Question

Let \(\begin{vmatrix}3x&-7\\1&4\end{vmatrix}=\begin{vmatrix}3&2\\ 4&x\end{vmatrix}\), then value of x is:

• A. \(\frac{-5}{3}\)
• B. 1
• C. -1
• D. \(\frac{2}{7}\)

Answer 1

The Correct Option is A

To solve the given equation, we need to evaluate the determinant of each matrix and set them equal to each other. The determinant of a 2x2 matrix \(\begin{vmatrix}a & b \\ c & d\end{vmatrix}\) is calculated using the formula \(ad - bc\).

Starting with the first matrix \(\begin{vmatrix}3x & -7 \\ 1 & 4\end{vmatrix}\), the determinant is calculated as follows:

\(3x \cdot 4 - (-7) \cdot 1 = 12x + 7\).

Now, the second matrix \(\begin{vmatrix}3 & 2 \\ 4 & x\end{vmatrix}\) has a determinant:

\(3 \cdot x - 2 \cdot 4 = 3x - 8\).

Equating the two determinants, we have:

\(12x + 7 = 3x - 8\).

To solve for \(x\), we first subtract \(3x\) from both sides:

\(12x - 3x + 7 = -8\).

Which simplifies to:

\(9x + 7 = -8\).

Subtract 7 from both sides:

\(9x = -8 - 7\).

Which gives:

\(9x = -15\).

Finally, divide both sides by 9:

\(x = \frac{-15}{9} = \frac{-5}{3}\).

The solution is therefore:

\(x = \frac{-5}{3}\).