Here's a more elaborate, clear, and educational solution:

Key Concepts Used:

1. Scalar Matrix: A scalar matrix is a diagonal matrix where all the diagonal elements are equal. It can be written as $kI$, where $k$ is a scalar and $I$ is the identity matrix. For a 2x2 matrix, it looks like $\begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$.
2. Inverse of a 2x2 Matrix: For a matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, its determinant is $|M| = ad - bc$. If $|M| \neq 0$, its inverse is $M^{-1} = \frac{1}{|M|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
3. Properties of Determinants: For an $n \times n$ matrix $A$ and a scalar $k$, the determinant of $kA$ is given by $|kA| = k^n |A|$. For a 2x2 matrix, this means $|kA| = k^2 |A|$.

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Step-by-Step Solution:

1. Understanding the Given Information

We are given that $A \cdot \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ is a scalar matrix. Let $B = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$. Since $A \cdot B$ is a scalar matrix, it must be of the form $\begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix}$ for some scalar $\lambda$. This is because a scalar matrix has identical entries on its main diagonal and zeros elsewhere. So, we can write: $A \cdot \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \quad (*)$

2. Finding Matrix A

To find matrix $A$, we need to isolate it from equation $(*)$. We can do this by multiplying both sides of the equation by the inverse of matrix $B$, i.e., $B^{-1}$, from the right. $A = \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \cdot B^{-1}$

First, let's calculate the inverse of $B = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$. The determinant of $B$ is $|B| = (1)(3) - (2)(0) = 3 - 0 = 3$. Now, we find $B^{-1}$: $B^{-1} = \frac{1}{|B|} \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{3}{3} & \frac{-2}{3} \\ \frac{0}{3} & \frac{1}{3} \end{bmatrix} = \begin{bmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{bmatrix}$

Now, substitute $B^{-1}$ back into the expression for $A$: $A = \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \cdot \begin{bmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{bmatrix}$ Perform the matrix multiplication: $A = \begin{bmatrix} (\lambda)(1) + (0)(0) & (\lambda)(-\frac{2}{3}) + (0)(\frac{1}{3}) \\ (0)(1) + (\lambda)(0) & (0)(-\frac{2}{3}) + (\lambda)(\frac{1}{3}) \end{bmatrix}$ $A = \begin{bmatrix} \lambda & -\frac{2}{3}\lambda \\ 0 & \frac{1}{3}\lambda \end{bmatrix}$ This is our matrix $A$ in terms of $\lambda$.

3. Using the Determinant Condition to Find $\lambda$

We are given that $|3A| = 108$. Since $A$ is a 2x2 matrix, we use the property $|kA| = k^n |A|$ where $n=2$. So, $|3A| = 3^2 |A| = 9|A|$. Therefore, $9|A| = 108$. Dividing by 9, we get $|A| = \frac{108}{9} = 12$.

Now, let's calculate the determinant of our matrix $A = \begin{bmatrix} \lambda & -\frac{2}{3}\lambda \\ 0 & \frac{1}{3}\lambda \end{bmatrix}$: $|A| = (\lambda)\left(\frac{1}{3}\lambda\right) - \left(-\frac{2}{3}\lambda\right)(0)$ $|A| = \frac{1}{3}\lambda^2 - 0$ $|A| = \frac{1}{3}\lambda^2$

Equating this to the value we found for $|A|$: $\frac{1}{3}\lambda^2 = 12$ Multiply both sides by 3: $\lambda^2 = 36$ Taking the square root: $\lambda = \pm 6$ We have two possible values for $\lambda$. We must check both.

4. Calculating $A^2$ for each value of $\lambda$

Case 1: $\lambda = 6$ Substitute $\lambda = 6$ into the expression for $A$: $A = \begin{bmatrix} 6 & -\frac{2}{3}(6) \\ 0 & \frac{1}{3}(6) \end{bmatrix} = \begin{bmatrix} 6 & -4 \\ 0 & 2 \end{bmatrix}$ Now, calculate $A^2 = A \cdot A$: $A^2 = \begin{bmatrix} 6 & -4 \\ 0 & 2 \end{bmatrix} \cdot \begin{bmatrix} 6 & -4 \\ 0 & 2 \end{bmatrix}$ $A^2 = \begin{bmatrix} (6)(6) + (-4)(0) & (6)(-4) + (-4)(2) \\ (0)(6) + (2)(0) & (0)(-4) + (2)(2) \end{bmatrix}$ $A^2 = \begin{bmatrix} 36 + 0 & -24 - 8 \\ 0 + 0 & 0 + 4 \end{bmatrix}$ $A^2 = \begin{bmatrix} 36 & -32 \\ 0 & 4 \end{bmatrix}$

Case 2: $\lambda = -6$ Substitute $\lambda = -6$ into the expression for $A$: $A = \begin{bmatrix} -6 & -\frac{2}{3}(-6) \\ 0 & \frac{1}{3}(-6) \end{bmatrix} = \begin{bmatrix} -6 & 4 \\ 0 & -2 \end{bmatrix}$ Now, calculate $A^2 = A \cdot A$: $A^2 = \begin{bmatrix} -6 & 4 \\ 0 & -2 \end{bmatrix} \cdot \begin{bmatrix} -6 & 4 \\ 0 & -2 \end{bmatrix}$ $A^2 = \begin{bmatrix} (-6)(-6) + (4)(0) & (-6)(4) + (4)(-2) \\ (0)(-6) + (-2)(0) & (0)(4) + (-2)(-2) \end{bmatrix}$ $A^2 = \begin{bmatrix} 36 + 0 & -24 - 8 \\ 0 + 0 & 0 + 4 \end{bmatrix}$ $A^2 = \begin{bmatrix} 36 & -32 \\ 0 & 4 \end{bmatrix}$

Both values of $\lambda$ yield the same $A^2$.

5. Final Answer

The matrix $A^2$ is $\begin{bmatrix} 36 & -32 \\ 0 & 4 \end{bmatrix}$.

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Tips and Common Mistakes:

• Understanding Matrix Types: Be clear on the definitions of diagonal, scalar, and identity matrices. A scalar matrix is a special case of a diagonal matrix, and an identity matrix is a special case of a scalar matrix (where $\lambda=1$).
• Matrix Inverse Formula: Memorize the formula for the inverse of a 2x2 matrix and be careful with the signs.
• Determinant Property $|kA|$: This is a crucial property for competitive exams. Remember that $|kA| = k^n |A|$ where $n$ is the order of the matrix. A common mistake is to write $|kA| = k|A|$.
• Solving for $\lambda$: When you have $\lambda^2 = \text{constant}$, always remember to consider both positive and negative roots (e.g., $\lambda = \pm 6$). Sometimes, both roots lead to the same final answer, but not always.
• Matrix Multiplication: Pay close attention to the order of multiplication ($AB \neq BA$ in general) and the dot product calculation for each element.

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Summary/Key Takeaway: This problem effectively tests your understanding of matrix definitions (scalar matrix), matrix operations (multiplication, inverse), and determinant properties. The key steps involved setting up the matrix equation, finding the inverse of a matrix, applying the determinant property $|kA|=k^n|A|$, solving for the unknown scalar, and finally performing matrix multiplication to find $A^2$. It's important to be thorough and check all possible solutions for the unknown scalar.

The final answer is (A) \left[ {\matrix{ 36 & { - 32} \cr 0 & {4} \cr } } \right].