# Prove that $\begin{bmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{bmatrix}=x^3$

Toolbox:
• (i) If two rows or columns are identical then the value of the determinant is zero
• (ii)Elementary transformation can be made by
• (a) interchanging two rows or columns
• (b) By adding or subtracting two or more rows or columns.
Step 1:
Let $\Delta=\begin{bmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{bmatrix}$
This can be split and written as
$\Delta=\begin{bmatrix} x & x & x \\ 5x & 4x & 2x \\ 10x & 8x & 3x \end{bmatrix}+\begin{bmatrix} y & x & x \\ 4y & 4x & 2x \\ 8y & 8x & 3x \end{bmatrix}$
$\Delta=\Delta_1+\Delta_2$
Consider $\Delta_1=\begin{bmatrix} x & x & x \\ 5x & 4x & 2x \\ 10x & 8x & 3x \end{bmatrix}$
Let us take x as the common factor from $C_1,C_2 and C_3$
$\Delta_1=x^3 \begin{bmatrix} 1 & 1 & 1 \\ 5 & 4 & 2 \\ 10 & 8 & 3 \end{bmatrix}$
Apply $C_2 \to C_2-C3$ and $C_3 \to C_3-C_1$
$\Delta_1=x^3 \begin{bmatrix} 1 & 0 & 0 \\ 5 & 2 & -3 \\ 10 & 5 & -7 \end{bmatrix}$
Now expanding along $R_1$ we get,
$\Delta_1=x^3[1(2 \times -7-5 \times -3)]$
$=x^3[-14+15]=x^3$
Step 2:
Now let us consider $\Delta_2=\begin{bmatrix} y & x & x \\ 4y & 4x & 2x \\ 8y & 8x & 3x \end{bmatrix}$
Let us take y from $C_1$ and x from $C_2$ and $C_3$ as common factors
$\Delta_2=xy \begin{bmatrix} 1 & 1 & 1 \\ 4 & 4 & 2 \\ 8 & 8 & 3 \end{bmatrix}$
Here two columns are identical,
Hence $\Delta _2=0$
Therefore $\Delta=\Delta_1+\Delta_2$
$\Delta=x^3+0$
$\Delta=x^3$
Solution : Hence proved