Problem 7 - Precalculus

Question

For each pair of matrices \(A\) and \(B\) find the following, if defined \- \(3 A\) \- \(-B\) \- \(A^{2}\) \- \(A-2 B\) \- \(A B\) \- \(B A\) $$ A=\left[\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 1 & -2 \\ -7 & 1 & -1 \end{array}\right], B=\left[\begin{array}{rrr} 1 & 2 & 1 \\ 17 & 33 & 19 \\ 10 & 19 & 11 \end{array}\right] $$

Step-by-Step Solution

Answer

3A, -B, A^2, A-2B, AB are defined; BA is the identity matrix.

Step 1: Calculate 3A

Multiply each element of matrix \( A \) by 3.\[3A = 3\times \begin{bmatrix} 2 & -3 & 5 \ 3 & 1 & -2 \ -7 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 6 & -9 & 15 \ 9 & 3 & -6 \ -21 & 3 & -3 \end{bmatrix}\]

Step 2: Calculate -B

Multiply each element of matrix \( B \) by -1.\[-B = -1\times \begin{bmatrix} 1 & 2 & 1 \ 17 & 33 & 19 \ 10 & 19 & 11 \end{bmatrix} = \begin{bmatrix} -1 & -2 & -1 \ -17 & -33 & -19 \ -10 & -19 & -11 \end{bmatrix}\]

Step 3: Calculate A squared \(A^2\)

First, perform the matrix multiplication \( A \times A \).\[A^2 = \begin{bmatrix} 2 & -3 & 5 \ 3 & 1 & -2 \ -7 & 1 & -1 \end{bmatrix} \times \begin{bmatrix} 2 & -3 & 5 \ 3 & 1 & -2 \ -7 & 1 & -1 \end{bmatrix}\]Perform the multiplication for each element:\[= \begin{bmatrix} (2)(2) + (-3)(3) + (5)(-7) & (2)(-3) + (-3)(1) + (5)(1) & (2)(5) + (-3)(-2) + (5)(-1) \ (3)(2) + (1)(3) + (-2)(-7) & (3)(-3) + (1)(1) + (-2)(1) & (3)(5) + (1)(-2) + (-2)(-1) \ (-7)(2) + (1)(3) + (-1)(-7) & (-7)(-3) + (1)(1) + (-1)(1) & (-7)(5) + (1)(-2) + (-1)(-1) \end{bmatrix}\]\[= \begin{bmatrix} -40 & -2 & 23 \ 25 & -8 & 13 \ -12 & 23 & -36 \end{bmatrix}\]

Step 4: Calculate A - 2B

Find \( 2B \) by multiplying each element of matrix \( B \) by 2 and then subtract from \( A \).\[2B = 2\times \begin{bmatrix} 1 & 2 & 1 \ 17 & 33 & 19 \ 10 & 19 & 11 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 2 \ 34 & 66 & 38 \ 20 & 38 & 22 \end{bmatrix}\]Now subtract:\[A - 2B = \begin{bmatrix} 2 & -3 & 5 \ 3 & 1 & -2 \ -7 & 1 & -1 \end{bmatrix} - \begin{bmatrix} 2 & 4 & 2 \ 34 & 66 & 38 \ 20 & 38 & 22 \end{bmatrix} = \begin{bmatrix} 0 & -7 & 3 \ -31 & -65 & -40 \ -27 & -37 & -23 \end{bmatrix}\]

Step 5: Calculate AB

Perform the matrix multiplication \( A \times B \) by computing the dot product of rows and columns.\[AB = \begin{bmatrix} 2 & -3 & 5 \ 3 & 1 & -2 \ -7 & 1 & -1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 1 \ 17 & 33 & 19 \ 10 & 19 & 11 \end{bmatrix}\]Calculate each element:\[= \begin{bmatrix} 2\times1 + (-3)\times17 + 5\times10 & 2\times2 + (-3)\times33 + 5\times19 & 2\times1 + (-3)\times19 + 5\times11 \ 3\times1 + 1\times17 + (-2)\times10 & 3\times2 + 1\times33 + (-2)\times19 & 3\times1 + 1\times19 + (-2)\times11 \ (-7)\times1 + 1\times17 + (-1)\times10 & (-7)\times2 + 1\times33 + (-1)\times19 & (-7)\times1 + 1\times19 + (-1)\times11 \end{bmatrix}\]\[= \begin{bmatrix} 0 & -94 & -40 \ 9 & -1 & -2 \ 0 & 0 & 0 \end{bmatrix}\]

Step 6: Calculate BA

Perform the matrix multiplication \( B \times A \) by computing the dot product of rows and columns.\[BA = \begin{bmatrix} 1 & 2 & 1 \ 17 & 33 & 19 \ 10 & 19 & 11 \end{bmatrix} \times \begin{bmatrix} 2 & -3 & 5 \ 3 & 1 & -2 \ -7 & 1 & -1 \end{bmatrix}\]Calculate each element:\[= \begin{bmatrix} 1\times2 + 2\times3 + 1\times(-7) & 1\times(-3) + 2\times1 + 1\times1 & 1\times5 + 2\times(-2) + 1\times(-1) \ 17\times2 + 33\times3 + 19\times(-7) & 17\times(-3) + 33\times1 + 19\times1 & 17\times5 + 33\times(-2) + 19\times(-1) \ 10\times2 + 19\times3 + 11\times(-7) & 10\times(-3) + 19\times1 + 11\times1 & 10\times5 + 19\times(-2) + 11\times(-1) \end{bmatrix}\]\[= \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]

Key Concepts

Matrix Multiplication

Matrix multiplication is a crucial operation in linear algebra and involves multiplying two matrices by taking the dot product of rows and columns. In matrix multiplication, the number of columns in the first matrix must match the number of rows in the second matrix.
This process can seem daunting, but it can be broken down into simple steps:

• Identify the dimensions of the matrices to ensure multiplication is possible. For example, if matrix \( A \) is 3x3 and matrix \( B \) is 3x3, then multiplication is feasible.
• Multiply each element of a row in the first matrix by each corresponding element of a column in the second matrix, then sum these products. This result is a single element in the product matrix.

For the given matrices \( A \) and \( B \), when calculating \( A \times B \), each element of the resulting matrix is computed as a sum of products. This is often denoted as:\[(A \times B)_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj}\]where \( i \) is the row number, \( j \) is the column number, and \( n \) is the number of elements in the row of matrix \( A \) (or column of matrix \( B \)).
Practice with real matrices to become more comfortable with this operation.

Scalar Multiplication

Scalar multiplication is one of the simplest matrix operations, where each entry in a matrix is multiplied by the same constant, known as a scalar. The beautiful thing about scalar multiplication is its directness and simplicity.
Here's how you can do it:

• Select your scalar. For instance, if you have the scalar 3, and you're multiplying by matrix \( A \), every element in \( A \) will be multiplied by 3.
• Multiply each element of the matrix by the scalar.
• The result is a matrix of the same dimensions as the original, but with each element scaled by the given factor.

Using the initial example, multiplying matrix \( A \) by 3 results in a new matrix where each element is three times that of the corresponding element in \( A \):\[\begin{bmatrix} 2 & -3 & 5 \ 3 & 1 & -2 \ -7 & 1 & -1 \end{bmatrix} \to \begin{bmatrix} 6 & -9 & 15 \ 9 & 3 & -6 \ -21 & 3 & -3 \end{bmatrix}\]Scalar multiplication is foundational for other matrix operations, acting as a building block for more complex calculations.

Matrix Subtraction

Matrix subtraction involves subtracting corresponding elements of two matrices. For subtraction to be possible, both matrices must have the same dimensions.
Here's a simple guide to follow for matrix subtraction:

• Ensure both matrices are of the same size, meaning they have the same number of rows and columns.
• Subtract each element of the second matrix from the corresponding element of the first matrix.
• The result is a new matrix of the same dimensions, where each element is the result of the subtraction.

In the problem above, the subtraction \( A - 2B \) was performed by first multiplying each element of matrix \( B \) by 2, then subtracting from matrix \( A \):\[A - 2B = \begin{bmatrix} 2 & -3 & 5 \ 3 & 1 & -2 \ -7 & 1 & -1 \end{bmatrix} - \begin{bmatrix} 2 & 4 & 2 \ 34 & 66 & 38 \ 20 & 38 & 22 \end{bmatrix} = \begin{bmatrix} 0 & -7 & 3 \ -31 & -65 & -40 \ -27 & -37 & -23 \end{bmatrix}\]Matrix subtraction is straightforward, offering clarity and simplicity in reducing matrices.

Matrix Addition

Matrix addition entails adding corresponding elements of two matrices. As with subtraction, both matrices must share the same dimensions for addition to be possible. Here's a straightforward guide:

• Verify both matrices are the same size, with an equal number of rows and columns.
• Add each element from one matrix with the corresponding element from another matrix.
• The result is a new matrix, of the same dimensions, where each element is the sum of the corresponding elements from the original matrices.

Even though matrix addition was not explicitly part of the provided exercise, understanding it complements matrix subtraction well, as both require the same preparatory steps regarding dimension checking.
Keep practicing with various matrices to ensure confidence and understanding of these fundamental operations, forming the foundation for advanced topics in linear algebra.