∣ T ∣ = ∣ x 2 2 x + 1 4 x + 4 6 x + 9 y 2 2 y + 1 4 y + 4 6 y + 9 z 2 2 z + 1 4 z + 4 6 z + 9 w 2 2 w + 1 4 w + 4 6 w + 9 ∣ |T|=\begin{vmatrix}
x^2 & 2x+1 & 4x+4 &6x + 9\\
y^2 & 2y+1 & 4y+4 &6y + 9\\
z^2 & 2z+1 & 4z+4 &6z + 9\\
w^2 & 2w+1 & 4w + 4&6w + 9
\end{vmatrix} ∣ T ∣ = ∣ ∣ x 2 y 2 z 2 w 2 2 x + 1 2 y + 1 2 z + 1 2 w + 1 4 x + 4 4 y + 4 4 z + 4 4 w + 4 6 x + 9 6 y + 9 6 z + 9 6 w + 9 ∣ ∣
if we add 2nd and 3rd columns, and subtract it from 4th column, we get:
∣ T ∣ = ∣ x 2 2 x + 1 4 x + 4 4 y 2 2 y + 1 4 y + 4 4 z 2 2 z + 1 4 z + 4 4 w 2 2 w + 1 4 w + 4 4 ∣ |T|=\begin{vmatrix}
x^2 & 2x+1 & 4x+4 & 4\\
y^2 & 2y+1 & 4y+4 &4\\
z^2 & 2z+1 & 4z+4 &4\\
w^2 & 2w+1 & 4w + 4&4
\end{vmatrix} ∣ T ∣ = ∣ ∣ x 2 y 2 z 2 w 2 2 x + 1 2 y + 1 2 z + 1 2 w + 1 4 x + 4 4 y + 4 4 z + 4 4 w + 4 4 4 4 4 ∣ ∣
if we multiply 2nd column by 2, and subtract it from 3rd column, we get:
∣ T ∣ = ∣ x 2 2 x + 1 2 4 y 2 2 y + 1 2 4 z 2 2 z + 1 2 4 w 2 2 w + 1 2 4 ∣ |T|=\begin{vmatrix}
x^2 & 2x+1 & 2 & 4\\
y^2 & 2y+1 & 2 &4\\
z^2 & 2z+1 & 2 &4\\
w^2 & 2w+1 & 2&4
\end{vmatrix} ∣ T ∣ = ∣ ∣ x 2 y 2 z 2 w 2 2 x + 1 2 y + 1 2 z + 1 2 w + 1 2 2 2 2 4 4 4 4 ∣ ∣
then, multiplying 3rd column by 2, we get:
2 ∣ T ∣ = ∣ x 2 2 x + 1 4 4 y 2 2 y + 1 4 4 z 2 2 z + 1 4 4 w 2 2 w + 1 4 4 ∣ = 0 2|T|=\begin{vmatrix}
x^2 & 2x+1 & 4 & 4\\
y^2 & 2y+1 & 4 &4\\
z^2 & 2z+1 & 4 &4\\
w^2 & 2w+1 & 4&4
\end{vmatrix}=0 2∣ T ∣ = ∣ ∣ x 2 y 2 z 2 w 2 2 x + 1 2 y + 1 2 z + 1 2 w + 1 4 4 4 4 4 4 4 4 ∣ ∣ = 0
since two columns of a determinant are identical.
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