If $$ \mathrm{A}=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right| $$ then \(\mathrm{A}^{2013}=\) \(\begin{array}{ll}\text { (a) } 3^{2013} \mathrm{~A} & \text { (b) }-3^{2012} \mathrm{I}\end{array}\) (c) \(3^{2012} \mathrm{~A}\) (d) \(3^{1006} \mathrm{~A}\)

Matrix multiplication is a fundamental operation in linear algebra where two matrices are multiplied to produce a third matrix. To multiply two matrices, you need to follow a specific set of rules.
For instance, consider two matrices, Matrix A and Matrix B. The entry in the i-th row and j-th column of the resulting Matrix AB, is calculated by taking the dot product of the i-th row of Matrix A and the j-th column of Matrix B. This means each element of AB can be calculated using the formula:
\( (AB)_{ij} = \text{sum of} (A_{ik} \times B_{kj}) \).
It's important that the number of columns in the first matrix should be equal to the number of rows in the second matrix for multiplication to be possible. In the exercise's context:
\[ A^2 = A \times A = \]
\[ \begin{bmatrix} 1&1&1\ 1&1&1\ 1&1&1\ \end{bmatrix} \begin{bmatrix} 1&1&1\ 1&1&1\ 1&1&1\ \end{bmatrix} \]
Performing this operation yields:
\[ A^2 = 3A \]
The resulting matrix is equivalent to Matrix A with each of its elements scaled by a factor of 3. This property simplifies the process when dealing with powers of matrices, as seen in the provided exercise.

When dealing with powers of matrices, such as \( A^n \), where 'n' is a positive integer, we are essentially multiplying matrix A by itself 'n-1' times.
In the example, we found that:
\[ A^2 = 3A \]\[ A^3 = A^2 \times A = 3A \times A = 9A \]\[ A^4 = A^3 \times A = 9A \times A = 27A \]
This pattern demonstrates that for any positive integer 'n', the power \( A^n \) will be equivalent to matrix A multiplied by 3 raised to the power of 'n-1'. So,
\[ A^n = 3^{(n-1)} \times A \]
This insight significantly reduces the complexity of calculating high powers of a matrix, as you needn't perform matrix multiplication repeatedly. Using this pattern, one can compute large powers like \( A^{2013} \) efficiently, as done in the solution.

The determinant of a matrix is a special scalar value that can be computed from its elements and provides important properties related to the matrix. For a 2x2 matrix, the determinant is calculated as:
\( \text{det} \begin{pmatrix} a & b \ c & d \end{pmatrix} = ad - bc \).
For larger square matrices, the determinant is calculated through more complex methods, such as expansion by minors or using the Leibniz formula. The determinant can tell us if the matrix is invertible, as a matrix with a determinant of zero does not have an inverse.
In the context of the exercise above, the determinant of a matrix is not directly involved in computing the powers of the matrix since we are dealing with the entire matrix rather than its determinant. However, understanding determinants is essential for other aspects of matrix algebra like matrix inversion and solving systems of linear equations. If the determinant of a matrix is zero, the matrix cannot be raised to negative powers, while a non-zero determinant indicates that it's theoretically possible to compute any integral power of the matrix.