If `[1 x] [(2,-1),(1,2)][(1),(3)]=[0]`, then x=

Options

A. `2/7`
B. `1/7`
C. `5/7`
D. `3/7`

Answer

To solve the equation \([1 \, x] \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = [0]\), we will follow these steps:

### Step 1: Matrix Multiplication
First, we need to multiply the first two matrices \([1 \, x]\) and \(\begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix}\).

The resulting matrix will be:
\[ [1 \cdot 2 + x \cdot 1 \quad 1 \cdot -1 + x \cdot 2] = [2 + x \quad -1 + 2x] \]

### Step 2: Multiply by the Third Matrix
Next, we multiply the result by the third matrix \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\).

Using the row-column multiplication:
\[ (2 + x) \cdot 1 + (-1 + 2x) \cdot 3 \]
This simplifies to:
\[ 2 + x - 3 + 6x = 7x - 1 \]

### Step 3: Set the Equation to Zero
According to the problem, this result equals \([0]\):
\[ 7x - 1 = 0 \]

### Step 4: Solve for x
Now, we solve for \(x\):
\[ 7x = 1 \implies x = \frac{1}{7} \]

### Final Answer
Thus, the value of \(x\) is:
\[ \boxed{\frac{1}{7}} \]

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