Question

Using the properties of determinants prove the following:

x & x+y & x+2y \\\ x+2y & x & x+y \\\ x+y & x+2y & x \\\ \end{matrix} \right|=9{{y}^{2}}\left( x+y \right)$$

Answer 1

We will try to make the elements of any one row or column to be zero, which will simplify our calculation. We will use row addition, column addition, row substitution, and column substitution operations to obtain the simplified matrix. In order to simplify the matrix, first, perform the following operation: ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$then take the common terms outside the matrix. We will keep simplifying until we get our first row as $0,0,1$. Then further we will apply the conventional method to calculate the determinant of the given matrix.

Complete step-by-step solution:
Let,${{R}_{1}},{{R}_{2}},{{R}_{3}}$ be the three rows and ${{C}_{1}},{{C}_{2}},{{C}_{3}}$be the three columns:
We will apply some operations on the matrix.
We will try and simplify the given matrix in order to find the determinant easily.
Operation1: ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$ (Row Addition and Substitution)
This will give us the following matrix :
$\left| \begin{matrix} x & x+y & x+2y \\\ x+2y & x & x+y \\\ x+y & x+2y & x \\\ \end{matrix} \right|\to \left| \begin{matrix} 3\left( x+y \right) & 3\left( x+y \right) & 3\left( x+y \right) \\\ x+2y & x & x+y \\\ x+y & x+2y & x \\\ \end{matrix} \right|$
After taking out $3(x+y)$ common, we get:
$3(x+y)\left| \begin{matrix} 1 & 1 & 1 \\\ x+2y & x & x+y \\\ x+y & x+2y & x \\\ \end{matrix} \right|$
Operation2: ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}$(Column Subtraction and Substitution)
$3(x+y)\left| \begin{matrix} 1 & 1 & 1 \\\ x+2y & x & x+y \\\ x+y & x+2y & x \\\ \end{matrix} \right|\to 3(x+y)\left| \begin{matrix} 0 & 1 & 1 \\\ 2y & x & x+y \\\ -y & x+2y & x \\\ \end{matrix} \right|$
Further, we will follow the final operation on the obtained matrix: ${{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$
$3(x+y)\left| \begin{matrix} 0 & 1 & 1 \\\ 2y & x & x+y \\\ -y & x+2y & x \\\ \end{matrix} \right|\to 3(x+y)\left| \begin{matrix} 0 & 0 & 1 \\\ 2y & y & x+y \\\ -y & 2y & x \\\ \end{matrix} \right|$
We have the final matrix as:
$3(x+y)\left| \begin{matrix} 0 & 0 & 1 \\\ 2y & -y & x+y \\\ -y & 2y & x \\\ \end{matrix} \right|$
Taking out the determinant of this matrix using the conventional method for a 3*3 matrix:
Let’s suppose a matrix be M: $\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|$
Now we know that,
Determinant of the above matrix will be: $\left| M \right|=a\left( ei-fh \right)-b\left( di-fg \right)+c\left( dh-eg \right)$
Applying the above formula onto the given matrix:

& 3(x+y)\left| \begin{matrix} 0 & 0 & 1 \\\ 2y & -y & x+y \\\ -y & 2y & x \\\ \end{matrix} \right| \\\ & 3\left( x+y \right)\left\\{ 0.\left[ \left( y.x \right)-\left( 2y.\left( x+y \right) \right) \right]-0.\left[ \left( 2y.x \right)-\left( -y.\left( x+y \right) \right) \right]+1.\left[ \left( 4{y^2} \right)-\left( {{y}^{2}} \right) \right] \right\\} \\\ & 3\left( x+y \right)\left( 0-0+3{{y}^{2}} \right)=3\left( x+y \right).3{{y}^{2}}=9{{y}^{2}}\left( x+y \right) \\\ \end{aligned}$$ Hence Proved. **Note:** We can also take out the determinants directly by applying the conventional method without simplifying but please note that that method will be lengthy and chances of committing mistakes are more. Remember to substitute the whole row with the same operation and let the other elements remain as it is.