Problem 7 - Algebra for College Students

Question

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rr} -1 & 0 \\ 2 & 3 \\ -5 & -4 \\ -7 & 11 \end{array}\right], \quad B=\left[\begin{array}{rr} 1 & 2 \\ -3 & 7 \\ 6 & -5 \\ 9 & -2 \end{array}\right] $$

Step-by-Step Solution

Answer

For Problem 1: \( A+B = \begin{bmatrix} 0 & 2 \\ -1 & 10 \\ 1 & -9 \\ 2 & 9 \end{bmatrix}, A-B = \begin{bmatrix} -2 & -2 \\ 5 & -4 \\ -11 & 1 \\ -16 & 13 \end{bmatrix}, 2A+3B = \begin{bmatrix} 1 & 6 \\ -5 & 27 \\ 8 & -23 \\ 13 & 16 \end{bmatrix}, 4A-2B = \begin{bmatrix} -6 & -4 \\ 14 & -2 \\ -32 & -6 \\ -46 & 48 \end{bmatrix}."

Step 1: Compute A + B

Add matrix \(A\) and matrix \(B\) by adding corresponding elements. \[A + B = \begin{bmatrix} -1+1 & 0+2 \ 2-3 & 3+7 \ -5+6 & -4-5 \ -7+9 & 11-2 \end{bmatrix} = \begin{bmatrix} 0 & 2 \ -1 & 10 \ 1 & -9 \ 2 & 9 \end{bmatrix}\]

Step 2: Compute A - B

Subtract matrix \(B\) from matrix \(A\) by subtracting corresponding elements. \[A - B = \begin{bmatrix} -1-1 & 0-2 \ 2+3 & 3-7 \ -5-6 & -4+5 \ -7-9 & 11+2 \end{bmatrix} = \begin{bmatrix} -2 & -2 \ 5 & -4 \ -11 & 1 \ -16 & 13 \end{bmatrix}\]

Step 3: Compute 2A + 3B

Calculate \(2A\) and \(3B\) first by multiplying each element of matrices \(A\) and \(B\) by 2 and 3, respectively, then add the results. \[2A = \begin{bmatrix} 2(-1) & 2(0) \ 2(2) & 2(3) \ 2(-5) & 2(-4) \ 2(-7) & 2(11)\end{bmatrix} = \begin{bmatrix} -2 & 0 \ 4 & 6 \ -10 & -8 \ -14 & 22 \end{bmatrix}\]\[3B = \begin{bmatrix} 3(1) & 3(2) \ 3(-3) & 3(7) \ 3(6) & 3(-5) \ 3(9) & 3(-2) \end{bmatrix} = \begin{bmatrix} 3 & 6 \ -9 & 21 \ 18 & -15 \ 27 & -6 \end{bmatrix}\]Now, add \(2A\) and \(3B\):\[2A + 3B = \begin{bmatrix} -2+3 & 0+6 \ 4-9 & 6+21 \ -10+18 & -8-15 \ -14+27 & 22-6 \end{bmatrix} = \begin{bmatrix} 1 & 6 \ -5 & 27 \ 8 & -23 \ 13 & 16 \end{bmatrix}\]

Step 4: Compute 4A - 2B

Calculate \(4A\) and \(2B\) first by multiplying each element of matrices \(A\) and \(B\) by 4 and 2, respectively, then subtract the results. \[4A = \begin{bmatrix} 4(-1) & 4(0) \ 4(2) & 4(3) \ 4(-5) & 4(-4) \ 4(-7) & 4(11) \end{bmatrix} = \begin{bmatrix} -4 & 0 \ 8 & 12 \ -20 & -16 \ -28 & 44 \end{bmatrix}\]\[2B = \begin{bmatrix} 2(1) & 2(2) \ 2(-3) & 2(7) \ 2(6) & 2(-5) \ 2(9) & 2(-2) \end{bmatrix} = \begin{bmatrix} 2 & 4 \ -6 & 14 \ 12 & -10 \ 18 & -4 \end{bmatrix}\]Now, subtract \(2B\) from \(4A\):\[4A - 2B = \begin{bmatrix} -4-2 & 0-4 \ 8+6 & 12-14 \ -20-12 & -16+10 \ -28-18 & 44+4 \end{bmatrix} = \begin{bmatrix} -6 & -4 \ 14 & -2 \ -32 & -6 \ -46 & 48 \end{bmatrix}\]

Key Concepts

Matrix Addition

Matrix addition is one of the fundamental operations you can perform on matrices. To add two matrices, they must be of the same dimension, meaning they have the same number of rows and columns. When adding matrices, each element in one matrix is added to the corresponding element in the other matrix.
For example, consider two matrices:

• Matrix A:\[\begin{bmatrix} -1 & 0 \ 2 & 3 \ -5 & -4 \ -7 & 11 \end{bmatrix}\]
• Matrix B:\[\begin{bmatrix} 1 & 2 \ -3 & 7 \ 6 & -5 \ 9 & -2 \end{bmatrix}\]

To perform matrix addition, we calculate: \[A + B = \begin{bmatrix} -1+1 & 0+2 \ 2-3 & 3+7 \ -5+6 & -4-5 \ -7+9 & 11-2 \end{bmatrix} = \begin{bmatrix} 0 & 2 \ -1 & 10 \ 1 & -9 \ 2 & 9 \end{bmatrix}\]
This is done by individually adding each element in matrix A with the corresponding element in matrix B.

Matrix Subtraction

Matrix subtraction is very similar to matrix addition. But instead of adding, you subtract corresponding elements from each other. Like with addition, the matrices must be of the same dimensions to be eligible for subtraction.
For instance, using matrices A and B:

• Matrix A:\[\begin{bmatrix} -1 & 0 \ 2 & 3 \ -5 & -4 \ -7 & 11 \end{bmatrix}\]
• Matrix B:\[\begin{bmatrix} 1 & 2 \ -3 & 7 \ 6 & -5 \ 9 & -2 \end{bmatrix}\]

To execute matrix subtraction, compute:\[A - B = \begin{bmatrix} -1-1 & 0-2 \ 2+3 & 3-7 \ -5-6 & -4+5 \ -7-9 & 11+2 \end{bmatrix} = \begin{bmatrix} -2 & -2 \ 5 & -4 \ -11 & 1 \ -16 & 13 \end{bmatrix}\]
This involves subtracting each element in matrix B from its corresponding element in matrix A.

Scalar Multiplication

Scalar multiplication involves multiplying each element of the matrix by a constant, known as a scalar. This operation results in a new matrix of the same dimensions, but with each element scaled proportionally.
For example, if we want to multiply matrix A by 2, the process is:\[2A = \begin{bmatrix} 2(-1) & 2(0) \ 2(2) & 2(3) \ 2(-5) & 2(-4) \ 2(-7) & 2(11) \end{bmatrix} = \begin{bmatrix} -2 & 0 \ 4 & 6 \ -10 & -8 \ -14 & 22 \end{bmatrix}\]
Similarly, multiplying matrix B by 3 gives us:\[3B = \begin{bmatrix} 3(1) & 3(2) \ 3(-3) & 3(7) \ 3(6) & 3(-5) \ 3(9) & 3(-2) \end{bmatrix} = \begin{bmatrix} 3 & 6 \ -9 & 21 \ 18 & -15 \ 27 & -6 \end{bmatrix}\]
Each element is simply multiplied by the scalar to achieve the transformed matrix.

Matrix Arithmetic Steps

Performing matrix arithmetic can involve multiple steps, particularly when combining operations such as scalar multiplication, addition, and subtraction. When you're tasked with an expression like \(2A+3B\) or \(4A-2B\), it's important to break it down into simpler parts.
First, tackle the scalar multiplication for each matrix involved. For example:

• Calculate \(2A\): \ \(\begin{bmatrix} -2 & 0 \ 4 & 6 \ -10 & -8 \ -14 & 22 \end{bmatrix}\)
• Calculate \(3B\): \ \(\begin{bmatrix} 3 & 6 \ -9 & 21 \ 18 & -15 \ 27 & -6 \end{bmatrix}\)

Next, execute matrix addition to solve for \(2A + 3B\):\[\begin{bmatrix} -2+3 & 0+6 \ 4-9 & 6+21 \ -10+18 & -8-15 \ -14+27 & 22-6 \end{bmatrix} = \begin{bmatrix} 1 & 6 \ -5 & 27 \ 8 & -23 \ 13 & 16 \end{bmatrix}\]
For subtraction like \(4A - 2B\), repeat the scalar multiplication, then perform subtraction:\[\begin{bmatrix} -4-2 & 0-4 \ 8+6 & 12-14 \ -20-12 & -16+10 \ -28-18 & 44+4 \end{bmatrix} = \begin{bmatrix} -6 & -4 \ 14 & -2 \ -32 & -6 \ -46 & 48 \end{bmatrix}\]
Breaking down the arithmetic into these steps makes it manageable and solvable.