Finding x in matrix algebra The question asked me to calculate $|A|$ for the given matrix below and state whether or not it is invertible.
$$
        \left(
        \begin{matrix}
        1 & 0 & 2 & 1\\
        0 & x & x & x\\
        x & x & 3x & 1\\
        0 & x & x & 0\\
        \end{matrix}
        \right)
$$
I got my answer to be $-2x^3 + 2x^3 = 0$, indicating that the matrix is not invertible because the determinant is a zero, regardless of the value of $x$.
However, I'm not certain whether I'm right after doing all the calculations. Could you help me verify?

Okay, here's every step:
$$\left|\begin{matrix} x & x & x \\ x & 3x & 1 \\ x & x & 0\end{matrix}\right|+2\left|\begin{matrix} 0 & x & x \\ x & x & 1 \\ 0 & x & 0\end{matrix}\right| - \left|\begin{matrix} 0 & x & x \\ x & x & 3x \\ 0 & x & x\end{matrix}\right| \\ = x^2\left|\begin{matrix} 1 & 1 & x \\ 1 & 3 & 1 \\ 1 & 1 & 0\end{matrix}\right|+2x^2\left|\begin{matrix} 0 & 1 & x \\ 1 & 1 & 1 \\ 0 & 1 & 0\end{matrix}\right| - x^3\left|\begin{matrix} 0 & 1 & 1 \\ 1 & 1 & 3 \\ 0 & 1 & 1\end{matrix}\right| \\ = x^2\left(\left|\begin{matrix} 3 & 1 \\ 1 & 0\end{matrix}\right| - \left|\begin{matrix} 1 & 1 \\ 1 & 0\end{matrix}\right| + x\left|\begin{matrix} 1 & 3 \\ 1 & 1\end{matrix}\right|\right) + 2x^2\left(- \left|\begin{matrix} 1 & 1 \\ 0 & 0\end{matrix}\right| + x\left|\begin{matrix} 1 & 1 \\ 0 & 1\end{matrix}\right|\right) -x^3\left(- \left|\begin{matrix} 1 & 3 \\ 0 & 1\end{matrix}\right| + \left|\begin{matrix} 1 & 1 \\ 0 & 1\end{matrix}\right|\right) \\ = -2x^3+2x^3-0 \\ =0$$
So you did it correctly.

Or, or course, for the easier way, just recognize that twice the first column plus the second column equals the third column of your matrix so the determinant must be $0$.
