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

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.

Solution 1:

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}

Solution 2:

Let $I$ be the $3\times 3$ identity matrix, $u=(1,1,1)^T$, $v=(a,b,c)^T$, and $w=(x,y,z)^T$. Then we wish to evaluate the determinant of $I+v u^T + u w^T$.

It is as easy to do this for dimension $n$ as dimension $3$ so let's do that. The shape of this matrix (and indeed the shape of the proposed answer) suggest that we should look at a larger matrix, one where the identity is 'bordered' in some way by $u,v,w$. After a little experimentation it seems we should look at the $(n+1+1)\times (n+1+1)$ block matrix $$M= \begin{bmatrix} I & u & v\\ w^T &-1 &0\\ u^T & 0 & -1\\ \end{bmatrix}. $$ We can add the $(n+1)$st row to each of the first $n$ rows without changing the value of the determinant, and likewise add the $(n+2)$nd column to each of the first $n$ columns without changing the value of the determinant. That is, $$ \det\begin{bmatrix} I & u & v\\ w^T &-1 &0\\ u^T & 0 & -1\\ \end{bmatrix} = \det\begin{bmatrix} I+u w^T+v u^T & 0 & v\\ w^T &-1 &0\\ 0 & 0 & -1\\ \end{bmatrix} = \det(I+u w^T+v u^T). $$

However we can evaluate $\det M$ in a different way. We can subtract the sum of the first $n$ columns from the $(n+1)$st column without changing the value of the determinant, and likewise we can subtract the sum of the first $n$ columns, each weighted by the corresponding entry of $v$ from the last column without changing the value of the determinant. That is,

$$ \det\begin{bmatrix} I & u & v\\ w^T &-1 &0\\ u^T & 0 & -1\\ \end{bmatrix} = \det\begin{bmatrix} I& 0 & 0\\ w^T &-1-u^T w &-v^T w\\ u^T & -u^T u & -1-v^T u\\ \end{bmatrix} = (1+u^T w)(1+w^T u)-n v^T w. $$

In the $3\times 3$ case that gives $$ \det \begin{bmatrix} 1 +a +x & a+y & a+z\\ b +x & 1+ b+y & b+z\\ c +x & c+y & 1+c+z\\ \end{bmatrix} = (1+x+y+z)(1+a+b+c)-3(ax+by+cz). $$

[I suspect this is just some variant of the Matrix Determinant Lemma.]