Find all values of x for which `|{:(2,-1,4),(3,0,5),(4,1,6):}|=|{:(x,4),(5,x):}|`.

Options

A. `pm3.78`
B. `pm4.47`
C. `pm5.12`
D. `pm6.19`

Explanation

To find all values of \( x \) for which

\[ \left| \begin{array}{ccc} 2 & -1 & 4 \\ 3 & 0 & 5 \\ 4 & 1 & 6 \end{array} \right| = \left| \begin{array}{cc} x & 4 \\ 5 & x \end{array} \right|, \]

we will calculate the determinants of both matrices and set them equal to each other.

### Step 1: Calculate the determinant of the first matrix \( D_1 \)

The determinant \( D_1 \) of the first matrix can be calculated using the formula for the determinant of a 3x3 matrix:

\[ D_1 = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

\[ \begin{array}{ccc} a = 2 & b = -1 & c = 4 \\ d = 3 & e = 0 & f = 5 \\ g = 4 & h = 1 & i = 6 \end{array} \]

Calculating \( D_1 \):

\[ D_1 = 2(0 \cdot 6 - 5 \cdot 1) - (-1)(3 \cdot 6 - 5 \cdot 4) + 4(3 \cdot 1 - 0 \cdot 4) \]

Calculating each term:

1. \( 2(0 - 5) = 2(-5) = -10 \)
2. \( -(-1)(18 - 20) = 1(-2) = 2 \)
3. \( 4(3 - 0) = 4 \cdot 3 = 12 \)

Now combine these results:

\[ D_1 = -10 + 2 + 12 = 4 \]

### Step 2: Calculate the determinant of the second matrix \( D_2 \)

The determinant \( D_2 \) of the second matrix is calculated as follows:

\[ D_2 = \left| \begin{array}{cc} x & 4 \\ 5 & x \end{array} \right| = x \cdot x - 4 \cdot 5 = x^2 - 20 \]

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

Now we set \( D_1 \) equal to \( D_2 \):

\[ 4 = x^2 - 20 \]

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

Rearranging the equation gives:

\[ x^2 - 20 - 4 = 0 \implies x^2 - 24 = 0 \]

Factoring or using the square root method:

\[ x^2 = 24 \implies x = \pm \sqrt{24} = \pm 2\sqrt{6} \]

### Final Values of \( x \)

Thus, the values of \( x \) are:

\[ x = 2\sqrt{6} \quad \text{and} \quad x = -2\sqrt{6} \]