The following matrices are $4 \times 4$ matrices.
$$A=\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1\\ 1 & 1& 1 &0\\ 1 &1 &0 &0 \end{bmatrix}\\ B= \begin{bmatrix} 0 & 1& 1& 1 \\ 1 & 1& 1& 1 \\ 1& 1& 1& 0 \\1 & 1& 0 & 0 \end{bmatrix} \\ C= \begin{bmatrix} x & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1&1&1&0 \\1&1&0&1 \end{bmatrix}$$
- Find $\det{A}$
- Find the cofactor $C_{11}$ of $B$ and then use that to find $\det{B}$
- Use the above results to find $\det{C}$ in terms of $A$ and $B$
I know the solutions, but I would like an explanation why.
$\det{A}=0$, $\det{B}=1$, $\det{C}= -(x-1)\det{B}$.
we know that aading the coefficient of one row to the other does not change the determinant so for matrix $A$ we can subtract the second row from the first row
$$\det A=\begin{vmatrix}1&1&1&1\\ 1&1&1&1\\ 1&1&1&0\\ 1&1&0&0\end{vmatrix}= \begin{vmatrix}1&1&1&1\\ 0&0&0&0\\ 1&1&1&0\\ 1&1&0&0\end{vmatrix}=0$$ From the fact that if all of the elements in a column or row is zero, the determinant will be zero
For matrix $B$, we first subtract second row from the first row in order to make the determinant simpler and to have just one nonzero element in the first row (This way the expansion of determinant w.r.t the first row will be much easier) so:
$$\det B=\begin{vmatrix}0&1&1&1\\ 1&1&1&1\\ 1&1&1&0\\ 1&1&0&0\end{vmatrix}= \begin{vmatrix}-1&0&0&0\\ 1&1&1&1\\ 1&1&1&0\\ 1&1&0&0\end{vmatrix}=(-1)\underbrace{(-1)^{1+1}\begin{vmatrix}1&1&1\\ 1&1&0\\ 1&0&0\end{vmatrix}}_{C_{11}\text{cofactor of B}}=(-1)(1)(-1)^{1+3}\begin{vmatrix}1&1\\ 1&0\end{vmatrix}=(-1)(-1)=1$$
For $C$ we have:
$$\det C=\begin{vmatrix}x&1&1&1\\ 1&1&1&1\\ 1&1&1&0\\ 1&1&0&1\end{vmatrix}= \begin{vmatrix}x-1&0&0&0\\ 1&1&1&1\\ 1&1&1&0\\ 1&1&0&1\end{vmatrix}=(x-1)(-1)^{1+1}\begin{vmatrix}1&1&1\\ 1&1&0\\ 1&0&1\end{vmatrix}$$
From the previous calculation we know that:
$$(-1)(-1)^{1+1}\begin{vmatrix}1&1&1\\ 1&1&0\\ 1&0&0\end{vmatrix}=\det B\Rightarrow (-1)^{1+1}\begin{vmatrix}1&1&1\\ 1&1&0\\ 1&0&0\end{vmatrix}=-\det B$$ So: $$\det C=(x-1)(-1)^{1+1}\begin{vmatrix}1&1&1\\ 1&1&0\\ 1&0&1\end{vmatrix}=-(x-1)\det B$$