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Prove this $3\times 3$ determinant using properties of determinant. $$\begin{vmatrix}y+z & x & x \\ y & z+x & y \\ z & z & x+y\end{vmatrix} = 4xyz$$

I have been trying to solve this one for over an hour now. I really can't even get started with it. The question specifically says I can't expand, and have to prove this using other properties of 3x3 determinants.

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    $\begingroup$ Hint: adding two rows will not change the determinant's value. $\endgroup$ – Sean Roberson Oct 3 '16 at 2:49
  • $\begingroup$ @SeanRoberson I don't get it still. I feel so slow.. $\endgroup$ – Eyad H. Oct 3 '16 at 2:54
  • $\begingroup$ Well have you at least tried following up on Sean's comment? What do you get when you add up certain rows? $\endgroup$ – TMM Oct 3 '16 at 3:06
  • $\begingroup$ @EyadH. When you subtract the second row from the first row, you get a new matrix that has the same determinant. Then, when you subtract the third row from your new first row you get another matrix with the very same determinant. Except now you have a matrix with a bunch of zeros, so calculating the determinant is now easy. I am sure you have seen this property before somewhere before you arrived at this problem. $\endgroup$ – imranfat Oct 3 '16 at 3:08
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The determinant $$ \det(A) = \det(a_1, a_2, a_3) $$ is an alternating multi linear form, this means $$ \det(a_1+ \alpha a_2, a_2, a_3) = \det(a_1, a_2, a_3) + \alpha \det(a_2, a_2, a_3) $$ because $\det$ is linear in the first argument as multi linear form.

Because it is an alternating form, if we exchange the first two arguments, the sign changes and we have $$ \det(a_2, a_2, a_3) = - \det(a_2, a_2, a_3) $$ which means $\det(a_2,a_2,a_3)$ vanishes. This leads to $$ \det(a_1+ \alpha a_2, a_2, a_3) = \det(a_1, a_2, a_3) $$ and similar for the other arguments.

We can use this property to simplify your determinant: \begin{align} d &= \begin{vmatrix} y + z & x & x \\ y & z + x & y \\ z & z & x + y \end{vmatrix} = \begin{vmatrix} 0 & -2z & -2y \\ y & z + x & y \\ z & z & x + y \end{vmatrix} = 2 \begin{vmatrix} 0 & -z & -y \\ y & z + x & y \\ z & z & x + y \end{vmatrix} \\ &= 2 \begin{vmatrix} 0 & -z & -y \\ y & x & 0 \\ z & 0 & x \end{vmatrix} = -2 \begin{vmatrix} 0 & z & y \\ y & x & 0 \\ z & 0 & x \end{vmatrix} = 2 \begin{vmatrix} 0 & y & z \\ y & 0 & x \\ z & x & 0 \end{vmatrix} \end{align} If $x = y = z = 0$ then $d = 0$ and the proposed formula holds. If two of the variables $x,y,z$ are zero, then one of the columns is zero and we have e.g. $$ \det(0, a_2, a_3) = \det(0 + a_2, a_2, a_3) = \det(a_2, a_2, a_3) = 0 $$ If e.g. $x = 0$ we have $$ d = 2 \begin{vmatrix} 0 & y & z \\ y & 0 & 0 \\ z & 0 & 0 \end{vmatrix} = 2 y z \begin{vmatrix} 0 & 1 & 1 \\ y & 0 & 0 \\ z & 0 & 0 \end{vmatrix} = 2 y z \begin{vmatrix} 0 & 0 & 1 \\ y & 0 & 0 \\ z & 0 & 0 \end{vmatrix} = -2 y z \begin{vmatrix} 0 & 0 & 1 \\ 0 & y & 0 \\ 0 & z & 0 \end{vmatrix} = 0 $$ and similar if just $y$ or $z$ vanishes.

If $x,y,z$ do not vanish each, we have \begin{align} d &= 2 \begin{vmatrix} 0 & y & z \\ y & 0 & x \\ z & x & 0 \end{vmatrix} = \frac{2}{xyz} \begin{vmatrix} 0 & xy & xz \\ yz & 0 & xz \\ yz & xy & 0 \end{vmatrix} = \frac{2}{xyz} \begin{vmatrix} 0 & xy & xz \\ yz & 0 & xz \\ yz & 0 & -xz \end{vmatrix} \\ &= \frac{2}{xyz} \begin{vmatrix} 0 & xy & xz \\ 0 & 0 & 2xz \\ yz & 0 & -xz \end{vmatrix} = \frac{4}{y} \begin{vmatrix} 0 & xy & xz \\ 0 & 0 & 1 \\ yz & 0 & -xz \end{vmatrix} = \frac{4}{y} \begin{vmatrix} 0 & xy & 0 \\ 0 & 0 & 1 \\ yz & 0 & 0 \end{vmatrix} \\ &= 4xyz \begin{vmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{vmatrix} = -4xyz \begin{vmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 4xyz \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ &= 4xyz \end{align}

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