Evaluate $\begin{vmatrix}1+a+x&a+y&a+z\\b+x&1+b+y&b+z\\c+x&c+y&1+c+z\end{vmatrix}$

$R_1\to R_1+R_2+R_3$ and letting $1+a+b+c=k$, I get $\begin{vmatrix}k+3x&k+3y&k+3z\\b+x&1+b+y&b+z\\c+x&c+y&1+c+z\end{vmatrix}$

Therefore, $k\begin{vmatrix}1&1&1\\b+x&1+b+y&b+z\\c+x&c+y&1+c+z\end{vmatrix}+3\begin{vmatrix}x&y&z\\b+x&1+b+y&b+z\\c+x&c+y&1+c+z\end{vmatrix}$

If I open the first determinant, I get $(1+a+b+c)(1-2x+y+z)$.

But the final answer given is $(1+a+b+c)(1+x+y+z)-3(ax+by+cz)$.

My answer for second determinant is also not matching with the answer.

## Best Answer

Checking with software, one gets

$$ \begin{vmatrix}k+3x&k+3y&k+3z\\b+x&1+b+y&b+z\\c+x&c+y&1+c+z\end{vmatrix} = (1+x+y+z)(1+a+b+c)-3(ax+by+cz)$$ and $$ k\begin{vmatrix}1&1&1\\b+x&1+b+y&b+z\\c+x&c+y&1+c+z\end{vmatrix} = (1+a+b+c)(1-2x+y+z)$$

so your error must be in calculating $$3\begin{vmatrix}x&y&z\\b+x&1+b+y&b+z\\c+x&c+y&1+c+z\end{vmatrix}.$$

That last determinant should be $3(1+a+b+c)x - 3(ax+by+cz).$ Indeed, one has $$(1 + a + b + c)(1-2x+y+z) + 3(1 + a + b + c)x - 3(ax+by+cz) = (1 + a + b + c)(1+x+y+z)-3(ax+by+cz).$$

To get that last calculation right, one can use row reduction:

\begin{align} \begin{vmatrix}x&y&z\\b+x&1+b+y&b+z\\c+x&c+y&1+c+z\end{vmatrix} &= \begin{vmatrix}x&y&z\\b&1+b&b\\c&c&1+c\end{vmatrix}\\ &= x((1+b)(1+c)-bc) - y(b(1+c)-bc) + z(bc-(1+b)c)\\ &= (1+b+c)x - by - cz \\ &= (1+a+b+c)x - (ax + by + cz). \end{align}