If `[ (2x, 3)] [(1,2),(-3,0)][(x),(8)]=0`, find 'x'

Answer

To solve the equation given by the product of the matrices \([(2x, 3)][(1, 2), (-3, 0)][(x), (8)] = 0\), we will follow these steps:

### Step 1: Multiply the first two matrices
We start by multiplying the first matrix \((2x, 3)\) with the second matrix \(\begin{pmatrix} 1 & 2 \\ -3 & 0 \end{pmatrix}\).

The multiplication of a \(1 \times 2\) matrix with a \(2 \times 2\) matrix results in a \(1 \times 2\) matrix.

\[ \begin{pmatrix} 2x & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -3 & 0 \end{pmatrix} = \begin{pmatrix} (2x \cdot 1 + 3 \cdot -3) & (2x \cdot 2 + 3 \cdot 0) \end{pmatrix} \]

Calculating the elements:
- First element: \(2x \cdot 1 + 3 \cdot -3 = 2x - 9\)
- Second element: \(2x \cdot 2 + 3 \cdot 0 = 4x\)

Thus, the result of the multiplication is:
\[ \begin{pmatrix} 2x - 9 & 4x \end{pmatrix} \]

### Step 2: Multiply the result with the third matrix
Now we multiply the resulting matrix \((2x - 9, 4x)\) with the third matrix \(\begin{pmatrix} x \\ 8 \end{pmatrix}\).

The multiplication of a \(1 \times 2\) matrix with a \(2 \times 1\) matrix results in a \(1 \times 1\) matrix (a single value).

\[ \begin{pmatrix} 2x - 9 & 4x \end{pmatrix} \begin{pmatrix} x \\ 8 \end{pmatrix} = (2x - 9) \cdot x + 4x \cdot 8 \]

Calculating this:
- First term: \((2x - 9) \cdot x = 2x^2 - 9x\)
- Second term: \(4x \cdot 8 = 32x\)

Thus, we have:
\[ 2x^2 - 9x + 32x = 2x^2 + 23x \]

### Step 3: Set the result equal to zero
According to the problem, this entire expression equals zero:
\[ 2x^2 + 23x = 0 \]

### Step 4: Factor the equation
We can factor out \(x\):
\[ x(2x + 23) = 0 \]

### Step 5: Solve for \(x\)
Setting each factor equal to zero gives us:
1. \(x = 0\)
2. \(2x + 23 = 0 \Rightarrow 2x = -23 \Rightarrow x = -\frac{23}{2}\)

### Final Solution
Thus, the values of \(x\) are:
\[ x = 0 \quad \text{or} \quad x = -\frac{23}{2} \]