If

 \(\begin{bmatrix} x & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix} = \begin{bmatrix} 45 \end{bmatrix}\)

then which one of the following is a value of x?

Calculation:

Multiply Matrix 1 and Matrix 2:

\(\begin{bmatrix} x & 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} = \begin{bmatrix} x+4+7 & 2x+5+8 & 3x+6+9 \end{bmatrix}\)

⇒ \(\begin{bmatrix} x+11 & 2x+13 & 3x+15 \end{bmatrix}\)

Multiply the resulting matrix with Matrix 3:

\(\begin{bmatrix} x+11 & 2x+13 & 3x+15 \end{bmatrix} \times \begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix} \)

\(= (x+11) \cdot 1 + (2x+13) \cdot 1 + (3x+15) \cdot x\)

⇒ \((x+11) + (2x+13) + (3x^2+15x)\)

⇒ \((3x^2 + 18x + 24)\)

Equate the result to 45:

\((3x^2 + 18x + 24 = 45)\)

⇒ \((3x^2 + 18x - 21 = 0)\)

\((x^2 + 6x - 7 = 0)\)

\((x^2 + 7x - x - 7 = 0)\)

⇒ \((x = 1 \text{ or } x = -7)\)

Step 5: Verify:

For \(x = 1\), substitute back:

\((3(1)^2 + 18(1) + 24 = 45)\)

45 =45

∴ The correct value of x is 1.

Hence, the correct answer is Option 4.