if `A=[[2,-3],[-4,1]]` then `(3A^2+12A)=?`

Options

A. `[(72,-63),(-84,51)]`
B. `[(72, -84),(-63,51)]`
C. `[(51,63),(84,72)]`
D. `[(51, 84),(63,72)]`

Explanation

To solve the problem \(3A^2 + 12A\) where \(A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}\), we will follow these steps:

### Step 1: Calculate \(A^2\)

To find \(A^2\), we need to multiply matrix \(A\) by itself.

\[ A^2 = A \times A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} \]

Calculating the elements of \(A^2\):

- First row, first column:
\[ (2 \times 2) + (-3 \times -4) = 4 + 12 = 16 \]

- First row, second column:
\[ (2 \times -3) + (-3 \times 1) = -6 - 3 = -9 \]

- Second row, first column:
\[ (-4 \times 2) + (1 \times -4) = -8 - 4 = -12 \]

- Second row, second column:
\[ (-4 \times -3) + (1 \times 1) = 12 + 1 = 13 \]

Thus, we have:
\[ A^2 = \begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix} \]

### Step 2: Calculate \(3A^2\)

Now we will multiply \(A^2\) by 3.

\[ 3A^2 = 3 \times \begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix} = \begin{bmatrix} 3 \times 16 & 3 \times -9 \\ 3 \times -12 & 3 \times 13 \end{bmatrix} = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix} \]

### Step 3: Calculate \(12A\)

Next, we will multiply matrix \(A\) by 12.

\[ 12A = 12 \times \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} = \begin{bmatrix} 12 \times 2 & 12 \times -3 \\ 12 \times -4 & 12 \times 1 \end{bmatrix} = \begin{bmatrix} 24 & -36 \\ -48 & 12 \end{bmatrix} \]

### Step 4: Add \(3A^2\) and \(12A\)

Now we will add \(3A^2\) and \(12A\).

\[ 3A^2 + 12A = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix} + \begin{bmatrix} 24 & -36 \\ -48 & 12 \end{bmatrix} \]

Calculating the elements:

- First row, first column:
\[ 48 + 24 = 72 \]

- First row, second column:
\[ -27 + (-36) = -63 \]

- Second row, first column:
\[ -36 + (-48) = -84 \]

- Second row, second column:
\[ 39 + 12 = 51 \]

Thus, we have:
\[ 3A^2 + 12A = \begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix} \]

### Final Answer

The final result is:
\[ 3A^2 + 12A = \begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix} \]

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