Question

Write the value $\left| \begin{matrix}
   x+y & y+z & z+x \\
   z & x & y \\
   -3 & -3 & -3 \\
\end{matrix} \right|$

Answer 1

Hint: Now we know using elementary row and column transformation does not change the value of determinant. Hence we can use row transformations to simplify this determinant. Now first we will ${{R}_{1}}={{R}_{1}}+{{R}_{3}}$ . This will give us a determinant in which the 1st row has the same elements. Hence we will take x + y + z common from the determinant. Now we will again use Row transformation ${{R}_{1}}\to {{R}_{1}}+\dfrac{1}{3}{{R}_{3}}$ and then solve the determinant.

Complete step-by-step answer:
Now consider the determinant $\left| \begin{matrix}
   x+y & y+z & z+x \\
   z & x & y \\
   -3 & -3 & -3 \\
\end{matrix} \right|$
Now we can use elementary row transformation to solve this determinant.
Row transformation can be adding one row to a row or adding multiple of a row to any row.
Here we will add row 2 to row 1.
Now here row transformation does not change the value of determinant.
Hence we can use them easily to simplify the determinant
Hence using ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}$ . We get the determinant as
$\left| \begin{matrix}
   x+y+z & y+z+x & z+x+y \\
   z & x & y \\
   -3 & -3 & -3 \\
\end{matrix} \right|$
Now rearranging the terms in row 1 we get.
$\left| \begin{matrix}
   x+y+z & x+y+z & x+y+z \\
   z & x & y \\
   -3 & -3 & -3 \\
\end{matrix} \right|$
Now we can see that all the terms in row 1 are equal. Now we can take this common and hence take it out of determinant.
\[\left( x+y+z \right)\left| \begin{matrix}
   1 & 1 & 1 \\
   z & x & y \\
   -3 & -3 & -3 \\
\end{matrix} \right|\]
Now again using the transformation ${{R}_{1}}\to {{R}_{1}}+\dfrac{1}{3}{{R}_{3}}$ . we get the determinant as
\[\left( x+y+z \right)\left| \begin{matrix}
   0 & 0 & 0 \\
   z & x & y \\
   -3 & -3 & -3 \\
\end{matrix} \right|\]
Now let us open the determinant, Hence we get
(x + y + z) [0(-3x – (- 3y)) – 0(- 3z - (-3y)) + 0(-3z – (-3y))].
= (x + y + z)[0 – 0 + 0]
=(x + y + z)[0]
=0
Hence we get the value of the determinant is 0.

Note: We have a property of determinant which says if any two rows or columns are the same then the value of the determinant is equal to 0. Hence if we apply row transformation \[{{R}_{1}}\to {{R}_{1}}-4{{R}_{1}}\] to the determinant \[\left| \begin{matrix}
   1 & 1 & 1 \\
   z & x & y \\
   -3 & -3 & -3 \\
\end{matrix} \right|\] we will get \[\left| \begin{matrix}
   -3 & -3 & -3 \\
   z & x & y \\
   -3 & -3 & -3 \\
\end{matrix} \right|\] and hence the value of determinant is 0.