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?

Recall that performing row replacement operations does not affect the determinant. Hence, if we replace row $3$ with (row $3$ + $(-x) \cdot $ row $1$), then replace row $3$ with (row $3$ - row$2$) and replace row $4$ with (row $4$ - row $2$), we obtain:
$$
        \begin{vmatrix}
        1 & 0 & 2 & 1\\
        0 & x & x & x\\
        x & x & 3x & 1\\
        0 & x & x & 0\\
        \end{vmatrix}
=
        \begin{vmatrix}
        1 & 0 & 2 & 1\\
        0 & x & x & x\\
        0 & x & x & 1-x\\
        0 & x & x & 0\\
        \end{vmatrix}
=
        \begin{vmatrix}
        1 & 0 & 2 & 1\\
        0 & x & x & x\\
        0 & 0 & 0 & 1-2x\\
        0 & 0 & 0 & -x\\
        \end{vmatrix}
$$
but since this last matrix is upper triangular, its determinant is the product of the diagonal entries. But one of the diagonal entries is a zero, so $\det A = 0$, as you computed.
