Answer :
Let's address the question step-by-step and find all the requested components.
### Step 1: Find the Transposes of [tex]\(X\)[/tex] and [tex]\(Y\)[/tex]
Given matrices:
[tex]\[ X = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} \][/tex]
[tex]\[ Y = \begin{pmatrix} 5 & 0 \\ -2 & 1 \end{pmatrix} \][/tex]
Transpose of [tex]\(X\)[/tex]:
[tex]\[ X^T = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} \][/tex]
Transpose of [tex]\(Y\)[/tex]:
[tex]\[ Y^T = \begin{pmatrix} 5 & -2 \\ 0 & 1 \end{pmatrix} \][/tex]
### Step 2: Find the Product [tex]\(XY\)[/tex]
To find [tex]\(XY\)[/tex], we perform matrix multiplication:
[tex]\[ XY = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ -2 & 1 \end{pmatrix} \][/tex]
Calculating each element:
- First row, first column: [tex]\( 2 \times 5 + (-1) \times (-2) = 10 + 2 = 12 \)[/tex]
- First row, second column: [tex]\( 2 \times 0 + (-1) \times 1 = 0 - 1 = -1 \)[/tex]
- Second row, first column: [tex]\( 1 \times 5 + 3 \times (-2) = 5 - 6 = -1 \)[/tex]
- Second row, second column: [tex]\( 1 \times 0 + 3 \times 1 = 0 + 3 = 3 \)[/tex]
Thus,
[tex]\[ XY = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix} \][/tex]
### Step 3: Find the Transpose of the Product [tex]\(XY\)[/tex]
Transpose of [tex]\(XY\)[/tex]:
[tex]\[ (XY)^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix}^T \][/tex]
[tex]\[ (XY)^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix} \][/tex]
### Step 4: Show That [tex]\((2X)^T = 2(X^T)\)[/tex]
First, calculate [tex]\(2X\)[/tex]:
[tex]\[ 2X = 2 \times \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 4 & -2 \\ 2 & 6 \end{pmatrix} \][/tex]
Now, compute the transpose of [tex]\(2X\)[/tex]:
[tex]\[ (2X)^T = \begin{pmatrix} 4 & 2 \\ -2 & 6 \end{pmatrix} \][/tex]
Next, compute [tex]\(2(X^T)\)[/tex]:
[tex]\[ 2(X^T) = 2 \times \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ -2 & 6 \end{pmatrix} \][/tex]
We see that:
[tex]\[ (2X)^T = 2(X^T) = \begin{pmatrix} 4 & 2 \\ -2 & 6 \end{pmatrix} \][/tex]
### Step 5: Show That [tex]\((XY)^T = Y^T \cdot X^T\)[/tex]
Now, we need to verify the transpose of the product rule:
[tex]\[ (XY)^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix} \][/tex]
Let's compute the product [tex]\(Y^T X^T\)[/tex]:
[tex]\[ Y^T \cdot X^T = \begin{pmatrix} 5 & -2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} \][/tex]
Calculating each element:
- First row, first column: [tex]\( 5 \times 2 + (-2) \times (-1) = 10 + 2 = 12 \)[/tex]
- First row, second column: [tex]\( 5 \times 1 + (-2) \times 3 = 5 - 6 = -1 \)[/tex]
- Second row, first column: [tex]\( 0 \times 2 + 1 \times (-1) = 0 - 1 = -1 \)[/tex]
- Second row, second column: [tex]\( 0 \times 1 + 1 \times 3 = 0 + 3 = 3 \)[/tex]
Thus,
[tex]\[ Y^T \cdot X^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix} \][/tex]
We see that:
[tex]\[ (XY)^T = Y^T \cdot X^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix} \][/tex]
### Summary
We have verified all given relationships and found the necessary transposes and products:
- [tex]\(X^T = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}\)[/tex]
- [tex]\(Y^T = \begin{pmatrix} 5 & -2 \\ 0 & 1 \end{pmatrix}\)[/tex]
- [tex]\((XY)^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix}\)[/tex]
- [tex]\((2X)^T = 2(X^T) = \begin{pmatrix} 4 & 2 \\ -2 & 6 \end{pmatrix}\)[/tex]
- [tex]\((XY)^T = Y^T \cdot X^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix}\)[/tex]
### Step 1: Find the Transposes of [tex]\(X\)[/tex] and [tex]\(Y\)[/tex]
Given matrices:
[tex]\[ X = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} \][/tex]
[tex]\[ Y = \begin{pmatrix} 5 & 0 \\ -2 & 1 \end{pmatrix} \][/tex]
Transpose of [tex]\(X\)[/tex]:
[tex]\[ X^T = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} \][/tex]
Transpose of [tex]\(Y\)[/tex]:
[tex]\[ Y^T = \begin{pmatrix} 5 & -2 \\ 0 & 1 \end{pmatrix} \][/tex]
### Step 2: Find the Product [tex]\(XY\)[/tex]
To find [tex]\(XY\)[/tex], we perform matrix multiplication:
[tex]\[ XY = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ -2 & 1 \end{pmatrix} \][/tex]
Calculating each element:
- First row, first column: [tex]\( 2 \times 5 + (-1) \times (-2) = 10 + 2 = 12 \)[/tex]
- First row, second column: [tex]\( 2 \times 0 + (-1) \times 1 = 0 - 1 = -1 \)[/tex]
- Second row, first column: [tex]\( 1 \times 5 + 3 \times (-2) = 5 - 6 = -1 \)[/tex]
- Second row, second column: [tex]\( 1 \times 0 + 3 \times 1 = 0 + 3 = 3 \)[/tex]
Thus,
[tex]\[ XY = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix} \][/tex]
### Step 3: Find the Transpose of the Product [tex]\(XY\)[/tex]
Transpose of [tex]\(XY\)[/tex]:
[tex]\[ (XY)^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix}^T \][/tex]
[tex]\[ (XY)^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix} \][/tex]
### Step 4: Show That [tex]\((2X)^T = 2(X^T)\)[/tex]
First, calculate [tex]\(2X\)[/tex]:
[tex]\[ 2X = 2 \times \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 4 & -2 \\ 2 & 6 \end{pmatrix} \][/tex]
Now, compute the transpose of [tex]\(2X\)[/tex]:
[tex]\[ (2X)^T = \begin{pmatrix} 4 & 2 \\ -2 & 6 \end{pmatrix} \][/tex]
Next, compute [tex]\(2(X^T)\)[/tex]:
[tex]\[ 2(X^T) = 2 \times \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ -2 & 6 \end{pmatrix} \][/tex]
We see that:
[tex]\[ (2X)^T = 2(X^T) = \begin{pmatrix} 4 & 2 \\ -2 & 6 \end{pmatrix} \][/tex]
### Step 5: Show That [tex]\((XY)^T = Y^T \cdot X^T\)[/tex]
Now, we need to verify the transpose of the product rule:
[tex]\[ (XY)^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix} \][/tex]
Let's compute the product [tex]\(Y^T X^T\)[/tex]:
[tex]\[ Y^T \cdot X^T = \begin{pmatrix} 5 & -2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} \][/tex]
Calculating each element:
- First row, first column: [tex]\( 5 \times 2 + (-2) \times (-1) = 10 + 2 = 12 \)[/tex]
- First row, second column: [tex]\( 5 \times 1 + (-2) \times 3 = 5 - 6 = -1 \)[/tex]
- Second row, first column: [tex]\( 0 \times 2 + 1 \times (-1) = 0 - 1 = -1 \)[/tex]
- Second row, second column: [tex]\( 0 \times 1 + 1 \times 3 = 0 + 3 = 3 \)[/tex]
Thus,
[tex]\[ Y^T \cdot X^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix} \][/tex]
We see that:
[tex]\[ (XY)^T = Y^T \cdot X^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix} \][/tex]
### Summary
We have verified all given relationships and found the necessary transposes and products:
- [tex]\(X^T = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}\)[/tex]
- [tex]\(Y^T = \begin{pmatrix} 5 & -2 \\ 0 & 1 \end{pmatrix}\)[/tex]
- [tex]\((XY)^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix}\)[/tex]
- [tex]\((2X)^T = 2(X^T) = \begin{pmatrix} 4 & 2 \\ -2 & 6 \end{pmatrix}\)[/tex]
- [tex]\((XY)^T = Y^T \cdot X^T = \begin{pmatrix} 12 & -1 \\ -1 & 3 \end{pmatrix}\)[/tex]
