If `A=[(0,1,2),(2,3,4),(4,5,6)],B=[(1,0,0),(0,1,0),(0,0,01)]` then `3A-4B=`

Options

A. `[(-4,3,6),(6,5,12),(12,15,14)]`
B. `[(0,1,2),(2,3,4),(4,5,6)]`
C. `[(-4,6,3),(6,5,12),(12,15,14)]`
D. `[(-4,3,6),(6,15,12),(12,5,14)]`

Answer

To solve the problem \(3A - 4B\) where \(A = \begin{pmatrix} 0 & 1 & 2 \\ 2 & 3 & 4 \\ 4 & 5 & 6 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\), we will follow these steps:

### Step 1: Calculate \(3A\)

To find \(3A\), we multiply each element of matrix \(A\) by 3.

\[ A = \begin{pmatrix} 0 & 1 & 2 \\ 2 & 3 & 4 \\ 4 & 5 & 6 \end{pmatrix} \]

Calculating \(3A\):

\[ 3A = 3 \times \begin{pmatrix} 0 & 1 & 2 \\ 2 & 3 & 4 \\ 4 & 5 & 6 \end{pmatrix} = \begin{pmatrix} 3 \times 0 & 3 \times 1 & 3 \times 2 \\ 3 \times 2 & 3 \times 3 & 3 \times 4 \\ 3 \times 4 & 3 \times 5 & 3 \times 6 \end{pmatrix} \]

Calculating each element:

\[ 3A = \begin{pmatrix} 0 & 3 & 6 \\ 6 & 9 & 12 \\ 12 & 15 & 18 \end{pmatrix} \]

### Step 2: Calculate \(4B\)

Next, we calculate \(4B\) by multiplying each element of matrix \(B\) by 4.

\[ B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating \(4B\):

\[ 4B = 4 \times \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 \times 1 & 4 \times 0 & 4 \times 0 \\ 4 \times 0 & 4 \times 1 & 4 \times 0 \\ 4 \times 0 & 4 \times 0 & 4 \times 1 \end{pmatrix} \]

Calculating each element:

\[ 4B = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} \]

### Step 3: Calculate \(3A - 4B\)

Now we subtract \(4B\) from \(3A\):

\[ 3A - 4B = \begin{pmatrix} 0 & 3 & 6 \\ 6 & 9 & 12 \\ 12 & 15 & 18 \end{pmatrix} - \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} \]

Subtracting corresponding elements:

\[ = \begin{pmatrix} 0 - 4 & 3 - 0 & 6 - 0 \\ 6 - 0 & 9 - 4 & 12 - 0 \\ 12 - 0 & 15 - 0 & 18 - 4 \end{pmatrix} \]

Calculating each element:

\[ = \begin{pmatrix} -4 & 3 & 6 \\ 6 & 5 & 12 \\ 12 & 15 & 14 \end{pmatrix} \]

### Final Answer

Thus, the result of \(3A - 4B\) is:

\[ \begin{pmatrix} -4 & 3 & 6 \\ 6 & 5 & 12 \\ 12 & 15 & 14 \end{pmatrix} \]