Answer :
Certainly! Let's go through the steps methodically to prove that [tex]$(AB)^T=B^T A^T$[/tex] for the given matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
### Step 1: Calculate the matrix product [tex]\(AB\)[/tex]
First, we multiply matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
Given:
[tex]\[ A = \begin{pmatrix} 2 & 4 \\ 3 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 3 \\ 0 & 4 \end{pmatrix} \][/tex]
[tex]\[ AB = \begin{pmatrix} 2 & 4 \\ 3 & 1 \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ 0 & 4 \end{pmatrix} \][/tex]
Carrying out the multiplication element-wise:
[tex]\[ AB = \begin{pmatrix} (2 \cdot 2 + 4 \cdot 0) & (2 \cdot 3 + 4 \cdot 4) \\ (3 \cdot 2 + 1 \cdot 0) & (3 \cdot 3 + 1 \cdot 4) \end{pmatrix} = \begin{pmatrix} 4 & 22 \\ 6 & 13 \end{pmatrix} \][/tex]
Thus,
[tex]\[ AB = \begin{pmatrix} 4 & 22 \\ 6 & 13 \end{pmatrix} \][/tex]
### Step 2: Transpose the product [tex]\(AB\)[/tex]
Next, we find the transpose of the matrix [tex]\(AB\)[/tex]:
[tex]\[ (AB)^T = \begin{pmatrix} 4 & 22 \\ 6 & 13 \end{pmatrix}^T = \begin{pmatrix} 4 & 6 \\ 22 & 13 \end{pmatrix} \][/tex]
### Step 3: Transpose matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]
Now we calculate the transposes of matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A^T = \begin{pmatrix} 2 & 4 \\ 3 & 1 \end{pmatrix}^T = \begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} \][/tex]
[tex]\[ B^T = \begin{pmatrix} 2 & 3 \\ 0 & 4 \end{pmatrix}^T = \begin{pmatrix} 2 & 0 \\ 3 & 4 \end{pmatrix} \][/tex]
### Step 4: Calculate the product [tex]\(B^T A^T\)[/tex]
Finally, we multiply the transposed matrices [tex]\(B^T\)[/tex] and [tex]\(A^T\)[/tex]:
[tex]\[ B^T A^T = \begin{pmatrix} 2 & 0 \\ 3 & 4 \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} \][/tex]
Carrying out the multiplication element-wise:
[tex]\[ B^T A^T = \begin{pmatrix} (2 \cdot 2 + 0 \cdot 4) & (2 \cdot 3 + 0 \cdot 1) \\ (3 \cdot 2 + 4 \cdot 4) & (3 \cdot 3 + 4 \cdot 1) \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 22 & 13 \end{pmatrix} \][/tex]
Thus,
[tex]\[ B^T A^T = \begin{pmatrix} 4 & 6 \\ 22 & 13 \end{pmatrix} \][/tex]
### Step 5: Conclude by comparing the results
We see that:
[tex]\[ (AB)^T = \begin{pmatrix} 4 & 6 \\ 22 & 13 \end{pmatrix} \][/tex]
and
[tex]\[ B^T A^T = \begin{pmatrix} 4 & 6 \\ 22 & 13 \end{pmatrix} \][/tex]
Therefore,
[tex]\[ (AB)^T = B^T A^T \][/tex]
We have shown that for the given matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex], the equality [tex]\((AB)^T = B^T A^T\)[/tex] holds true.
### Step 1: Calculate the matrix product [tex]\(AB\)[/tex]
First, we multiply matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
Given:
[tex]\[ A = \begin{pmatrix} 2 & 4 \\ 3 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 3 \\ 0 & 4 \end{pmatrix} \][/tex]
[tex]\[ AB = \begin{pmatrix} 2 & 4 \\ 3 & 1 \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ 0 & 4 \end{pmatrix} \][/tex]
Carrying out the multiplication element-wise:
[tex]\[ AB = \begin{pmatrix} (2 \cdot 2 + 4 \cdot 0) & (2 \cdot 3 + 4 \cdot 4) \\ (3 \cdot 2 + 1 \cdot 0) & (3 \cdot 3 + 1 \cdot 4) \end{pmatrix} = \begin{pmatrix} 4 & 22 \\ 6 & 13 \end{pmatrix} \][/tex]
Thus,
[tex]\[ AB = \begin{pmatrix} 4 & 22 \\ 6 & 13 \end{pmatrix} \][/tex]
### Step 2: Transpose the product [tex]\(AB\)[/tex]
Next, we find the transpose of the matrix [tex]\(AB\)[/tex]:
[tex]\[ (AB)^T = \begin{pmatrix} 4 & 22 \\ 6 & 13 \end{pmatrix}^T = \begin{pmatrix} 4 & 6 \\ 22 & 13 \end{pmatrix} \][/tex]
### Step 3: Transpose matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]
Now we calculate the transposes of matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A^T = \begin{pmatrix} 2 & 4 \\ 3 & 1 \end{pmatrix}^T = \begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} \][/tex]
[tex]\[ B^T = \begin{pmatrix} 2 & 3 \\ 0 & 4 \end{pmatrix}^T = \begin{pmatrix} 2 & 0 \\ 3 & 4 \end{pmatrix} \][/tex]
### Step 4: Calculate the product [tex]\(B^T A^T\)[/tex]
Finally, we multiply the transposed matrices [tex]\(B^T\)[/tex] and [tex]\(A^T\)[/tex]:
[tex]\[ B^T A^T = \begin{pmatrix} 2 & 0 \\ 3 & 4 \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} \][/tex]
Carrying out the multiplication element-wise:
[tex]\[ B^T A^T = \begin{pmatrix} (2 \cdot 2 + 0 \cdot 4) & (2 \cdot 3 + 0 \cdot 1) \\ (3 \cdot 2 + 4 \cdot 4) & (3 \cdot 3 + 4 \cdot 1) \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 22 & 13 \end{pmatrix} \][/tex]
Thus,
[tex]\[ B^T A^T = \begin{pmatrix} 4 & 6 \\ 22 & 13 \end{pmatrix} \][/tex]
### Step 5: Conclude by comparing the results
We see that:
[tex]\[ (AB)^T = \begin{pmatrix} 4 & 6 \\ 22 & 13 \end{pmatrix} \][/tex]
and
[tex]\[ B^T A^T = \begin{pmatrix} 4 & 6 \\ 22 & 13 \end{pmatrix} \][/tex]
Therefore,
[tex]\[ (AB)^T = B^T A^T \][/tex]
We have shown that for the given matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex], the equality [tex]\((AB)^T = B^T A^T\)[/tex] holds true.
