(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
(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.
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