- The scalar multiplication of a matrix \(R\) by a constant \(k\) is achieved by multiplying every element of \(R\) by \(k\).
- Applying this to \(R = \begin{bmatrix} 3 & 7 \ -2 & -11 \end{bmatrix}\) and \(k = 4\), we calculate each element: \(4 \times 3 = 12\), \(4 \times 7 = 28\), \(4 \times (-2) = -8\), and \(4 \times (-11) = -44\).
- This results in the matrix \(\begin{bmatrix} 12 & 28 \ -8 & -44 \end{bmatrix}\).
- Comparing with the given choices, the result is the fourth option: \(\boxed{\left[\begin{array}{cc}12 & 28 \ -8 & -44\end{array}\right]}\).
Explanation
- Problem Analysis and Definition of Scalar Multiplication
We are given the matrix \(R = \begin{bmatrix} 3 & 7 \ -2 & -11 \end{bmatrix}\) and we need to compute the product of this matrix by the scalar 4, denoted as \(4R\). The scalar multiplication of a matrix is defined by multiplying each individual element of the matrix by the given scalar value. Given that the matrix \(R\) is a \(2 \times 2\) matrix, its elements are organized in two rows and two columns. Specifically, the element in the first row and first column is 3, the element in the first row and second column is 7, the element in the second row and first column is -2, and the element in the second row and second column is -11.
- Execution of Matrix Multiplication
To calculate \(4R\), we apply the scalar 4 to each entry within the matrix:
\[4R = 4 \times \begin{bmatrix} 3 & 7 \ -2 & -11 \end{bmatrix} = \begin{bmatrix} 4 \times 3 & 4 \times 7 \ 4 \times (-2) & 4 \times (-11) \end{bmatrix}\]
Now, we perform the arithmetic for each position:
- For the first row, first column: \(4 \times 3 = 12\)
- For the first row, second column: \(4 \times 7 = 28\)
- For the second row, first column: \(4 \times (-2) = -8\)
- For the second row, second column: \(4 \times (-11) = -44\)
Combining these results, we get:
\[4R = \begin{bmatrix} 12 & 28 \ -8 & -44 \end{bmatrix}\]
- Comparison and Final Result
Comparing our calculated matrix \(\begin{bmatrix} 12 & 28 \ -8 & -44 \end{bmatrix}\) with the given options:
- Option 1: \(\begin{bmatrix} 7 & 11 \ 2 & -7 \end{bmatrix}\)
- Option 2: \(\begin{bmatrix} 12 & 7 \ -2 & -44 \end{bmatrix}\)
- Option 3: \(\begin{bmatrix} 7 & 28 \ 2 & -44 \end{bmatrix}\)
- Option 4: \(\begin{bmatrix} 12 & 28 \ -8 & -44 \end{bmatrix}\)
Our calculated result corresponds exactly to the fourth option. Therefore, \(4R = \begin{bmatrix} 12 & 28 \ -8 & -44 \end{bmatrix}\).
Examples
Scalar multiplication of matrices is widely used in computer graphics to scale or transform 2D objects. If you have a set of points defined as vectors (or a matrix of coordinates), multiplying this matrix by a scalar factor allows you to uniformly resize the shape. For example, if a small digital image is represented by a matrix of pixel color intensity values, multiplying the matrix by a scale factor could be used to adjust the overall brightness or contrast of the image. Understanding these basic operations is fundamental for any work involving data structures, image processing, or linear algebra in engineering.