Problem 10 - Algebra for College Students

Question

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ 3 A-4 E $$

Step-by-Step Solution

Answer

The resulting matrix is \(\begin{bmatrix} -5 & -26 \\ -19 & 0 \end{bmatrix}.\)

Step 1: Scalar Multiplication of Matrix A

First, multiply each element of matrix \(A\) by 3. Given \(A = \begin{bmatrix} 1 & -2 \ 3 & 4 \end{bmatrix}\), the result is: \[3A = 3 \times \begin{bmatrix} 1 & -2 \ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 \times 1 & 3 \times -2 \ 3 \times 3 & 3 \times 4 \end{bmatrix} = \begin{bmatrix} 3 & -6 \ 9 & 12 \end{bmatrix}.\]

Step 2: Scalar Multiplication of Matrix E

Next, multiply each element of matrix \(E\) by 4. Given \(E = \begin{bmatrix} 2 & 5 \ 7 & 3 \end{bmatrix}\), the result is: \[4E = 4 \times \begin{bmatrix} 2 & 5 \ 7 & 3 \end{bmatrix} = \begin{bmatrix} 4 \times 2 & 4 \times 5 \ 4 \times 7 & 4 \times 3 \end{bmatrix} = \begin{bmatrix} 8 & 20 \ 28 & 12 \end{bmatrix}.\]

Step 3: Subtract Matrix 4E from Matrix 3A

Now, subtract the resulting matrix from Step 2 from the resulting matrix from Step 1: \[3A - 4E = \begin{bmatrix} 3 & -6 \ 9 & 12 \end{bmatrix} - \begin{bmatrix} 8 & 20 \ 28 & 12 \end{bmatrix} = \begin{bmatrix} 3 - 8 & -6 - 20 \ 9 - 28 & 12 - 12 \end{bmatrix} = \begin{bmatrix} -5 & -26 \ -19 & 0 \end{bmatrix}.\]

Key Concepts

Scalar Multiplication

In matrix algebra, scalar multiplication is a simple yet fundamental operation. It involves multiplying each element of a matrix by a scalar, which is just a fancy word for a real number. In the exercise provided, the matrix \(A\) needed to be scaled up by a factor of 3. This meant that every element in matrix \(A\) was multiplied by 3.

Let's illustrate this with matrix \(A\), which is given as: \[A = \begin{bmatrix} 1 & -2 \ 3 & 4 \end{bmatrix}\] Multiplying each element by 3 yields: \[3A = \begin{bmatrix} 3 \times 1 & 3 \times -2 \ 3 \times 3 & 3 \times 4 \end{bmatrix} = \begin{bmatrix} 3 & -6 \ 9 & 12 \end{bmatrix}\]

Notice how each element in the new matrix, \(3A\), is simply three times the corresponding element in the original matrix. This operation is straightforward and sets the foundation for more complex matrix calculations, like matrix subtraction and addition.

Matrix Subtraction

Matrix subtraction is another critical operation in linear algebra. It allows us to find the difference between two matrices of the same dimension. To subtract one matrix from another, simply subtract each element of the second matrix from the corresponding element of the first matrix. In the solution given, the result of \(4E\) was subtracted from that of \(3A\).

Here’s a step-by-step look: Given
\[3A = \begin{bmatrix} 3 & -6 \ 9 & 12 \end{bmatrix} \] and
\[4E = \begin{bmatrix} 8 & 20 \ 28 & 12 \end{bmatrix}\]

The subtraction process is: \[3A - 4E = \begin{bmatrix} 3 - 8 & -6 - 20 \ 9 - 28 & 12 - 12 \end{bmatrix} = \begin{bmatrix} -5 & -26 \ -19 & 0 \end{bmatrix}\]

This result is a new matrix that contains the differences of the respective elements from \(3A\) and \(4E\). For subtraction to be valid, the matrices involved must have the same dimensions.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. The operations of scalars and matrices, such as scalar multiplication and matrix subtraction, are fundamental concepts in linear algebra. These operations are crucial for solving systems of linear equations, performing transformations, and understanding higher-dimensional spaces.

One key aspect of linear algebra is how it simplifies the representation of complex systems by using matrices. It allows for:

• Simplified computations through concise notation
• Efficient handling of large systems of equations
• Understanding and visualization of multi-dimensional spaces

Linear algebra forms the basis of modern computer algorithms, graphics, machine learning, and even areas as diverse as quantum mechanics and economic modeling. Mastering these fundamental operations like scalar multiplication and matrix subtraction is essential for delving deeper into more advanced studies in this field.