$\begin{vmatrix} 1 & a &bc \\ 1& b & ac\\ 1&c & ab \end{vmatrix}=\begin{vmatrix} 1 & a &a^2 \\ 1& b&b^2 \\ 1& b & c^2 \end{vmatrix}$

Question

Prove that \begin{align}\begin{vmatrix} 1 & a &bc \\ 1& b & ac\\ 1&c & ab \end{vmatrix}&=\begin{vmatrix} 1 & a &a^2 \\ 1& b&b^2 \\ 1& b & c^2 \end{vmatrix}\\&=(c-a)(b-a)(c-b)\begin{vmatrix} 1 & a & a^2\\ 0& 1 &b+a \\ 0& 0 & 1 \end{vmatrix}\\\\ &=(c-a)(b-a)(c-b)\end{align}

I got the last two equalities but I did't get first two. I need help with first two

I believe it should be $c$ instead of $b$ in the third row and second column. — Itay4, Jul 18 '17 at 07:52
I guest the (3,2) entry in the second matrix is $c$, not $b$. — C. Ding, Jul 18 '17 at 07:53
is there a possibility that Itay4 and C.Ding are right but the question is wrong? — Siong Thye Goh, Jul 18 '17 at 07:55
If $a=c=0$ and $b=1$, then the first determinant is $0$ and the second one is $-1$. — José Carlos Santos, Jul 18 '17 at 07:59
Possible duplicate of Linear Algebra determinant reduction which shows a different method to prove this question. — Frenzy Li, Jul 18 '17 at 08:04

score 2 · Answer 1 · edited Jul 18 '17 at 08:14

2

\begin{align}\begin{vmatrix} 1 & a &bc \\ 1& b & ac\\ 1&c & ab \end{vmatrix}&=\begin{vmatrix} 1 & a &a^2 \\ 1& b&b^2 \\ 1& c & c^2 \end{vmatrix}\\\\ \begin{vmatrix} 1 & a &bc \\ 0& b-a & ac-bc\\ 0&c-a & ab-bc \end{vmatrix}&=\begin{vmatrix} 1 & a &a^2 \\ 0& b-a&b^2-a^2 \\ 0& c-a & c^2-a^2 \end{vmatrix}\\\\ (b-a)(c-a)\begin{vmatrix} 1 & a &bc \\ 0& 1& -c\\ 0&1 &-b \end{vmatrix}&=(b-a)(c-a)\begin{vmatrix} 1 & a &a^2 \\ 0& 1&b+a \\ 0& 1 & c+a \end{vmatrix}\\\\ (b-a)(c-a)\begin{vmatrix} 1 & a &bc \\ 0& 1& -c\\ 0&0 &c-b \end{vmatrix}&=(b-a)(c-a)\begin{vmatrix} 1 & a &a^2 \\ 0& 1&b+a \\ 0& 0& c-b \end{vmatrix}\end{align}

edited Jul 18 '17 at 08:14

lioness99a

4,943

answered Jul 18 '17 at 08:08

4

-1: using the awful and misguided "working on both sides of the equality" technique that so often leads to error. – Martin Argerami Jul 18 '17 at 13:23
2

@MartinArgerami: thanks for recalling that subjectivity still rules. – Jul 18 '17 at 13:31

score 0 · Answer 2 · answered Nov 24 '22 at 08:08

0

From the last column, we remove the common factor $abc$;

Then we multiply the first line by $a$, the second line by $b$ and the last line by $c$.

answered Nov 24 '22 at 08:08

medicu

4,482

$\begin{vmatrix} 1 & a &bc \\ 1& b & ac\\ 1&c & ab \end{vmatrix}=\begin{vmatrix} 1 & a &a^2 \\ 1& b&b^2 \\ 1& b & c^2 \end{vmatrix}$

2 Answers2