1
$\begingroup$

$$\begin{vmatrix} 1 & a^2+bc & a^3\\ 1 & b^2+ca & b^3\\ 1 & c^2+ab & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a^2+b^2+c^2)$$

we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calculation to apply so was hoping for an hint.

Edit: just tried the problem and here is how I have done it

$$\begin{vmatrix} 1 & a^2+bc & a^3\\ 1 & -(a^2+b^2)+c(a-b) & -(a^3-b^3)\\ 1 & c^2-a^2-b(c-a) & c^3-a^3 \end{vmatrix}$$

then

$$\begin{vmatrix} 1 & a^2+bc & a^3\\ 0 & -((a-b)(a+b))+c(a-b) & -((a-b)(a^2+ab+b^2))\\ 0 & (c-a)(c+a)-b(c-a) & (c-a)(c^2+ca+a^2) \end{vmatrix}$$

then

$$(a-b)(c-a)\begin{vmatrix} 1 & a^2+bc & a^3\\ 0 & a+b+c & -(a^2+ab+b^2)\\ 0 & a-b+c & (c^2+ca+a^2) \end{vmatrix}$$

then $$(a-b)(c-a)\begin{vmatrix} a+b+c & -(a^2+ab+b^2)\\ a-b+c & (c^2+ca+a^2) \end{vmatrix}$$

then

$$(a-b)(c-a) * ( (a+b+c)(c^2+ca+a^2)-(a-b+c)(-(a^2+ab+b^2)) )$$

Here I am Confused on how to multiply them and get the answer

note: typed because I own a very bad handwriting

Thank you every one for your help

$\endgroup$
  • 3
    $\begingroup$ It's just a computation, have you made any progress? $\endgroup$ – Najib Idrissi Aug 30 '14 at 10:26
  • 1
    $\begingroup$ no actuall we have to show the both are equal without expansion $\endgroup$ – Raza Hassan Aug 30 '14 at 11:04
  • $\begingroup$ Subtract the first row from the second and the third row. This produces two zeros and you only have to calculate a $2\ \times \ 2$ - determinant. $\endgroup$ – Peter Aug 30 '14 at 11:13
  • $\begingroup$ That is the kind of information you are supposed to put in your question in the first place; all important information belongs in your question. Next, can you show any thoughts, ideas, work or attempts? $\endgroup$ – blue Aug 30 '14 at 11:14
2
$\begingroup$

The first trick is to get as much zeroes as you can in the first row. That makes multiplication easier.

$\begin{vmatrix} 1 & a^2+bc & a^3 \\ 1 & b^2+ca & b^3\\ 1 & c^2+ab &c^3 \end{vmatrix}$

subtracting second row from first row and third row from second row:

$(a-b)(b-c)\begin{vmatrix} 0 & a+b-c & a^2+ab+b^2 \\ 0 & b+c-a & b^2+bc+c^2\\ 1 & c^2+ab &c^3 \end{vmatrix}$

Subtracting second row from first row:

$-(a-b)(b-c)(c-a)\begin{vmatrix} 0 & 2 & a+b+c \\ 0 & b+c-a & b^2+bc+c^2\\ 1 & c^2+ab &c^3 \end{vmatrix}$

exchanging row and column:

$(a-b)(b-c)(c-a)\begin{vmatrix} 0 & 0 & 1 \\ 2 & b+c-a & c^2+ab\\ a+b+c & b^2+bc+c^2 &c^3 \end{vmatrix}$

Now take determinant and get the result.

$\endgroup$
3
$\begingroup$

Seeing the column of $1$'s, a first thought would be to subtract the first row from the other two. This turns out to give you a factor $b-a$ in the second row and a factor $c-a$ in the third. $$ \begin{align} \begin{vmatrix} 1 & a^2+bc & a^3\\ 1 & b^2+ca & b^3\\ 1 & c^2+ab & c^3 \end{vmatrix} &= \begin{vmatrix} 1 & a^2+bc & a^3\\ 0 & b^2-a^2+c(a-b) & b^3-a^3\\ 0 & c^2-a^2+b(a-c) & c^3-a^3 \end{vmatrix} \\{}& =(b-a)(c-a) \begin{vmatrix} 1 & a^2+bc & a^3\\ 0 & (b+a)-c & a^2+ab+b^2\\ 0 & (c+a)-b & a^2+ac+c^2 \end{vmatrix} \end{align} $$ Now you can subtract the second row from the third, which makes the latter divisible by $b-c$, so you can factor again. I shouldn't be too difficult to complete the computation.

$\endgroup$
  • $\begingroup$ but according to the question the factoe should be (a-b) and not (b-a) $\endgroup$ – Raza Hassan Aug 30 '14 at 13:36
  • $\begingroup$ @RazaHassan: $(b-a)=-(a-b)$, you can take care of the sign later. It happened that factoring out $b-a$ gave a more pleasant result, so I preferred this to reduce the chance of making an error doing the mental computation. $\endgroup$ – Marc van Leeuwen Aug 30 '14 at 13:57
2
$\begingroup$

The solution-attempt in the question is excellent (after your edit)!

But unfortunately you made a small mistake. After taking out the factor $(a-b)(c-a)$, the middle element should be $c-a-b$, not $a+b+c$. If you fix that, the remaining determinant can be expanded and easily factorised.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.