Let A = `[{:(x + 2, 3x),(3,x + 2):}], B = [{:(x , 0),(5 , x + 2):}]`. Then all solutions of the equation det (AB ) = 0 is

Options

A. 1, - 1, 0, 2
B. 1, 4, 0, -2
C. 1, - 1, 4, 3
D. `-1, 4, 0, 3 `

Answer

To solve the equation \( \text{det}(AB) = 0 \) where \( A = \begin{pmatrix} x + 2 & 3x \\ 3 & x + 2 \end{pmatrix} \) and \( B = \begin{pmatrix} x & 0 \\ 5 & x + 2 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the product \( AB \)

To find \( AB \), we multiply the matrices \( A \) and \( B \):

\[ AB = A \cdot B = \begin{pmatrix} x + 2 & 3x \\ 3 & x + 2 \end{pmatrix} \cdot \begin{pmatrix} x & 0 \\ 5 & x + 2 \end{pmatrix} \]

Calculating the elements of the resulting matrix:

- First row, first column:
\[ (x + 2) \cdot x + 3x \cdot 5 = x^2 + 2x + 15x = x^2 + 17x \]

- First row, second column:
\[ (x + 2) \cdot 0 + 3x \cdot (x + 2) = 3x^2 + 6x \]

- Second row, first column:
\[ 3 \cdot x + (x + 2) \cdot 5 = 3x + 5x + 10 = 8x + 10 \]

- Second row, second column:
\[ 3 \cdot 0 + (x + 2) \cdot (x + 2) = (x + 2)^2 = x^2 + 4x + 4 \]

Thus, the product \( AB \) is:

\[ AB = \begin{pmatrix} x^2 + 17x & 3x^2 + 6x \\ 8x + 10 & (x + 2)^2 \end{pmatrix} \]

### Step 2: Calculate the determinant \( \text{det}(AB) \)

The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \).

For our matrix \( AB \):

\[ \text{det}(AB) = (x^2 + 17x)(x^2 + 4x + 4) - (3x^2 + 6x)(8x + 10) \]

### Step 3: Expand the determinant expression

Expanding \( (x^2 + 17x)(x^2 + 4x + 4) \):

\[ = x^4 + 4x^3 + 4x^2 + 17x^3 + 68x^2 + 68x \]
\[ = x^4 + 21x^3 + 72x^2 + 68x \]

Expanding \( (3x^2 + 6x)(8x + 10) \):

\[ = 24x^3 + 30x^2 + 48x^2 + 60x \]
\[ = 24x^3 + 78x^2 + 60x \]

### Step 4: Set the determinant to zero

Now, substituting back into the determinant equation:

\[ x^4 + 21x^3 + 72x^2 + 68x - (24x^3 + 78x^2 + 60x) = 0 \]

Combining like terms:

\[ x^4 + (21x^3 - 24x^3) + (72x^2 - 78x^2) + (68x - 60x) = 0 \]
\[ x^4 - 3x^3 - 6x^2 + 8x = 0 \]

### Step 5: Factor the polynomial

Factoring out \( x \):

\[ x(x^3 - 3x^2 - 6x + 8) = 0 \]

This gives us one solution:

\[ x = 0 \]

### Step 6: Solve the cubic equation

Now we need to solve \( x^3 - 3x^2 - 6x + 8 = 0 \). We can use the Rational Root Theorem or synthetic division to find the roots. Testing \( x = 2 \):

\[ 2^3 - 3(2^2) - 6(2) + 8 = 8 - 12 - 12 + 8 = -8 \quad \text{(not a root)} \]

Testing \( x = 1 \):

\[ 1^3 - 3(1^2) - 6(1) + 8 = 1 - 3 - 6 + 8 = 0 \quad \text{(is a root)} \]

Now, we can factor \( x - 1 \) out of \( x^3 - 3x^2 - 6x + 8 \) using synthetic division, which gives us:

\[ x^3 - 3x^2 - 6x + 8 = (x - 1)(x^2 - 2x - 8) \]

### Step 7: Solve the quadratic equation

Now we solve \( x^2 - 2x - 8 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} \]

This gives us:

\[ x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{-4}{2} = -2 \]

### Final Solutions

The solutions to the equation \( \text{det}(AB) = 0 \) are:

\[ x = 0, \quad x = -2, \quad x = 1, \quad x = 4 \]