Question

By using properties of determinants, show that:
(i)$\begin{vmatrix} x+4 & 2x & 2x\\ 2x & x+4 & 2x \\2x &2x&x+4\end{vmatrix}$=(5x+4)(4-x)² 
(II)$\begin{vmatrix} y+k & y & y\\ y & y+k & y \\y &y&y+k\end{vmatrix}$=k²(3y+k)

Answer 1

(i)△=$\begin{vmatrix} x+4 & 2x & 2x\\ 2x & x+4 & 2x \\2x &2x&x+4\end{vmatrix}$

Applying R₁ → R₁ + R₂ + R₃, we have: 

△=$\begin{vmatrix} 5x+4 & 5x+4 & 5x+4\\ 2x & x+4 & 2x \\2x &2x&x+4\end{vmatrix}$ 

=(5x+4)$\begin{vmatrix} 1 & 1 & 1\\ 2x & x+4 & 2x \\2x &2x&x+4\end{vmatrix}$

 Applying C₂ → C₂ − C₁, C₃ → C₃ − C₁, we have: 

△=(5x+4)I$\begin{vmatrix} 1 & 0 & 0\\ 2x & -x+4 & 0 \\2x &0&-x+4\end{vmatrix}$

=(5x+4)(4-x)(4-x)$\begin{vmatrix} 1 & 0 & 0\\ 2x & 1 & 0 \\2x &0&1\end{vmatrix}$

Expanding along C₃, we have: 
△=(5x+4)(4-x)²$\begin{vmatrix} 1 & 0 \\ 2x & 1 \end{vmatrix}$
=(5x+4)(4-x)²

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(ii)△=$\begin{vmatrix} y+k & y & y\\ y & y+k & y \\y &y&y+k\end{vmatrix}$

Applying R₁ → R₁ + R₂ + R_3, we have: 

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

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

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we have: 

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

=k²(3y+k)I$\begin{vmatrix}1 & 0 & 0\\ y & 1 & 0 \\y &0&1\end{vmatrix}$

Expanding along C₃, we have: 

△=k²(3y+k)$\begin{vmatrix} 1 & 0 \\ y & 1 \end{vmatrix}$=k²(3y+k) 

Hence, the given result is proved.