To find the value of [tex]\( x \)[/tex] such that [tex]\( P Q = Q P \)[/tex] for the given matrices [tex]\( P \)[/tex] and [tex]\( Q \)[/tex], we need to perform matrix multiplication on both [tex]\( P Q \)[/tex] and [tex]\( Q P \)[/tex] and then equate the corresponding elements. Below is the detailed step-by-step solution.
### Matrices Given:
[tex]\[ P = \begin{bmatrix} 1 & x \\ 3 & 4 \end{bmatrix} \][/tex]
[tex]\[ Q = \begin{bmatrix} 5 & x^2 \\ 3x & 11 \end{bmatrix} \][/tex]
### Step 1: Compute [tex]\( P Q \)[/tex]
Calculate the product [tex]\( P Q \)[/tex]:
[tex]\[ P Q = \begin{bmatrix} 1 & x \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & x^2 \\ 3x & 11 \end{bmatrix} \][/tex]
[tex]\[
P Q = \begin{bmatrix}
1 \cdot 5 + x \cdot 3x & 1 \cdot x^2 + x \cdot 11 \\
3 \cdot 5 + 4 \cdot 3x & 3 \cdot x^2 + 4 \cdot 11
\end{bmatrix}
\][/tex]
[tex]\[
P Q = \begin{bmatrix}
5 + 3x^2 & x^2 + 11x \\
15 + 12x & 3x^2 + 44
\end{bmatrix}
\][/tex]
### Step 2: Compute [tex]\( Q P \)[/tex]
Calculate the product [tex]\( Q P \)[/tex]:
[tex]\[ Q P = \begin{bmatrix} 5 & x^2 \\ 3x & 11 \end{bmatrix} \begin{bmatrix} 1 & x \\ 3 & 4 \end{bmatrix} \][/tex]
[tex]\[
Q P = \begin{bmatrix}
5 \cdot 1 + x^2 \cdot 3 & 5 \cdot x + x^2 \cdot 4 \\
3x \cdot 1 + 11 \cdot 3 & 3x \cdot x + 11 \cdot 4
\end{bmatrix}
\][/tex]
[tex]\[
Q P = \begin{bmatrix}
5 + 3x^2 & 5x + 4x^2 \\
3x + 33 & 3x^2 + 44
\end{bmatrix}
\][/tex]
### Step 3: Equate the Corresponding Elements of [tex]\( P Q \)[/tex] and [tex]\( Q P \)[/tex]
For [tex]\( P Q = Q P \)[/tex], the corresponding elements must be equal:
1. [tex]\( 5 + 3x^2 \)[/tex] should equal [tex]\( 5 + 3x^2 \)[/tex] (this is always true).
2. [tex]\( x^2 + 11x \)[/tex] should equal [tex]\( 5x + 4x^2 \)[/tex].
3. [tex]\( 15 + 12x \)[/tex] should equal [tex]\( 3x + 33 \)[/tex].
4. [tex]\( 3x^2 + 44 \)[/tex] should equal [tex]\( 3x^2 + 44 \)[/tex] (this is always true).
### Step 4: Solve the Equations
#### Equation 2:
[tex]\[ x^2 + 11x = 5x + 4x^2 \][/tex]
Rearrange to form a standard quadratic equation:
[tex]\[ x^2 + 11x - 5x - 4x^2 = 0 \][/tex]
[tex]\[ -3x^2 + 6x = 0 \][/tex]
[tex]\[ 3x(2 - x) = 0 \][/tex]
Thus, we have:
[tex]\[ x = 0 \quad \text{or} \quad x = 2 \][/tex]
#### Equation 3:
[tex]\[ 15 + 12x = 3x + 33 \][/tex]
Rearrange:
[tex]\[ 12x - 3x = 33 - 15 \][/tex]
[tex]\[ 9x = 18 \][/tex]
[tex]\[ x = 2 \][/tex]
### Step 5: Check for Consistency
From both equations (Equation 2 and Equation 3), the consistent solution for [tex]\( x \)[/tex] is:
[tex]\[ x = 2 \][/tex]
Thus, the value of [tex]\( x \)[/tex] is [tex]\( 2 \)[/tex] so that [tex]\( P Q = Q P \)[/tex].
