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$x≠y≠z$ $\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix} = 0$
Then xy+yz+zx = | A. x+y+z | B. $xyz$ | C. $xyz\over(x+y+z)$ | D. $(x+y+z)\over xyz$ |

Given Ans - D

What I did first was R1->R1-R3 & R2->R2-R3 and throwing (x-z) and (y-z) to the 0.....but this way the opened determinant is still too complex

What I did second was putting values of x and y but with that I was only able to eliminate option A & B

I need help with the correct approach (the correct row transformation) or any other method I can try.

Mitul Agrawal
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3 Answers3

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Hint:

$$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix}=\begin{vmatrix}x&x^3&x^4\\y&y^3&y^4\\z&z^3&z^4\end{vmatrix}-\begin{vmatrix}x&x^3&1\\y&y^3&1\\z&z^3&1\end{vmatrix}=xyz\begin{vmatrix}1&x^2&x^3\\1&y^2&y^3\\1&z^2&z^3\end{vmatrix}-\begin{vmatrix}1&x&x^3\\1&y&y^3\\1&z&z^3\end{vmatrix}$$

Can you end it from here?

$$\begin{vmatrix}1&x^2&x^3\\1&y^2&y^3\\1&z^2&z^3\end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)$$ $$\begin{vmatrix}1&x&x^3\\1&y&y^3\\1&z&z^3\end{vmatrix}=(x-y)(y-z)(z-x)(x+y+z)$$

LHF
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  • I think we know that $\det(A+B)\neq \det(A)+\det(B)$, determinant is not linear, it is $n-$linear. – K.K.McDonald Mar 14 '20 at 08:01
  • @K.K.McDonald, note that the first matrix is not $A+B$ (that you would be with $(x-x), (x^3-x^3), x^4-1$ on the first row). In a determinant, if each element in any row (or column) consists of the sum of two terms, then the determinant can be expressed as sum of two determinants of same order – LHF Mar 14 '20 at 08:03
  • Ummm, thank you for your comment, I am still struggling with $$\begin{vmatrix}x&x^3&x^4-1\y&y^3&y^4-1\z&z^3&z^4-1\end{vmatrix}=\begin{vmatrix}x&x^3&x^4\y&y^3&y^4\z&z^3&z^4\end{vmatrix}-\begin{vmatrix}x&x^3&1\y&y^3&1\z&z^3&1\end{vmatrix}$$ Can you please edit your answer to show how we got this identity. Thanks in advance. – K.K.McDonald Mar 14 '20 at 08:08
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    Thank You @Atticus this didnt click me earlier, let me try to get the answer now. – Mitul Agrawal Mar 14 '20 at 08:10
  • @K.K.McDonald, It'a determinant property. $\det A$ is an additive function of a fixed row. This means that if $A, B$, and $C$ are identical except that $rowi(A) = rowi(B) + rowi(C)$, then $\det(A) = \det(B) + \det(C)$. I don't remember the proof, i think it used the determinant definition. Let me look it up. – LHF Mar 14 '20 at 08:11
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    Ok, thanks, I got it now, you decomposed the last column! Thanks you, nice trick! – K.K.McDonald Mar 14 '20 at 08:14
  • @K.K.McDonald, Only now I realise I explained it really poorly in the last comments. What I did was decomposing the column, but I was talking about row decomposition (since it's the same property) instead of calling it column decomposition. Sorry for the confusion. – LHF Mar 14 '20 at 08:16
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    Got the answer :) – Mitul Agrawal Mar 14 '20 at 08:18
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    That's ok, it was a really nice solution. Taught an old dog a new trick C: – K.K.McDonald Mar 14 '20 at 08:18
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    @Atticus I have provided an alternative answer. – Jean Marie Mar 14 '20 at 11:24
2

A solution in a different spirit :

We are going to assume that none of the values $x,y$ or $z$ is zero (otherwise, the expression (D) would be meaningless).

Let

$$M=\begin{pmatrix}x^4-1&x^3&x\\y^4-1&y^3&y\\z^4-1&z^3&z\end{pmatrix}$$

(I have modified the order of columns, WLOG).

As $\det(M)=0$, there exists a linear dependency on the columns of $M$, i.e., there exists $a,b,c$ such that

$$\begin{cases}a(x^4-1)+bx^3+cx&=&0\\a(y^4-1)+by^3+cy&=&0\\a(z^4-1)+bz^3+cz&=&0\end{cases}$$

Let us assume $a \neq 0$ (see remark below). WLOG, we can assume that $a=1$.

As a consequence, 4th degree polynomial :

$$P(t):=t^4+bt^3+0t^2+ct-1$$

has $x,y,z$ for its roots ; let us denote by $r$ the fourth root.

Now, let us write the second and the last Viète formulas (those which do not involve an unknown letter) :

$$\left\{\begin{array}{rcr}xy+yz+zx+rx+ry+rz&=&0\\xyzr&=&-1\end{array}\right.$$

$$r=- \dfrac{xy+yz+zx}{x+y+z}=- \dfrac{1}{xyz}$$

therefore proving result D.

Remark : In fact $a$ cannot be $0$. Otherwise, it would mean that there is a linear dependency between the two first columns. Should column 2 be a multiple of column 1, their resp. entries would be proportional, i.e.,

$$\dfrac{x^3}{x}=\dfrac{y^3}{y}=\dfrac{z^3}{z}$$

$$\iff \ \ x^2=y^2=z^2 $$

which is not possible for different $x,y,z$ (2 at least would be equal...).

Jean Marie
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1

We can take advantage of this being a multiple choice question.

Expanding out the determinant $$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix}$$ will give products that look like $xy^3(z^4-1)$, which give us six degree-$8$ terms (such as $xy^3z^4$) and six degree-$4$ terms (such as $-xy^3$).

We know the determinant is divisible by $(x-y)(y-z)(x-z)$, because if any two of $x,y,z$ are equal, the determinant is $0$. Factoring that out, we should get the difference of a degree-$5$ polynomial and a degree-$1$ polynomial.

The four answers are predicting that the polynomial we have left is a multiple of:

\begin{array}{cc} (A) & xy + yz + zx - x - y - z \\ (B) & xy + yz + zx - xyz \\ (C) & (xy + yz + zx)(x + y + z) - xyz \\ (D) & (xy + yz + zx)(xyz) - x - y - z \end{array} Only option (D) is the difference of a degree-$5$ polynomial and a degree-$1$ polynomial, so it is the only possibility.

Misha Lavrov
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