Finding the determinant of a $3\times 3 $ matrix with row vectors of the form $[ak\quad a^2+k^2\quad 1]$.

Question

I am currently solving the matrix (I need to find determinant):

\begin{bmatrix} ab & a^2 + b^2 & 1 \\[0.3em] ac & a^2 + c^2 & 1 \\[0.3em] ad & a^2 + d^2 & 1 \end{bmatrix}

I got down to the equation:

$a(cd^2 - c^2d + bd^2 - b^2d +bc^2 - b^2c)$

I know that the answer is $a(b-d)(c-d)(c-b)$, however I have hard time getting there. It seems that I am missing some simplification trick or something.

I already tried factoring out some of the variables but did not progressed much.

Could you give me a hint how should I approach such an expression?

Answer 1

You have

$$\det\begin{bmatrix} ab & a^2 + b^2 & 1 \\[0.3em] ac & a^2 + c^2 & 1 \\[0.3em] ad & a^2 + d^2 & 1 \end{bmatrix} = a \det\begin{bmatrix} b & a^2 + b^2 & 1 \\[0.3em] c-b & c^2 - b^2 & 0 \\[0.3em] d-b & d^2 - b^2 & 0 \end{bmatrix} $$ $$=a(c-b)(d-b) \det\begin{bmatrix} 1 & c+ b\\[0.3em] 1 & d+b \end{bmatrix} = a (c-b)(d-b)(d-c)$$

Answer 2

The problem might be that your factorized formula is incorrect. In fact, it must have a factor $a$, i.e., $$ \det(A)=-a\cdot (b - c)(b - d)(c - d). $$ Your formula doesn't depend on $a$ at all. Edit You have corrected it.

The formula then follows from $$ (b-c)(b-d)(c-d)= b^2c - b^2d - bc^2 + bd^2 + c^2d - cd^2$$ what you had already, by grouping terms with $b-c$, namely $(b-c)(bc-dc-bd-d^2).$