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Matrix multiplication is a fundamental operation where two matrices are multiplied together to produce a new matrix.
To perform matrix multiplication, each element of the resulting matrix is computed as the sum of products of elements from the rows of the first matrix and columns of the second matrix.
It's important to remember that matrix multiplication is not commutative; thus, the order in which you multiply the matrices matters. The formula for multiplying two matrices, \(A\) and \(B\), where \(A\) is of size \(m \times n\) and \(B\) is of size \(n \times p\), is given by:
\( (AB)_{ij} = \text{sum of } (A_{ik} \times B_{kj}) \text{ for } k=1 \text{ to } n \)

For example, multiplying \[ \begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix} \] and \[ \begin{bmatrix} 6 & 7 \ 8 & 9 \end{bmatrix} \] involves combining the elements as follows: \[ \begin{bmatrix} (2 \times 6 + 3 \times 8) & (2 \times 7 + 3 \times 9) \ (4 \times 6 + 5 \times 8) & (4 \times 7 + 5 \times 9) \end{bmatrix} = \begin{bmatrix} 36 & 41 \ 62 & 69 \end{bmatrix} \]
This showcases the complexity and structure that matrix multiplication brings to linear algebra.

Scalar multiplication involves multiplying each element of a matrix by a single number, known as a scalar.
This operation is straightforward but crucial in matrix algebra. If you have a matrix \(A\) and a scalar \(k\), the result of the scalar multiplication \(kA\) is obtained by multiplying every element of \(A\) by \(k\).
For instance, consider matrix \(A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}\) and scalar \(k = 3\). The resulting matrix would be:\[ 3A = \begin{bmatrix} 3 \times 1 & 3 \times 2 \ 3 \times 3 & 3 \times 4 \end{bmatrix} = \begin{bmatrix} 3 & 6 \ 9 & 12 \end{bmatrix} \]
This simplicity makes scalar multiplication a fundamental tool in various matrix equations and transformations.
In our exercise, we computed -2A and 4B by scalar multiplication. For \(A = \begin{bmatrix} -2 & 4 \ 0 & 3 \end{bmatrix}\) and scalar -2, the result is:\[ -2A = \begin{bmatrix} -2 \times -2 & 4 \times -2 \ 0 \times -2 & 3 \times -2 \end{bmatrix} = \begin{bmatrix} 4 & -8 \ 0 & -6 \end{bmatrix} \].
Similarly, for \(B = \begin{bmatrix} -6 & 2 \ 4 & 0 \end{bmatrix}\) and scalar 4:\[ 4B = \begin{bmatrix} 4 \times -6 & 4 \times 2 \ 4 \times 4 & 4 \times 0 \end{bmatrix} = \begin{bmatrix} -24 & 8 \ 16 & 0 \end{bmatrix} \].

Matrix addition is the process of adding two matrices by adding the corresponding entries together. To perform matrix addition, the two matrices must have the same dimensions.
If matrix \(A\) is of size \(m \times n\) and matrix \(B\) is also of size \(m \times n\), their sum matrix \(C = A + B\) will also be of size \(m \times n\). The \((i,j)\)-th entry of \(C\) is given by: \( C_{ij} = A_{ij} + B_{ij} \).
For example, adding matrices \[ \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \] and \[ \begin{bmatrix} 5 & 6 \ 7 & 8 \end{bmatrix} \] results in: \[ \begin{bmatrix} 1 + 5 & 2 + 6 \ 3 + 7 & 4 + 8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \ 10 & 12 \end{bmatrix} \].
Returning to our exercise, after multiplying the matrices by scalars, we added \( -2A \) and \( 4B \). The resulting matrices were:\[ \begin{bmatrix} 4 & -8 \ 0 & -6 \end{bmatrix} \] and \[ \begin{bmatrix} -24 & 8 \ 16 & 0 \end{bmatrix} \].
Adding these element-wise, we performed:\( -2A + 4B = \begin{bmatrix} 4 + (-24) & -8 + 8 \ 0 + 16 & -6 + 0 \end{bmatrix} = \begin{bmatrix} -20 & 0 \ 16 & -6 \end{bmatrix} \).
This final matrix is the solution to the problem.