To solve the problem of finding [tex]\(3A - 4B^{\prime}\)[/tex], we need to follow these steps:
1. Compute [tex]\(3A\)[/tex]: This involves multiplying each element of matrix [tex]\(A\)[/tex] by 3.
2. Compute [tex]\(B^{\prime}\)[/tex]: This is the transpose of matrix [tex]\(B\)[/tex], where we switch the rows and columns of [tex]\(B\)[/tex].
3. Compute [tex]\(4B^{\prime}\)[/tex]: This involves multiplying each element of the transposed matrix [tex]\(B^{\prime}\)[/tex] by 4.
4. Compute [tex]\(3A - 4B^{\prime}\)[/tex]: This is done by subtracting the corresponding elements of [tex]\(4B^{\prime}\)[/tex] from [tex]\(3A\)[/tex].
Let's go through each of these steps in detail:
### Step 1: Compute [tex]\(3A\)[/tex]
Matrix [tex]\(A\)[/tex] is:
[tex]\[
A = \begin{pmatrix}
1 & 5 & 3 \\
2 & 4 & 0 \\
3 & -1 & -5
\end{pmatrix}
\][/tex]
Multiplying each element of [tex]\(A\)[/tex] by 3 gives:
[tex]\[
3A = 3 \times \begin{pmatrix}
1 & 5 & 3 \\
2 & 4 & 0 \\
3 & -1 & -5
\end{pmatrix} = \begin{pmatrix}
3 & 15 & 9 \\
6 & 12 & 0 \\
9 & -3 & -15
\end{pmatrix}
\][/tex]
### Step 2: Compute the transpose of [tex]\(B\)[/tex]
Matrix [tex]\(B\)[/tex] is:
[tex]\[
B = \begin{pmatrix}
2 & -1 & 0 \\
0 & -2 & 5 \\
1 & 2 & 0
\end{pmatrix}
\][/tex]
The transpose of [tex]\(B\)[/tex], denoted as [tex]\(B^{\prime}\)[/tex], is:
[tex]\[
B^{\prime} = \begin{pmatrix}
2 & 0 & 1 \\
-1 & -2 & 2 \\
0 & 5 & 0
\end{pmatrix}
\][/tex]
### Step 3: Compute [tex]\(4B^{\prime}\)[/tex]
Multiplying each element of [tex]\(B^{\prime}\)[/tex] by 4 gives:
[tex]\[
4B^{\prime} = 4 \times \begin{pmatrix}
2 & 0 & 1 \\
-1 & -2 & 2 \\
0 & 5 & 0
\end{pmatrix} = \begin{pmatrix}
8 & 0 & 4 \\
-4 & -8 & 8 \\
0 & 20 & 0
\end{pmatrix}
\][/tex]
### Step 4: Compute [tex]\(3A - 4B^{\prime}\)[/tex]
Subtracting the corresponding elements of [tex]\(4B^{\prime}\)[/tex] from [tex]\(3A\)[/tex] gives:
[tex]\[
3A - 4B^{\prime} = \begin{pmatrix}
3 & 15 & 9 \\
6 & 12 & 0 \\
9 & -3 & -15
\end{pmatrix} - \begin{pmatrix}
8 & 0 & 4 \\
-4 & -8 & 8 \\
0 & 20 & 0
\end{pmatrix} = \begin{pmatrix}
3 - 8 & 15 - 0 & 9 - 4 \\
6 - (-4) & 12 - (-8) & 0 - 8 \\
9 - 0 & -3 - 20 & -15 - 0
\end{pmatrix}
\][/tex]
Simplifying this, we get:
[tex]\[
3A - 4B^{\prime} = \begin{pmatrix}
-5 & 15 & 5 \\
10 & 20 & -8 \\
9 & -23 & -15
\end{pmatrix}
\][/tex]
Hence, the final result is:
[tex]\[
3A - 4B^{\prime} = \begin{pmatrix}
-5 & 15 & 5 \\
10 & 20 & -8 \\
9 & -23 & -15
\end{pmatrix}
\][/tex]
