&y \\ y&y+k&y \\ y&y&y+k \end{vmatrix} = k^2 (3y+k)$

Note: This is part 2 of a 2 part question, split as 2 separate questions here.

Toolbox:
• (ii) If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any one column of the determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\bigtriangleup=\begin{vmatrix}y+k& y & y\\y & y+k & y\\y &y & y+k\end{vmatrix}$

By adding $R_1,R_2,R_3$ we get $R_1\rightarrow R_1+R_2+R_3$

Let $\bigtriangleup=\begin{vmatrix}3y+k& 3y+k & 3y+k\\y & y+k & y\\y &y & y+k\end{vmatrix}$

Take (3y+k) as the common factor from $R_1$

$\bigtriangleup=(3y+k)\begin{vmatrix}1& 1& 1\\y & y+k & y\\y &y & y+k\end{vmatrix}$

By applying $C_1\rightarrow C_1-C_2$ and $C_2\rightarrow C_2-C_3$

$\bigtriangleup=(3y+k)\begin{vmatrix}0& 0& 1\\-k & k & y\\0 &-k & y+k\end{vmatrix}$

By applying $C_1\rightarrow C_1+C_2$

$\bigtriangleup=(3y+k)\begin{vmatrix}0& 0& 1\\0 & k & y\\-k&-k & y+k\end{vmatrix}$

Now expanding along $R_1$ we get,

$\bigtriangleup=(3y+k)1(0\times -k-k\times -k)$

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

Hence $\begin{vmatrix}y+k & y & y\\y & y+k & y\\y & y & y+k\end{vmatrix}=k^2(3y+k)$

Hence proved.