Here's an elaborate, clear, and educational solution to the problem.

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Understanding the Key Concepts

This problem involves several fundamental operations with matrices, culminating in finding the adjoint of a matrix expression. Let's first recall the essential definitions and formulas:

1. Matrix Multiplication: If $P = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $Q = \begin{bmatrix} e & f \\ g & h \end{bmatrix}$, then their product $PQ$ is given by: $PQ = \begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix}$ Remember, matrix multiplication is generally not commutative ($PQ \neq QP$).

2. Scalar Multiplication of a Matrix: If $k$ is a scalar and $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then $kM$ is obtained by multiplying every element of $M$ by $k$: $kM = \begin{bmatrix} ka & kb \\ kc & kd \end{bmatrix}$

3. Matrix Addition: If $P = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $Q = \begin{bmatrix} e & f \\ g & h \end{bmatrix}$, then their sum $P+Q$ is obtained by adding corresponding elements: $P+Q = \begin{bmatrix} a+e & b+f \\ c+g & d+h \end{bmatrix}$ Matrices must be of the same order to be added.

4. Adjoint of a $2 \times 2$ Matrix: For a general $2 \times 2$ matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, its adjoint, denoted as adj($M$), is found by swapping the diagonal elements and changing the signs of the off-diagonal elements: $\text{adj}(M) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Our goal is to first calculate the matrix $3A^2 + 12A$ and then find its adjoint using these rules.

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Step-by-Step Solution

1. Given Matrix A: We are given the matrix $A$: $A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$

2. Calculate $A^2$ To find $3A^2 + 12A$, we first need to compute $A^2$. Why this step? Because $A^2$ is a fundamental component of the expression we need to evaluate. $A^2 = A \cdot A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$ Now, we perform matrix multiplication:

• First row, first column element: $(2)(2) + (-3)(-4) = 4 + 12 = 16$
• First row, second column element: $(2)(-3) + (-3)(1) = -6 - 3 = -9$
• Second row, first column element: $(-4)(2) + (1)(-4) = -8 - 4 = -12$
• Second row, second column element: $(-4)(-3) + (1)(1) = 12 + 1 = 13$

So, $A^2$ is: $A^2 = \begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix}$

3. Calculate $3A^2$ Next, we multiply $A^2$ by the scalar 3. Why this step? This is one of the terms in the target expression $3A^2 + 12A$. $3A^2 = 3 \begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix}$ Multiply each element of $A^2$ by 3:

• $3 \times 16 = 48$
• $3 \times (-9) = -27$
• $3 \times (-12) = -36$
• $3 \times 13 = 39$

So, $3A^2$ is: $3A^2 = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix}$

4. Calculate $12A$ Similarly, we multiply the original matrix $A$ by the scalar 12. Why this step? This is the other term in the expression $3A^2 + 12A$. $12A = 12 \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$ Multiply each element of $A$ by 12:

• $12 \times 2 = 24$
• $12 \times (-3) = -36$
• $12 \times (-4) = -48$
• $12 \times 1 = 12$

So, $12A$ is: $12A = \begin{bmatrix} 24 & -36 \\ -48 & 12 \end{bmatrix}$

5. Calculate $3A^2 + 12A$ Now we add the two matrices we just calculated, $3A^2$ and $12A$. Why this step? This completes the evaluation of the matrix expression whose adjoint we need to find. $3A^2 + 12A = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix} + \begin{bmatrix} 24 & -36 \\ -48 & 12 \end{bmatrix}$ Add corresponding elements:

• First row, first column: $48 + 24 = 72$
• First row, second column: $(-27) + (-36) = -27 - 36 = -63$
• Second row, first column: $(-36) + (-48) = -36 - 48 = -84$
• Second row, second column: $39 + 12 = 51$

Let $M = 3A^2 + 12A$. Then: $M = \begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$

6. Calculate adj($3A^2 + 12A$) Finally, we find the adjoint of the matrix $M = \begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$. Why this step? This is the final operation requested by the problem statement. Using the formula for the adjoint of a $2 \times 2$ matrix, $\text{adj}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$:

• Swap the diagonal elements: $72$ and $51$ become $51$ and $72$.
• Change the signs of the off-diagonal elements: $-63$ becomes $63$, and $-84$ becomes $84$.

So, adj($3A^2 + 12A$) is: $\text{adj}(M) = \begin{bmatrix} 51 & -(-63) \\ -(-84) & 72 \end{bmatrix} = \begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$

7. Compare with Options Comparing our result with the given options: (A) \left[ {\matrix{ {51} & {63} \cr {84} & {72} \cr } } \right] (B) \left[ {\matrix{ {51} & {84} \cr {63} & {72} \cr } } \right] (C) \left[ {\matrix{ {72} & {-63} \cr {-84} & {51} \cr } } \right] (D) \left[ {\matrix{ {72} & {-84} \cr {-63} & {51} \cr } } \right]

Our calculated adjoint matches option (A).

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Tips and Common Mistakes to Avoid

• Order of Operations: Always follow the standard order of operations (PEMDAS/BODMAS). Here, $A^2$ must be calculated before scalar multiplication or addition.
• Matrix Multiplication: Be extremely careful with the row-by-column multiplication. A common mistake is to multiply elements position-wise (like scalar multiplication), which is incorrect.
• Signs: Pay close attention to negative signs, especially during multiplication and when taking the adjoint. A single sign error can lead to a completely wrong answer.
• Adjoint Formula for $2 \times 2$: Remember that for $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. The diagonal elements are swapped, and the off-diagonal elements have their signs changed. Do not swap the off-diagonal elements as well, or you might end up with the transpose of the cofactor matrix.

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Summary and Key Takeaway

This problem is a comprehensive test of basic matrix operations: multiplication, scalar multiplication, and addition, followed by finding the adjoint of a $2 \times 2$ matrix. The solution requires careful step-by-step calculation and adherence to the definitions of each operation. Always break down complex matrix expressions into simpler, manageable steps to minimize errors.

The final answer is \boxed{\left[ {\matrix{ {51} & {63} \cr {84} & {72} \cr } } \right]}.