If $\left[\begin{array}{ll}2 x & 3\end{array}\right]\left[\begin{array}{rr}1 & 2 \\ -3 & 0\end{array}\right]\left[\begin{array}{l}x \\ 8\end{array}\right]=0$, then the value of $\mathrm{x}$ is

Step 1 ：First, we need to multiply the matrices: \(\begin{bmatrix} 2x & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} \begin{bmatrix} x \\ 8 \end{bmatrix}\)

Step 2 ：Next, we simplify the matrix multiplication: \(\begin{bmatrix} x(2x - 9) + 32x \end{bmatrix}\)

Step 3 ：Now, we set the result equal to 0: \(x(2x - 9) + 32x = 0\)

Step 4 ：Finally, we solve for x: \(\boxed{x = 0 \text{ or } x = -\frac{23}{2}}\)