All right, let's solve the problem step by step.
We are given two matrices [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] and we need to find the determinant of the matrix [tex]\(\frac{1}{2}P + Q\)[/tex].
Matrix [tex]\( P \)[/tex] is:
[tex]\[ P = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} \][/tex]
Matrix [tex]\( Q \)[/tex] is:
[tex]\[ Q = \begin{pmatrix} 2 & 4 \\ 6 & 2 \end{pmatrix} \][/tex]
First, let's compute the determinant of [tex]\( P \)[/tex].
### Step 1: Determinant of [tex]\( P \)[/tex]
For a 2x2 matrix [tex]\( P = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], the determinant is given by:
[tex]\[ \text{det}(P) = ad - bc \][/tex]
Substituting the values from matrix [tex]\( P \)[/tex]:
[tex]\[ \text{det}(P) = (2 \cdot 1) - (0 \cdot 1) = 2 - 0 = 2 \][/tex]
So, the determinant of [tex]\( P \)[/tex] is:
[tex]\[ \text{det}(P) = 2 \][/tex]
### Step 2: Determinant of [tex]\( Q \)[/tex]
Similarly, we compute the determinant of [tex]\( Q \)[/tex].
For matrix [tex]\( Q = \begin{pmatrix} 2 & 4 \\ 6 & 2 \end{pmatrix} \)[/tex]:
[tex]\[ \text{det}(Q) = (2 \cdot 2) - (4 \cdot 6) = 4 - 24 = -20 \][/tex]
So, the determinant of [tex]\( Q \)[/tex] is:
[tex]\[ \text{det}(Q) = -20 \][/tex]
### Step 3: Calculate Matrix [tex]\(\frac{1}{2}P + Q\)[/tex]
First, we need to compute [tex]\(\frac{1}{2}P\)[/tex].
[tex]\[
\frac{1}{2}P = \frac{1}{2} \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}
\][/tex]
Now, add this result to matrix [tex]\( Q \)[/tex].
[tex]\[
\frac{1}{2}P + Q = \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} + \begin{pmatrix} 2 & 4 \\ 6 & 2 \end{pmatrix} = \begin{pmatrix} 1 + 2 & 0 + 4 \\ \frac{1}{2} + 6 & \frac{1}{2} + 2 \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 6.5 & 2.5 \end{pmatrix}
\][/tex]
### Step 4: Determinant of [tex]\(\frac{1}{2}P + Q\)[/tex]
Now we compute the determinant of the new matrix [tex]\( \frac{1}{2}P + Q \)[/tex].
For matrix [tex]\( \begin{pmatrix} 3 & 4 \\ 6.5 & 2.5 \end{pmatrix} \)[/tex]:
[tex]\[
\text{det}\left(\frac{1}{2}P + Q\right) = (3 \cdot 2.5) - (4 \cdot 6.5) = 7.5 - 26 = -18.5
\][/tex]
### Final Results:
- The determinant of [tex]\( P \)[/tex] is:
[tex]\[ \text{det}(P) = 2 \][/tex]
- The determinant of [tex]\( Q \)[/tex] is:
[tex]\[ \text{det}(Q) = -20 \][/tex]
- The determinant of matrix [tex]\( \frac{1}{2}P + Q \)[/tex] is:
[tex]\[ \text{det}\left(\frac{1}{2}P + Q\right) = -18.5 \][/tex]
These are the required determinants.
