Calculating the determinant of a matrices

Question

Show that

$ det \left[ {\begin{array}{cc} 0 & 1 & 1 & 1\\ 1 & 0& x & x\\ 1 & x & 0 & x\\ 1 & x & x & 0 \end{array} } \right] $ = $-3x^2$

I have calculated a 2x2 determinant and 3x3 but never 4x4 so im quite confused on this

Answer 1

You can use the Laplace formula directly:

$$ \det\begin{bmatrix} 0 & 1 & 1 & 1\\ 1 & 0& x & x\\ 1 & x & 0 & x\\ 1 & x & x & 0 \end{bmatrix} = -\det\begin{bmatrix}1& x & x\\1 & 0 & x\\ 1 & x & 0\end{bmatrix} +\det\begin{bmatrix}1& 0 & x\\1 & x & x\\ 1 & x & 0\end{bmatrix} -\det\begin{bmatrix}1& 0 & x\\1 & x & 0\\ 1 & x & x\end{bmatrix} $$ Switching two rows negates the determinant. We see that we can get the first determinant by switching the rows of the second determinant once and also by switching the rows of the third determinand twice. We are left with

$$ {}=-3\det\begin{bmatrix}1& x & x\\1 & 0 & x\\ 1 & x & 0\end{bmatrix}$$

Applying the formula for $3\times3$ matrices yields

$${} = -3 (0 + x^2 + x^2 - 0 - x^2 - 0) = -3x^2. $$

Answer 2

after $-R_2+R_3\to R_3$, $-R_2+R_4\to R_4$, $-xR_1+R_3\to R_3$, $-xR_1+R_4\to R_4$, $R_2+R_3\to R_3$, $\frac{-1}{2}R_3+R_4\to R_4$ and $R_1\leftrightarrow R_2$ we have triangular below matrix thus $ det \left[ {\begin{array}{cc} 1 & 0& x & x\\ 0 & 1 & 1 & 1\\ 0 & 0 & -2x & -x\\ 0 & 0 & 0 & \frac{-3}{2}x \end{array} } \right] $ = $3x^2$ hence $$ det \left[ {\begin{array}{cc} 0 & 1 & 1 & 1\\ 1 & 0& x & x\\ 1 & x & 0 & x\\ 1 & x & x & 0 \end{array} } \right] = -3x^2$$ Note that $\color{red}{-}$ is because of $R_1\leftrightarrow R_2$