Problem 17 - Algebra and Trigonometry Real Mathematics, Real People

Question

Operations with Matrices Find, if possible, \((a) A+B,(b) A-B,(c) 3 A,\) and \((d) 3 A-2 B.\) Use the matrix capabilities of a graphing utility to verify your results. $$\begin{aligned} &A=\left[\begin{array}{lllll} 4 & 5 & -1 & 3 & 4 \\ 1 & 2 & -2 & -1 & 0 \end{array}\right]\\\ &B=\left[\begin{array}{rrrrr} 1 & 0 & -1 & 1 & 0 \\ -6 & 8 & 2 & -3 & -7 \end{array}\right] \end{aligned}$$

Step-by-Step Solution

Answer

a) A+B = \[\left[\begin{array}{ccccc} 5 & 5 & -2 & 4 & 4 \ -5 & 10 & 0 & -4 & -7 \end{array}\right]\], b) A-B = \[\left[\begin{array}{ccccc} 3 & 5 & 0 & 2 & 4 \ 7 & -6 & -4 & 2 & 7 \end{array}\right]\], c) 3A = \[\left[\begin{array}{ccccc} 12 & 15 & -3 & 9 & 12 \ 3 & 6 & -6 & -3 & 0 \end{array}\right]\], d) 3A-2B = \[\left[\begin{array}{ccccc} 10 & 15 & -1 & 7 & 12 \ 15 & -10 & -10 & 5 & 14 \end{array}\right]\].

Step 1: Matrix Addition (A + B)

To add two matrices, simply add the corresponding entries from each matrix. That is, you add the element in the first row and first column of matrix A to the element in the first row and first column of matrix B, and so on. For these matrices we get:\[ A+B=\left[\begin{array}{ccccc} 4+1 & 5+0 & -1-1 & 3+1 & 4+0 \ 1-6 & 2+8 & -2+2 & -1-3 & 0-7 \end{array}\right]=\left[\begin{array}{ccccc} 5 & 5 & -2 & 4 & 4 \ -5 & 10 & 0 & -4 & -7 \end{array}\right]\]

Step 2: Matrix Subtraction (A - B)

To subtract matrix B from A, subtract the corresponding entries. The operation is similar to matrix addition: you subtract the element in the first row and first column of matrix B from the similar element from matrix A, and so on. The result is as following: \[A-B=\left[\begin{array}{ccccc} 4-1 & 5-0 & -1+1 & 3-1 & 4-0 \ 1+6 & 2-8 & -2-2 & -1+3 & 0+7 \end{array}\right] =\left[\begin{array}{ccccc} 3 & 5 & 0 & 2 & 4 \ 7 & -6 & -4 & 2 & 7 \end{array}\right]\]

Step 3: Scalar multiplication (3A)

To multiply a matrix by a scalar, you simply multiply every element in the matrix by that scalar. For the matrix A and scalar 3, this operation is performed like this: \[3A=\left[\begin{array}{ccccc} 3*4 & 3*5 & 3*(-1) & 3*3 & 3*4 \ 3*1 & 3*2 & 3*(-2) & 3*(-1) & 3*0 \end{array}\right]=\left[\begin{array}{ccccc} 12 & 15 & -3 & 9 & 12 \ 3 & 6 & -6 & -3 & 0 \end{array}\right]\]

Step 4: Combined Operations: Scalar Multiplication and Subtraction (3A - 2B)

Last operation involves both scalar multiplication and matrix subtraction. First, each matrix is multiplied by its respective scalar. Then the results are subtracted from each other. Applying these operations: \[(3A-2B)=3A-2B=\left[\begin{array}{ccccc} 3*4-2*1 & 3*5-2*0 & 3*(-1)-2*(-1) & 3*3-2*1 & 3*4-2*0 \ 3*1-2*(-6) & 3*2-2*8 & 3*(-2)-2*2 & 3*(-1)-2*(-3) & 3*0-2*(-7) \end{array}\right] =\left[\begin{array}{ccccc} 10 & 15 & -1 & 7 & 12 \ 15 & -10 & -10 & 5 & 14 \end{array}\right]\]

Key Concepts

Matrix Addition

Matrix addition is the process of adding two matrices by adding their corresponding elements. For two matrices to be added, they must have the same dimensions, meaning the same number of rows and columns.

Given matrices A and B:

• A = \( \left[\begin{array}{ccccc} 4 & 5 & -1 & 3 & 4 \ 1 & 2 & -2 & -1 & 0 \\end{array}\right] \)
• B = \( \left[\begin{array}{ccccc} 1 & 0 & -1 & 1 & 0 \ -6 & 8 & 2 & -3 & -7 \\end{array}\right] \)

To perform matrix addition \(A + B\), we add the elements from the same positions, resulting in:

• First position: \(4 + 1 = 5\)
• Second position: \(5 + 0 = 5\)
• Continue this process for all elements...

The resulting matrix from \(A + B\) is: \[\left[\begin{array}{ccccc}5 & 5 & -2 & 4 & 4\-5 & 10 & 0 & -4 & -7\\end{array}\right]\]

Matrix Subtraction

Matrix subtraction operates similarly to matrix addition but involves subtracting the corresponding elements of two matrices. As with addition, the matrices must have the same dimensions for subtraction to occur.

Consider matrices A and B mentioned earlier. In matrix subtraction \(A - B\), we subtract each element of B from the corresponding element in A.

• First position: \(4 - 1 = 3\)
• Second position: \(5 - 0 = 5\)
• And continue this process for all corresponding elements...

The resulting matrix from \(A - B\) is:\[\left[\begin{array}{ccccc}3 & 5 & 0 & 2 & 4\7 & -6 & -4 & 2 & 7\\end{array}\right]\]

Scalar Multiplication

Scalar multiplication involves multiplying every element of a matrix by a constant called a scalar. This operation changes the magnitude of the matrix but not its structure or dimension.

Let's apply scalar multiplication to matrix A with scalar 3:

• Multiply each element of A by 3:
• \(3 \times 4 = 12\)
• \(3 \times 5 = 15\)
• ... and so on for each element of A

The resulting matrix from \(3A\) is:\[\left[\begin{array}{ccccc}12 & 15 & -3 & 9 & 12\3 & 6 & -6 & -3 & 0\\end{array}\right]\]

Combined Operations

Combined operations involve performing more than one matrix operation sequentially. Here, we look at scalar multiplication combined with matrix subtraction, like in the operation \(3A - 2B\).

First, perform scalar multiplication on each matrix:

• \(3A\) means multiply each element in A by 3.
• \(2B\) means multiply each element in B by 2.

Next, subtract the resulting matrices:

• Subtract each element of \(2B\) from the corresponding element of \(3A\).
• For example, the first element: \(3 \times 4 - 2 \times 1 = 10\)
• Continue this for all elements...

The resulting matrix for the operation \(3A - 2B\) is:\[\left[\begin{array}{ccccc}10 & 15 & -1 & 7 & 12\15 & -10 & -10 & 5 & 14\\end{array}\right]\]