so I got this task from my professor and wanted to ask for help I have this following matrices
(a)
$$A = \begin{pmatrix} -3 & -11 & -11 & 45 \\ 1 & 11 & 10 & -83 \\ 1 & -6 & -5 & 81 \\ 0 & -3 & -3 & 42 \end{pmatrix}$$
I did this one with the Laplace expansion stuff $4 \times 4$ and got $42$, though I don't know if it's right yet.
Now here comes the real problem
(b)
$$ B = \begin{pmatrix} 1+a_1 & a_2 & \dots & a_n \\ a_1 & 1+a_2 & \dots & a_n \\ \dots & \dots & \dots & \dots \\ a_1 & a_2 & \dots & 1+a_n \end{pmatrix}$$
with $a_1, \dots , a_n$ are elements of $\mathbb{R}$.
So how do I do this one ? also with the Laplace expansion ? What kind of value should come out ? a something ? If there were more numbers I could solve it more easily and I don't know how big it is.. like $4\times 4$ or something.. but it's $n\times n$ so how am I doing this one
And at last
(c) $$C = (c_{ij})$$ with $c_{ij} = 0$ if $i = j$ $1$ if $i != j$ for $1<i,j<n$
I have the same problem with this one (similiar to b)
It would be nice if someone could give me hints/a solution/etc Thanks for reading
What I meant for $b$ was:
$$\left|\begin{matrix}1+a & b & c \\ a & 1+b & c \\a & b & 1+c\end{matrix}\right|=\left|\begin{matrix}1 & -1 & 0 \\ a & 1+b & c \\a & b & 1+c \end{matrix}\right|$$
$$=\left|\begin{matrix}1+b & c \\ b & 1+c\end{matrix}\right|+\left|\begin{matrix}a & c \\ a & 1+c\end{matrix}\right|$$
$$=\left|\begin{matrix}1 & -1 \\ b & 1+c\end{matrix}\right|+a\left|\begin{matrix}1 & c \\ 1 & 1+c\end{matrix}\right|$$
$$=([1+c]+[b])+a([1+c]-[c])=a+b+c+1$$
If you know induction, it will be a very quick proof (if you can see the general solution from the above). Achille Hui's comment gives a much faster route to the solution.
Now for question $c$, it is almost a special case of $(b)$. Notice if we choose $a_i=-1$ for all $i$ then the matrix becomes
$$\left(\begin{matrix}0 & -1 & \cdots & -1\\ -1 & 0 & \cdots & -1\\ \vdots & \vdots & \ddots & -1\\ -1 & -1 & \cdots & 0\end{matrix}\right)$$
From the result in $(b)$ the determinant of this matrix is $1-n$. The matrix in $(c)$ is the negative of this one. So factoring out a $(-1)$ from each row will finish it off.