Find all values of x that make `|(2,-1,4),(3,0,5),(4,1,6)|=|(x,4),(5,x)|`

Options

A. 0
B. `+-1.43`
C. `+-3`
D. `+-4.47`

Explanation

To solve the problem, we need to find all values of \( x \) that make the determinant of the 3x3 matrix equal to the determinant of the 2x2 matrix.

### Step 1: Calculate the determinant of the 3x3 matrix

The given 3x3 matrix is:
\[ \begin{pmatrix} 2 & -1 & 4 \\ 3 & 0 & 5 \\ 4 & 1 & 6 \end{pmatrix} \]

The determinant of a 3x3 matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix:
- \( a = 2, b = -1, c = 4 \)
- \( d = 3, e = 0, f = 5 \)
- \( g = 4, h = 1, i = 6 \)

Calculating the determinant:
\[ \text{det}(A) = 2(0 \cdot 6 - 5 \cdot 1) - (-1)(3 \cdot 6 - 5 \cdot 4) + 4(3 \cdot 1 - 0 \cdot 4) \]
\[ = 2(0 - 5) + 1(18 - 20) + 4(3 - 0) \]
\[ = 2(-5) + 1(-2) + 4(3) \]
\[ = -10 - 2 + 12 = 0 \]

### Step 2: Calculate the determinant of the 2x2 matrix

The given 2x2 matrix is:
\[ \begin{pmatrix} x & 4 \\ 5 & x \end{pmatrix} \]

The determinant of a 2x2 matrix is calculated as:
\[ \text{det}(B) = ax - by \]
where the matrix is:
\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

For our matrix:
- \( a = x, b = 4 \)
- \( c = 5, d = x \)

Calculating the determinant:
\[ \text{det}(B) = x \cdot x - 4 \cdot 5 = x^2 - 20 \]

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

We have:
\[ \text{det}(A) = 0 \quad \text{and} \quad \text{det}(B) = x^2 - 20 \]

Setting them equal:
\[ 0 = x^2 - 20 \]

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

Rearranging gives:
\[ x^2 = 20 \]

Taking the square root of both sides:
\[ x = \pm \sqrt{20} \]
\[ x = \pm 2\sqrt{5} \]

### Final Values of \( x \)

Thus, the values of \( x \) that satisfy the equation are:
\[ x = 2\sqrt{5} \quad \text{and} \quad x = -2\sqrt{5} \]