Answer to Question #269890 in Linear Algebra for john

Question

Question 4 Use row operation to show that

det T = 0

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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

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Answer 1

$|T|=\begin{vmatrix} x^2 & 2x+1 & 4x+4 &6x + 9\\ y^2 & 2y+1 & 4y+4 &6y + 9\\ z^2 & 2z+1 & 4z+4 &6z + 9\\ w^2 & 2w+1 & 4w + 4&6w + 9 \end{vmatrix}$

if we add 2nd and 3rd columns, and subtract it from 4th column, we get:

$|T|=\begin{vmatrix} x^2 & 2x+1 & 4x+4 & 4\\ y^2 & 2y+1 & 4y+4 &4\\ z^2 & 2z+1 & 4z+4 &4\\ w^2 & 2w+1 & 4w + 4&4 \end{vmatrix}$

if we multiply 2nd column by 2, and subtract it from 3rd column, we get:

$|T|=\begin{vmatrix} x^2 & 2x+1 & 2 & 4\\ y^2 & 2y+1 & 2 &4\\ z^2 & 2z+1 & 2 &4\\ w^2 & 2w+1 & 2&4 \end{vmatrix}$

then, multiplying 3rd column by 2, we get:

$2|T|=\begin{vmatrix} x^2 & 2x+1 & 4 & 4\\ y^2 & 2y+1 & 4 &4\\ z^2 & 2z+1 & 4 &4\\ w^2 & 2w+1 & 4&4 \end{vmatrix}=0$

since two columns of a determinant are identical.