Q : 10          By using 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)    

Answers (1)

Given determinant:

                                      \triangle = \begin{vmatrix} y+k & y & y\\ y & y+k &y \\ y & y & y+k \end{vmatrix}

Applying row transformation R_{1} \rightarrow R_{1} +R_{2}+R_{3}  we get;

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

=(3y+k) \begin{vmatrix}1 & 1 & 1\\ y & y+k &y \\ y & y & y+k \end{vmatrix}                        [taking common (3y + k) factor]

Now, applying column transformation C_{1} \rightarrow C_{1} - C_{2}  and C_{2} \rightarrow C_{2} - C_{3}

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

=(3y+k)(k^2) \begin{vmatrix}0 & 0 & 1\\ -1 & 1 &y \\ 0 & -1 & y+k \end{vmatrix}

=k^2 (3y+k)

Hence proved.

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