Using the properties of determinants in evaluate:

Let

Expanding |A| along C₁, we get

= (a + x) [(a + y)(a + z) – yz] – y [(x)(a + z) – xz] + z [xy – x (a + y)]

= (a + x) [a² + az + ya + yz – yz] – y [ax + xz – xz] + z [xy – xa + xy]

= (a + x) [a² + az + ya] – y [ax] + z [– xa]

= a(a² + az + ya) + x(a² + az + ya) – yax – zxa

= a³ + a²z + ya² + xa² + xaz + xya – yax – zxa

= a³ + a²z + ya² + xa²

= a² (a + z + y + x)