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 + R3, 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.