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Chapter 6: Problem 16

If possible, find each of the following. (a) \(A+B\) (b) \(3 A\) (c) \(2 A-3 B\) $$A=\left[\begin{array}{rrr}1 & -2 & 5 \\\3 & -4 & -1\end{array}\right]$$ $$B=\left[\begin{array}{rrr}0 & -1 & -5 \\\\-3 & 1 & 2\end{array}\right]$$

Short Answer

Expert verified
(a) \(\begin{bmatrix} 1 & -3 & 0 \\ 0 & -3 & 1 \end{bmatrix}\), (b) \(\begin{bmatrix} 3 & -6 & 15 \\ 9 & -12 & -3 \end{bmatrix}\), (c) \(\begin{bmatrix} 2 & -1 & 25 \\ 15 & -11 & -8 \end{bmatrix}\).

Step by step solution

01

Addition of Matrices (A+B)

To find the sum of two matrices, add the corresponding elements from each matrix. Matrix \(A\) and \(B\) have the same dimensions, so we can perform the addition. \[A+B = \begin{bmatrix} 1 & -2 & 5 \ 3 & -4 & -1 \end{bmatrix} + \begin{bmatrix} 0 & -1 & -5 \ -3 & 1 & 2 \end{bmatrix} = \begin{bmatrix} (1+0) & (-2-1) & (5-5) \ (3-3) & (-4+1) & (-1+2) \end{bmatrix} = \begin{bmatrix} 1 & -3 & 0 \ 0 & -3 & 1 \end{bmatrix} \]The sum of matrices \(A\) and \(B\) is \(\begin{bmatrix} 1 & -3 & 0 \ 0 & -3 & 1 \end{bmatrix}\).
02

Scalar Multiplication of Matrix (3A)

To find \(3A\), multiply each element of matrix \(A\) by the scalar 3.\[3A = 3 \cdot \begin{bmatrix} 1 & -2 & 5 \ 3 & -4 & -1 \end{bmatrix} = \begin{bmatrix} 3 \cdot 1 & 3 \cdot (-2) & 3 \cdot 5 \ 3 \cdot 3 & 3 \cdot (-4) & 3 \cdot (-1) \end{bmatrix} = \begin{bmatrix} 3 & -6 & 15 \ 9 & -12 & -3 \end{bmatrix} \]The result of scalar multiplication is \(\begin{bmatrix} 3 & -6 & 15 \ 9 & -12 & -3 \end{bmatrix}\).
03

Linear Combination of Matrices (2A - 3B)

To find \(2A - 3B\), calculate \(2A\) by multiplying each element of matrix \(A\) by 2, and \(3B\) by multiplying each element of matrix \(B\) by 3. Then subtract \(3B\) from \(2A\).First calculate \(2A\):\[2A = 2 \cdot \begin{bmatrix} 1 & -2 & 5 \ 3 & -4 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -4 & 10 \ 6 & -8 & -2 \end{bmatrix}\]Next calculate \(3B\):\[3B = 3 \cdot \begin{bmatrix} 0 & -1 & -5 \ -3 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & -3 & -15 \ -9 & 3 & 6 \end{bmatrix}\]Now compute: \[2A - 3B = \begin{bmatrix} 2 & -4 & 10 \ 6 & -8 & -2 \end{bmatrix} - \begin{bmatrix} 0 & -3 & -15 \ -9 & 3 & 6 \end{bmatrix} = \begin{bmatrix} (2-0) & (-4+3) & (10+15) \ (6+9) & (-8-3) & (-2-6) \end{bmatrix} = \begin{bmatrix} 2 & -1 & 25 \ 15 & -11 & -8 \end{bmatrix}\]The result of the linear combination is \(\begin{bmatrix} 2 & -1 & 25 \ 15 & -11 & -8 \end{bmatrix}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Addition
Matrix addition is a straightforward process that involves adding corresponding elements of the matrices involved. For the matrices to be added, they need to have the same dimensions, meaning they have the same number of rows and columns.

For instance, if we have two matrices, \( A \) and \( B \), both are \( 2 \times 3 \) matrices, you would add each element in \( A \) with each corresponding element in \( B \).

Consider:
  • The element in the first row, first column of \( A \) is added to the first row, first column of \( B \).
  • This is repeated for elements in subsequent positions, such as second row, first column, and so forth.
This results in a new matrix of the same dimension, showing the sum of each pair of elements from \( A \) and \( B \). It is important to remember that if the matrices do not align in terms of dimensions, matrix addition cannot be performed. This step ensures that the data from each matrix can be combined directly and meaningfully.
Scalar Multiplication
Scalar multiplication involves multiplying each element of a matrix by a single number, known as a scalar. It's as though you were multiplying the entire matrix by that number.

For example, to compute \( 3A \) for a matrix \( A \), take each element in \( A \) and multiply it by the scalar \( 3 \). This process affects all elements uniformly:
  • Suppose an element of \( A \) is \( 1 \). After multiplication by \( 3 \), it becomes \( 3 \times 1 = 3 \).
  • All elements undergo the same operation, whether positive, negative, or zero.
Scalar multiplication is a key operation because it scales the matrix up or down, shrinking or stretching its values while maintaining its structure. Every element's position remains unchanged, which is crucial for performing further matrix operations such as addition or multiplication with other matrices.
Linear Combination
A linear combination of matrices involves two or more matrices and typically includes both scalar multiplication and addition or subtraction.

To obtain a linear combination like \( 2A - 3B \), follow these steps:
  • First, apply scalar multiplication to multiply each element of matrix \( A \) by \( 2 \) and each element of matrix \( B \) by \( 3 \).
  • You'll get new matrices for \( 2A \) and \( 3B \).
  • Subtract the resultant \( 3B \) from \( 2A \), meaning subtract each element of \( 3B \) from the corresponding element of \( 2A \).
This method allows you to combine multiple matrices and scalars systematically. Linear combinations are foundational in fields like linear algebra, where they are used to express solutions of linear systems, transformations, and more. Importantly, the processes involved still rely on having matrices of compatible sizes, ensuring that each scalar-multiplied matrix can be combined with all others.

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Most popular questions from this chapter

Minimizing Cost Two substances, \(\mathbf{X}\) and \(\mathbf{Y}\), are found in pet food. Each substance contains the ingrodients A and B. Substance \(X\) is \(20 \%\) ingredient \(A\) and \(50 \%\) ingredient B. Substance \(Y\) is \(50 \%\) ingredient \(A\) and \(30 \%\) ingredient \(\mathbf{B}\). The cost of substance \(\mathbf{X}\) is \(\$ 2\) per pound, and the cost of substance \(Y\) is \(\$ 3\) per pound. The pet store needs at least 251 pounds of ingredient \(A\) and at least 200 pounds of ingredient \(B\). If cost is to be minimal, how many pounds of each substance should be ordered? Find the minimum cost.

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