Question

Using properties of determinants, prove that $\left| \begin{matrix} x+y & x & x \\\ 5x+4y & 4x & 2x \\\ 10x+8y & 8x & 3x \\\ \end{matrix} \right|={{x}^{3}}$

Answer 1

Start by simplification of the determinant given in the question using elementary row transformation such that you get 2 zeroes in the same row or column. If needed you can use elementary column transformations as well. Once you get two zeroes in the same row/ column, open the determinant and solve to get the answer.

Complete step by step answer:
Let us start the simplification of the determinant using elementary row transformation. We know that if we subtract a row or column from the other row or column of a determinant, the result of the determinant is never changed.
$\left| \begin{matrix} x+y & x & x \\\ 5x+4y & 4x & 2x \\\ 10x+8y & 8x & 3x \\\ \end{matrix} \right|$
We will subtract 2 times of Row 2 from Row 3. On doing so, we get
$\left| \begin{matrix} x+y & x & x \\\ 5x+4y & 4x & 2x \\\ 0 & 0 & -x \\\ \end{matrix} \right|$
Now we will subtract 4 times of Row 1 from row 2. On doing so, we get
$\left| \begin{matrix} x+y & x & x \\\ x & 0 & -2x \\\ 0 & 0 & -x \\\ \end{matrix} \right|$
Now, we will open the determinant about the third Row. On doing so, we get
$0\left( x\times \left( -2x \right)-0\times x \right)-0\left( \left( x+y \right)\left( -2x \right)-x\times x \right)+\left( -x \right)\left( \left( x+y \right)\times 0-x\times x \right)$
$=0-0-x\left( 0-{{x}^{2}} \right)$
$={{x}^{3}}$
So, the left-hand side of the equation is equal to the right hand side of the equation given in the question. Hence, we can say that we have proved $\left| \begin{matrix} x+y & x & x \\\ 5x+4y & 4x & 2x \\\ 10x+8y & 8x & 3x \\\ \end{matrix} \right|={{x}^{3}}$ .

Note: The most important thing is to be wise while using the row transformations, if you make the correct row transformations the determinant can be simplified to a large extent but the wrong transformations may lead to complicating the problems. Also, be careful about the signs while opening the determinant.