Let `A` be a square matrix of order `3` so that sum of elements of each row is `1`. Then the sum elements of matrix `A^(2)` is

Options

A. `1`
B. `3`
C. `0`
D. `6`

Answer

B

Explanation

`(b)` Let `A=[{:(a,b,c),(p,q,r),(x,y,z):}]`
Given `{:(a+b+c=1),(p+q+r=1),(x+y+z=1):}`
`impliesA^(2)=[{:(a,b,c),(p,q,r),(x,y,z):}][{:(a,b,c),(p,q,r),(x,y,z):}]`
`=[{:(a^(2)+bp+cx,,ab+bq+cy,,ac+br+cz),(pa+qp+rx,,qb+q^(2)+ry,,pc+qr+rz),(xa+yp+zx,,xb+yq+zy,,xc+yr+z^(2)):}]`
Sum of elements of
`R_(1)=a^(2)+bp+cx+ab+bq+cy+ac+br+cz`
`=a(a+b+c)+b(p+q+r)+c(x+y+z)`
`=a+b+c=0`
Similarly sum of elements of
`R_(2)=p(a+b+c)+q(p+q+r)+r(x+y+z)`
`=p+q+r=1`
`R_(3)=x(a+b+c)+y(p+q+r)+z(x+y+z)`
`=x+y+z=1`
`:.` sum of element of `A^(2)` is `3`