Problem 1 - Algebra for College Students

Question

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rrr} 2 & -1 & 4 \\ -2 & 0 & 5 \end{array}\right], \quad B=\left[\begin{array}{rrr} -1 & 4 & -7 \\ 5 & -6 & 2 \end{array}\right] $$

Step-by-Step Solution

Answer

1) \( A+B = \begin{bmatrix} 1 & 3 & -3 \\ 3 & -6 & 7 \end{bmatrix} \); 2) \( A-B = \begin{bmatrix} 3 & -5 & 11 \\ -7 & 6 & 3 \end{bmatrix} \); 3) \( 2A+3B = \begin{bmatrix} 1 & 10 & -13 \\ 11 & -18 & 16 \end{bmatrix} \); 4) \( 4A-2B = \begin{bmatrix} 10 & -12 & 30 \\ -18 & 12 & 16 \end{bmatrix} \)."

Step 1: Compute A+B

To find \( A + B \), add the corresponding elements of matrices \( A \) and \( B \). This means: \[ (A+B) = \left[\begin{array}{ccc} 2 + (-1) & -1 + 4 & 4 + (-7) \ -2 + 5 & 0 + (-6) & 5 + 2 \end{array}\right] \]Calculating this gives:\( A+B = \left[\begin{array}{ccc} 1 & 3 & -3 \ 3 & -6 & 7 \end{array}\right] \).

Step 2: Compute A-B

To find \( A - B \), subtract the corresponding elements of matrix \( B \) from matrix \( A \):\[ (A-B) = \left[\begin{array}{ccc} 2 - (-1) & -1 - 4 & 4 - (-7) \ -2 - 5 & 0 - (-6) & 5 - 2 \end{array}\right] \]Calculating this gives:\( A-B = \left[\begin{array}{ccc} 3 & -5 & 11 \ -7 & 6 & 3 \end{array}\right] \).

Step 3: Compute 2A

To find \( 2A \), multiply every element of matrix \( A \) by 2:\[ 2A = \left[\begin{array}{ccc} 2 \cdot 2 & 2 \cdot (-1) & 2 \cdot 4 \ 2 \cdot (-2) & 2 \cdot 0 & 2 \cdot 5 \end{array}\right] = \left[\begin{array}{ccc} 4 & -2 & 8 \ -4 & 0 & 10 \end{array}\right] \].

Step 4: Compute 3B

To find \( 3B \), multiply every element of matrix \( B \) by 3:\[ 3B = \left[\begin{array}{ccc} 3 \cdot (-1) & 3 \cdot 4 & 3 \cdot (-7) \ 3 \cdot 5 & 3 \cdot (-6) & 3 \cdot 2 \end{array}\right] = \left[\begin{array}{ccc} -3 & 12 & -21 \ 15 & -18 & 6 \end{array}\right] \].

Step 5: Compute 2A + 3B

To find \( 2A + 3B \), add the corresponding elements of matrices \( 2A \) and \( 3B \):\[ (2A+3B) = \left[\begin{array}{ccc} 4 + (-3) & -2 + 12 & 8 + (-21) \ -4 + 15 & 0 + (-18) & 10 + 6 \end{array}\right] \]Calculating this gives:\( 2A+3B = \left[\begin{array}{ccc} 1 & 10 & -13 \ 11 & -18 & 16 \end{array}\right] \).

Step 6: Compute 4A

To find \( 4A \), multiply every element of matrix \( A \) by 4:\[ 4A = \left[\begin{array}{ccc} 4 \cdot 2 & 4 \cdot (-1) & 4 \cdot 4 \ 4 \cdot (-2) & 4 \cdot 0 & 4 \cdot 5 \end{array}\right] = \left[\begin{array}{ccc} 8 & -4 & 16 \ -8 & 0 & 20 \end{array}\right] \].

Step 7: Compute 2B

To find \( 2B \), multiply every element of matrix \( B \) by 2:\[ 2B = \left[\begin{array}{ccc} 2 \cdot (-1) & 2 \cdot 4 & 2 \cdot (-7) \ 2 \cdot 5 & 2 \cdot (-6) & 2 \cdot 2 \end{array}\right] = \left[\begin{array}{ccc} -2 & 8 & -14 \ 10 & -12 & 4 \end{array}\right] \].

Step 8: Compute 4A - 2B

To find \( 4A - 2B \), subtract the corresponding elements of matrix \( 2B \) from matrix \( 4A \):\[ (4A-2B) = \left[\begin{array}{ccc} 8 - (-2) & -4 - 8 & 16 - (-14) \ -8 - 10 & 0 - (-12) & 20 - 4 \end{array}\right] \]Calculating this yields:\( 4A-2B = \left[\begin{array}{ccc} 10 & -12 & 30 \ -18 & 12 & 16 \end{array}\right] \).

Key Concepts

Matrix Addition

Matrix Addition is a process of combining two matrices of the same dimension by adding corresponding elements together. This is similar to standard numerical addition, where each position in the matrix pair is considered separately. For instance, if we have two matrices, say \( A \) and \( B \), matrix addition involves taking the element from row 1, column 1 in \( A \) and adding it to the element from row 1, column 1 in \( B \), and so on for every corresponding element in the matrices.

Let's look at an example using matrices from our step-by-step solution above:

• First matrix: \( A = \begin{bmatrix} 2 & -1 & 4 \ -2 & 0 & 5 \end{bmatrix} \)
• Second matrix: \( B = \begin{bmatrix} -1 & 4 & -7 \ 5 & -6 & 2 \end{bmatrix} \)

To find \( A + B \), we perform the operation:
\[A + B = \begin{bmatrix} 2 + (-1) & -1 + 4 & 4 + (-7) \ -2 + 5 & 0 + (-6) & 5 + 2 \end{bmatrix} = \begin{bmatrix} 1 & 3 & -3 \ 3 & -6 & 7 \end{bmatrix}.\]
This results in a new matrix where each element is the sum of corresponding elements from \( A \) and \( B \).
Matrix addition is only possible when both matrices are of the same size, meaning they have the same number of rows and columns.

Matrix Subtraction

Matrix Subtraction is similar to matrix addition, but instead of adding corresponding elements, we subtract them. It involves taking the element from matrix \( A \) and subtracting the corresponding element from matrix \( B \).

Consider the matrices from the original problem:

• Matrix \( A = \begin{bmatrix} 2 & -1 & 4 \ -2 & 0 & 5 \end{bmatrix} \)
• Matrix \( B = \begin{bmatrix} -1 & 4 & -7 \ 5 & -6 & 2 \end{bmatrix} \)

To find \( A - B \), we perform the operation:
\[A - B = \begin{bmatrix} 2 - (-1) & -1 - 4 & 4 - (-7) \ -2 - 5 & 0 - (-6) & 5 - 2 \end{bmatrix} = \begin{bmatrix} 3 & -5 & 11 \ -7 & 6 & 3 \end{bmatrix}.\]
This results in a new matrix with elements that are the differences of the elements in \( A \) and \( B \). Like matrix addition, subtraction can only be performed on matrices of the same dimension.

Scalar Multiplication

Scalar Multiplication involves multiplying each element of a matrix by a single constant, known as a scalar. This scalar can be any real number. This operation essentially "scales" the matrix by the given factor, altering each element by the same amount.

Take the matrix \( A = \begin{bmatrix} 2 & -1 & 4 \ -2 & 0 & 5 \end{bmatrix} \) as an example. If we want to multiply this matrix by a scalar of 2, each element in \( A \) is multiplied by 2:
\[2A = \begin{bmatrix} 2 \cdot 2 & 2 \cdot (-1) & 2 \cdot 4 \ 2 \cdot (-2) & 2 \cdot 0 & 2 \cdot 5 \end{bmatrix} = \begin{bmatrix} 4 & -2 & 8 \ -4 & 0 & 10 \end{bmatrix}.\]
Scalar multiplication is useful in various matrix operations such as scaling transformations and in computations like finding linear combinations.

Linear Combinations of Matrices

Linear Combinations of Matrices involve combining several matrices by applying addition and scalar multiplication. This constitutes forming a new matrix by adding together matrices that have been multiplied by coefficients (scalars).

The concept of linear combinations can be seen in the expression \( 2A + 3B \). Here, we first perform scalar multiplication on each matrix:

• \( 2A = \begin{bmatrix} 4 & -2 & 8 \ -4 & 0 & 10 \end{bmatrix} \)
• \( 3B = \begin{bmatrix} -3 & 12 & -21 \ 15 & -18 & 6 \end{bmatrix} \)

Then, by adding these results, we achieve:\[2A + 3B = \begin{bmatrix} 4 + (-3) & -2 + 12 & 8 + (-21) \ -4 + 15 & 0 + (-18) & 10 + 6 \end{bmatrix} = \begin{bmatrix} 1 & 10 & -13 \ 11 & -18 & 16 \end{bmatrix}.\]
Linear combinations are a fundamental concept in linear algebra and are essential for understanding vector spaces, solving systems of linear equations, and in applications like data transformations.