Question

using properties of determinants prove that...
1 1 1+3x
1+3y 1 1
1 1+3z 1

= 9(3xyz+xy+yz+zx)

Answer 1

hii.... here is ur proof

Answer 2

Answer:

Proved below.

Step-by-step explanation:

Given : $\left|\begin{array}{ccc}1&1&1+3x\:\\ 1+3y&1&1\\ 1&1+3z&1\end{array}\right|$

We have to prove $\left|\begin{array}{ccc}1&1&1+3x\:\\ 1+3y&1&1\\ 1&1+3z&1\end{array}\right|=9(3xyz+xy+yz+zx)$

Consider the given matrix,

$\left|\begin{array}{ccc}1&1&1+3x\:\\ 1+3y&1&1\\ 1&1+3z&1\end{array}\right|$

apply column addition,

$R_2\rightarrow R_2-R_1$

$\left|\begin{array}{ccc}1&1&1+3x\:\\3y&0&-3x\\1&1+3z&1\end{array}\right|$

now apply , $R_3\rightarrow R_3-R_1$

$\left|\begin{array}{ccc}1&1&1+3x\:\\3y&0&-3x\\0&3z&-3x\end{array}\right|$

Expand along $R_1$, we have,

$D=1(0+9xz)-1(-9xy-0)+1((1+3x)(9yz-0))$

Simplify, we get,

$D=9(3xyz+xy+yz+zx)$

Hence, proved.