Find the determinant using colum or row operations

Question

I find problem in simplification. When I tried to simplify I ended up doing the regular process of finding the determinant value. The matrix is $\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ ab & bc & ca \end{pmatrix}$.

You need to find a determinant. Then you learn that you end up "doing the regular process of finding the determinant value". I don't see any issues here. — user1551, Jul 08 '16 at 09:28
What makes you think there is a way to find the determinant that's any simpler than "the regular process"? — Gerry Myerson, Jul 08 '16 at 09:34

score 1 · Answer 1 · answered Jul 08 '16 at 09:33

1

Subtract the first column from the other two, to get $\begin{vmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ ab & bc - ab & ca - ab \end{vmatrix}$. Now compute the determinant, developing it along the first row, and obtain $\begin{vmatrix} b-a & c-a \\ bc-ab & ca-ab \end{vmatrix}$. This can further be calculated (with the usual formula for $2\times2$ determinants) to get $a(b-a)(c-b) - b(c-a)^2$, which you can't simplify anymore.

answered Jul 08 '16 at 09:33

Alex M.

35,207

Dear @Majid, please stop vandalizing my answer. Both your edit attempts got rejected, please stop trying to push them through. Also, it is considered rude to change the formatting that some other user chose to use simply because you would have done it differently. – Alex M. Jul 09 '16 at 08:31

score 1 · Answer 2 · answered Jul 08 '16 at 20:51

1

What about computing the determinant directly based on the first row?

$$\det\left(\begin{array}{ccc}1&1&1\\a&b&c\\ab&bc&ca\end {array}\right)=(bca-bc^2)-(ca^2-abc)+(abc-ab^2)=3abc-bc^2-ca^2-ab^2$$

answered Jul 08 '16 at 20:51

Majid

3,357

score 0 · Answer 3 · 2016-07-08T09:51:06.150

0

$$\det\left(\begin{array}{ccc}1&1&1\\a&b&c\\ab&bc&ca\end {array}\right)=abc\det\left(\begin{array}{ccc}\frac{1}{a}&\frac{1}{b}&\frac{1}{c}\\1&1&1\\b&c&a\end {array}\right)=abc\det\left(\begin{array}{ccc}\frac{1}{a}-\frac{1}{b}&\frac{1}{b}&\frac{1}{c}-\frac{1}{b}\\0&1&0\\b-c&c&a-c\end {array}\right)=$$ $$abc\det\left(\begin{array}{ccc}\frac{1}{a}-\frac{1}{b}&\frac{1}{c}-\frac{1}{b}\\b-c&a-c\end {array}\right)=c(b-a)(a-c)-a(b-c)^{2}$$

edited Jul 08 '16 at 09:51

answered Jul 08 '16 at 09:27

No, what you get is (pay attention to the first row) $abc \det \begin{pmatrix} \frac 1 a & \frac 1 b & \frac 1 c \ 1 & 1 & 1 \ b & c & a \end{pmatrix}$. – Alex M. Jul 08 '16 at 09:35
wait aminute, you can't factorize a term for a column??????? – Jul 08 '16 at 09:36
@juliogodoy: When you factorize by $a$, you must divide all the elements of that column by $a$. In particular, you must also divide $1$ by $a$, getting $\frac 1 a$. – Alex M. Jul 08 '16 at 09:37
@AlexM. you right – Jul 08 '16 at 09:37
sorry i make mistake – Jul 08 '16 at 09:39
i fix the problem – Jul 08 '16 at 09:51

score 0 · Answer 4 · answered Jul 08 '16 at 21:12

Maybe it's unlikely that you would have learned of this method, but via Dodgson condensation, you can reduce this to a problem of computing the determinants of several $2\times2$ matrices, which you might find easier to compute strictly via row operations. $$\begin{vmatrix}1&1&1\\a&b&c\\ab&bc&ca\end{vmatrix}=\begin{vmatrix}\dfrac{\begin{vmatrix}1&1\\a&b\end{vmatrix}}{b}&\dfrac{\begin{vmatrix}1&1\\b&c\end{vmatrix}}{b}\\[1ex] \dfrac{\begin{vmatrix}a&b\\ab&bc\end{vmatrix}}{b}&\dfrac{\begin{vmatrix}b&c\\bc&ca\end{vmatrix}}{b} \end{vmatrix}$$ The first determinant (in the $(1,1)$ position), for instance, would be $$\begin{vmatrix}1&1\\a&b\end{vmatrix}=\begin{vmatrix}1&1\\0&b-a\end{vmatrix}=\begin{vmatrix}1&0\\0&b-a\end{vmatrix}=(b-a)\begin{vmatrix}1&0\\0&1\end{vmatrix}=b-a$$

Find the determinant using colum or row operations

4 Answers4