Question

SOLUTION Let the value of the given determinant be ( Delta ). Then, ( Delta=left|begin{array}{ccc}a+x & y & z x & a+y & z x & y & a+zend{array}right| )
( =left|begin{array}{ccc}a+x+y+z & y & z a+x+y+z & a+y & z a+x+y+z & y & a+z 1 & y & z 1 & a+y & z 1 & y & a+zend{array}right| )
( =(a+x+y+z) cdotleft|begin{array}{cc}x+y & y z+y & zend{array}right| )
( left[operatorname{taking}(a+x+y+z) operatorname{common}left(operatorname{rom} C_{1}right]right. )
( =(a+x+y+z) cdotleft|begin{array}{ccc}1 & y & z 0 & a & 0 0 & 0 & aend{array}right| quadleft[R_{2} rightarrow R_{2}-R_{1} text { and } R_{3} rightarrow R_{3}-R_{1}right] )
( left.=(a+x+y+z) cdot 1 cdotleft|begin{array}{ll}a & 0 0 & aend{array}right| quad text { [expanded by } C_{1}right] )
( =a^{2}(a+x+y+z) )
Hence, ( Delta=a^{2}(a+x+y+z) )

# EXAMPLE 10 Using properties of determinants, show that a+X Y Z X a+y z = a(a + x + y +z). x y a +2

Solution