Question

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

Answer 1

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.

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.