Wright Simpson
08/13/2023 · Senior High School
Let \( D=\left[\begin{array}{cc}1 & 1 \\ -8 & 3\end{array}\right] \) and \( P=\left[\begin{array}{cc}-2 & 2 \\ 6 & -5\end{array}\right] \) Calculate \( (D P)^{T} \) without calculating \( D P \).
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To calculate \( (D P)^T \) without directly computing \( D P \), we can use the property of transposes that states:
\[
(AB)^T = B^T A^T
\]
for any matrices \( A \) and \( B \). In our case, we have \( A = D \) and \( B = P \). Therefore, we can express \( (D P)^T \) as:
\[
(D P)^T = P^T D^T
\]
Now, we need to compute \( P^T \) and \( D^T \).
First, we calculate \( P^T \):
\[
P = \begin{bmatrix} -2 & 2 \\ 6 & -5 \end{bmatrix}
\]
Taking the transpose, we get:
\[
P^T = \begin{bmatrix} -2 & 6 \\ 2 & -5 \end{bmatrix}
\]
Next, we calculate \( D^T \):
\[
D = \begin{bmatrix} 1 & 1 \\ -8 & 3 \end{bmatrix}
\]
Taking the transpose, we get:
\[
D^T = \begin{bmatrix} 1 & -8 \\ 1 & 3 \end{bmatrix}
\]
Now we can compute \( (D P)^T \) using the transposes we found:
\[
(D P)^T = P^T D^T = \begin{bmatrix} -2 & 6 \\ 2 & -5 \end{bmatrix} \begin{bmatrix} 1 & -8 \\ 1 & 3 \end{bmatrix}
\]
Now we perform the matrix multiplication:
1. The element in the first row, first column:
\[
-2 \cdot 1 + 6 \cdot 1 = -2 + 6 = 4
\]
2. The element in the first row, second column:
\[
-2 \cdot -8 + 6 \cdot 3 = 16 + 18 = 34
\]
3. The element in the second row, first column:
\[
2 \cdot 1 + -5 \cdot 1 = 2 - 5 = -3
\]
4. The element in the second row, second column:
\[
2 \cdot -8 + -5 \cdot 3 = -16 - 15 = -31
\]
Putting it all together, we have:
\[
(D P)^T = \begin{bmatrix} 4 & 34 \\ -3 & -31 \end{bmatrix}
\]
Thus, the final result is:
\[
\boxed{\begin{bmatrix} 4 & 34 \\ -3 & -31 \end{bmatrix}}
\]
Quick Answer
Calculate \( (D P)^T \) by taking the transpose of \( P \) and \( D \) separately, then multiplying them. The result is \( \begin{bmatrix} 4 & 34 \\ -3 & -31 \end{bmatrix} \).
Answered by UpStudy AI and reviewed by a Professional Tutor
UpStudy ThothAI
Self-Developed and Ever-Improving
Thoth AI product is constantly being upgraded and optimized.
Covers All Major Subjects
Capable of handling homework in math, chemistry, biology, physics, and more.
Instant and Accurate
Provides immediate and precise solutions and guidance.
Try Now
Ask Tutors
Ask AI
10x
Fastest way to Get Answers & Solutions
By text
Enter your question here…
By image
Re-Upload
Submit