If `A=|{:(,2,-3),(,-4,1):}|` then adj `(3A^(2)+12A)` is equal to

Options

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

Answer

To solve the problem, we need to find the adjoint of the matrix \(3A^2 + 12A\) where \(A = \begin{pmatrix} 2 & -3 \\ -4 & 1 \end{pmatrix}\).

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

First, we calculate \(A^2\):
\[ A^2 = A \cdot A = \begin{pmatrix} 2 & -3 \\ -4 & 1 \end{pmatrix} \cdot \begin{pmatrix} 2 & -3 \\ -4 & 1 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \(2 \cdot 2 + (-3) \cdot (-4) = 4 + 12 = 16\)
- First row, second column: \(2 \cdot (-3) + (-3) \cdot 1 = -6 - 3 = -9\)
- Second row, first column: \(-4 \cdot 2 + 1 \cdot (-4) = -8 - 4 = -12\)
- Second row, second column: \(-4 \cdot (-3) + 1 \cdot 1 = 12 + 1 = 13\)

Thus,
\[ A^2 = \begin{pmatrix} 16 & -9 \\ -12 & 13 \end{pmatrix} \]

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

Next, we calculate \(3A^2\):
\[ 3A^2 = 3 \cdot \begin{pmatrix} 16 & -9 \\ -12 & 13 \end{pmatrix} = \begin{pmatrix} 48 & -27 \\ -36 & 39 \end{pmatrix} \]

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

Now, we calculate \(12A\):
\[ 12A = 12 \cdot \begin{pmatrix} 2 & -3 \\ -4 & 1 \end{pmatrix} = \begin{pmatrix} 24 & -36 \\ -48 & 12 \end{pmatrix} \]

### Step 4: Calculate \(3A^2 + 12A\)

Now, we add \(3A^2\) and \(12A\):
\[ 3A^2 + 12A = \begin{pmatrix} 48 & -27 \\ -36 & 39 \end{pmatrix} + \begin{pmatrix} 24 & -36 \\ -48 & 12 \end{pmatrix} \]
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,
\[ 3A^2 + 12A = \begin{pmatrix} 72 & -63 \\ -84 & 51 \end{pmatrix} \]

### Step 5: Calculate the Adjoint

To find the adjoint of the matrix \(B = \begin{pmatrix} 72 & -63 \\ -84 & 51 \end{pmatrix}\), we first find the minors and then the cofactors.

**Minors:**
- \(M_{11} = 51\)
- \(M_{12} = -84\)
- \(M_{21} = -63\)
- \(M_{22} = 72\)

**Cofactors:**
- \(C_{11} = (-1)^{1+1} M_{11} = 51\)
- \(C_{12} = (-1)^{1+2} M_{12} = 84\)
- \(C_{21} = (-1)^{2+1} M_{21} = 63\)
- \(C_{22} = (-1)^{2+2} M_{22} = 72\)

Thus, the cofactor matrix is:
\[ \begin{pmatrix} 51 & 84 \\ 63 & 72 \end{pmatrix} \]

Now, we take the transpose of the cofactor matrix to find the adjoint:
\[ \text{adj}(B) = \begin{pmatrix} 51 & 63 \\ 84 & 72 \end{pmatrix} \]

### Final Result

The adjoint of \(3A^2 + 12A\) is:
\[ \text{adj}(3A^2 + 12A) = \begin{pmatrix} 51 & 63 \\ 84 & 72 \end{pmatrix} \]