If `|{:(2x,5),(8,x):}|=|{:(6,-2),(7,3):}|`, then the value of x is

Options

A. 3
B. `+-3`
C. `+-6`
D. `6`

Answer

To solve the equation \( | \begin{pmatrix} 2x & 5 \\ 8 & x \end{pmatrix} | = | \begin{pmatrix} 6 & -2 \\ 7 & 3 \end{pmatrix} | \), we will first calculate the determinants of both matrices.

### Step 1: Calculate the determinant of the first matrix

The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by the formula:
\[ |A| = ad - bc \]

For the first matrix \( \begin{pmatrix} 2x & 5 \\ 8 & x \end{pmatrix} \):
- \( a = 2x \)
- \( b = 5 \)
- \( c = 8 \)
- \( d = x \)

Using the determinant formula:
\[ | \begin{pmatrix} 2x & 5 \\ 8 & x \end{pmatrix} | = (2x)(x) - (5)(8) = 2x^2 - 40 \]

### Step 2: Calculate the determinant of the second matrix

Now, let's calculate the determinant of the second matrix \( \begin{pmatrix} 6 & -2 \\ 7 & 3 \end{pmatrix} \):
- \( a = 6 \)
- \( b = -2 \)
- \( c = 7 \)
- \( d = 3 \)

Using the determinant formula:
\[ | \begin{pmatrix} 6 & -2 \\ 7 & 3 \end{pmatrix} | = (6)(3) - (-2)(7) = 18 + 14 = 32 \]

### Step 3: Set the determinants equal to each other

Now we set the two determinants equal to each other:
\[ 2x^2 - 40 = 32 \]

### Step 4: Solve for \( x \)

To solve for \( x \), first add 40 to both sides:
\[ 2x^2 = 32 + 40 \]
\[ 2x^2 = 72 \]

Next, divide both sides by 2:
\[ x^2 = \frac{72}{2} = 36 \]

Now, take the square root of both sides:
\[ x = \pm 6 \]

### Final Answer

Thus, the values of \( x \) are:
\[ x = 6 \quad \text{or} \quad x = -6 \]