If [tex]$A=\left[\begin{array}{rrr}1 & 5 & 3 \\ 2 & 4 & 0 \\ 3 & -1 & -5\end{array}\right]$[/tex] and [tex][tex]$B=\left[\begin{array}{rrr}2 & -1 & 0 \\ 0 & -2 & 5 \\ 1 & 2 & 0\end{array}\right]$[/tex][/tex], then find [tex]$3A - 4B^{\prime}$[/tex].



Answer :

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]