Question

Prove by determinant properties:

[(ax+by, ay+bz, az+bx)(ay+bz, az+bx, ax+by)(az+bx, ax+by, ay+bz)] = (a³+b³)[(x,y,z)(y,z,x)(z,x,y)]

\(\begin{vmatrix}ax+by&ay+bz&az+bx\\ay+bz&az+bx&ax+by\\az+bx&ax+by&ay+bz\end{vmatrix}\)

Answer 1

\(\begin{vmatrix}ax+by&ay+bz&az+bx\\ay+bz&az+bx&ax+by\\az+bx&ax+by&ay+bz\end{vmatrix}\)

Applying R₁ → R₁ + R₂ + R₃

_{\(=\begin{vmatrix}a(x+y+z)+b(x+y+z)&a(x+y+z)+b(x+y+z)&a(x+y+z)+b(x+y+z)\\ay+bz&az+bx&ax+by\\az+bx&ax+by&ay+bz\end{vmatrix}\)}

_{\(=\begin{vmatrix}(a+b)(x+y+z))&(a+b)(x+y+z)&(a+b)(x+y+z)\\ay+bz&az+bx&ax+by\\az+bx&ax+by&ay+bz\end{vmatrix}\)} 

Applying R₁ → \(\frac{R_1}{(a+b)(x+y+z)}\)

\(=(a+b)(x+y+z)\begin{vmatrix}1&1&1\\ay+bz&az+bx&ax+by\\az+bx&ax+by&ay+bz\end{vmatrix}\)

Applying C₂ → C₂ - C₃, C₃ → C₃ - C₁

 = (a + b) (x + y + z)\(\begin{vmatrix}1 & 0&0\\ay+bz&a(z-y)+b(x-z)&a(x-y)+b(y-z)\\az+bx&a(x-z)+b(y-x)&a(y-z)+b(z-x)\end{vmatrix}\) 

Expanding determinant along row R₁

 = (a + b) (x + y + z)(a²(2yz - y² - z²) + ab(x-z)(y-z) + ab(z-x)(z-y)+b² (2xz - x² - y²) - a²(x² - xz - xy + yz) - ab(x - y)(y - x) - ab (y - z)(x - z) - b²(y² - xy - yz + xz))

 = (a + b)(x + y + z) (a² (2yz - y² - z² - x² + xz + xy - yz) + b²(2xz - x² - y² - y² + xy + yz - xz) + ab(xy-yz - xz +z² - xz - yz + xy - xy + y² + xy - xy + xz + yz - z²))

 = (a + b) (x + y + z) (a²(xy + yz + xz - x² - y² - z²) + b²(xy + yz + xz - x² - y² - z²) + ab(-2xy - xz - yz + z² + x² + y²))

 = (a + b) (a² + b² - ab) (x + y + z) (xy + yz + xz - x² - y² - z²)

 = (a³ + b³) (x + y + z) (xy + yz + xz - x² - y² - z²)----(1)

Now, \(\begin{vmatrix}x&y&z\\y&z&x\\z&x&y\end{vmatrix}\) = (x + y + z)\(\begin{vmatrix}1&1&1\\y&z&x\\z&x&y\end{vmatrix}\) 

(By applying R₁ → \(\frac1{x+y+z}\)(R₁ + R₂ + R₃))

 = (x + y + z) \(\begin{vmatrix}1&0&0\\y&z-y&x-y\\z&x-z&y-z\end{vmatrix}\) 

By applying

(C₂ → C₂ - C₁, C₃ → C₃ - C₁)

= (x + y + z) [(z - y) (y - z) - (x - z) (x - y)]

(On expanding determinant along row R₁)

 =  (x + y + z) (yz - y² - z²+ yz - x² + xy + xz - yz)

 = (x + y + z) (xy + yz + xz - x² - y² - z²)----(2)

Hence, from (1) and (2), we can conclude

\(\begin{vmatrix}ax+by&ay+bz&az+bx\\ay+bz&az+bx&ax+by\\az+bx&ax+by&ay+bz\end{vmatrix}\) = a³ + b³\(\begin{vmatrix}x&y&z\\y&z&x\\z&x&y\end{vmatrix}\)