Determinant of the sum of some special matrix

Question

$A,B$ are $3\times 3$ matrices. It is known that:

1. $\det(A)=0$
2. $\forall i,j: b_{ij}=1$, where $b_{ij}$ is an element of matrix $B$
3. $\det(A+B)=1$

Find $\det(A+2014B)$

I don't know what to do. I found example in smaller dimensional of matrix $A = \begin{bmatrix} 6&3\\4&2\end{bmatrix}$, but that's all.

Please give me a hint.

Answer 1

Let $C$ be any invertible matrix and $\lambda$ any real (or complex) number. We can rewrite our $B$ as an outer product of matrices:

$$B = u \otimes u^T$$ where $u = (1,1,\ldots,1)^T$ is the $n \times 1$ column matrix with all entries $1$. We have:

$$\det(C + \lambda u \otimes u^T) = \det(C(I_n + \lambda C^{-1}u\otimes u^T)) = \det(C)\det(I_n + \lambda C^{-1}u \otimes u^T)$$ Because of the special form of the last expression, it is equal to $$\det(C) (1 + \text{tr}(\lambda C^{-1}u \otimes u^T)) = \det(C) (1 + \lambda u^T C^{-1}u)$$

Recall when $C$ is invertible, $\det(C) C^{-1} = \text{adj}(C)$ where $\text{adj}(C)$ is the adjugate matrix of $C$. This implies

$$\det(C + \lambda u\otimes u^T) = \det(C) + \lambda u^T\cdot (\text{adj}(C) u)\tag{*1}$$

Notice the entries of $\text{adj}(C)$ are polynomials in entries of $C$. Since $(*1)$ is valid for invertible $C$ and the set of invertible matrices are dense in the set of real (or complex) matrices. $(*1)$ is valid even when $C$ is not invertible!

Apply this to our matrix $A$ and set $\lambda$ to $2014$, we get

$$\begin{align} \det(A+2014B) &= \det(A) + 2014u^T\cdot( \text{adj}(A) u)\\ &= -2013\det(A) + 2014(\det(A) + u^T\cdot( \text{adj}(A) u))\\ &= -2013\det(A) + 2014\det(A+B)\\ &= 2014 \end{align} $$

Answer 2

Possible method of approach:

We can write that $B = xx^T$ where $x = (1,1,1)^T$. We can then use the matrix determinant lemma to state $$ \det(A + 2014B) = \\ \det((A + B) + 2013 xx^T) =\\ (1 + 2013 \,x^T(A+B)^{-1}x) \det(A+B) =\\ 1 + 2013 \,x^T(A+B)^{-1}x $$ I'm not sure if we can get the answer from here (or how to do so).

Answer 3

Hint. We may (why?) assume that $A=(u,\ v,\ au+bv)$ for some vectors $u,v$ and some scalars $a,b$. Let $w$ be an arbitrary vector. By multilinearity of the determinant function, one can show that $$ \det(u+w,\ v+w,\ au+bv+w) = (1-a-b)\det(u,v,w) $$ and therefore the determinant is linear in $w$.