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Chapter 1: Problem 3

Let \\[ A=\left(\begin{array}{rr} 1 & 1 \\ 2 & -1 \end{array}\right) \quad \text { and } \quad B=\left(\begin{array}{ll} 2 & 1 \\ 1 & 3 \end{array}\right) \\] (a) Calculate \(A \mathbf{b}_{1}\) and \(A \mathbf{b}_{2}\) (b) Calculate \(\overrightarrow{\mathbf{a}}_{1} B\) and \(\overrightarrow{\mathbf{a}}_{2} B\) (c) Multiply \(A B,\) and verify that its column vectors are the vectors in part (a) and its row vectors are the vectors in part (b).

Short Answer

Expert verified
In summary, we have calculated the following: (a) \(A \cdot b_{1} = \left(\begin{array}{rr} 3 \\ 3 \end{array}\right)\) and \(A \cdot b_{2} = \left(\begin{array}{rr} 4 \\ -1 \end{array}\right)\) (b) \(a_{1} \cdot B = \left(\begin{array}{ll} 3 & 4 \end{array}\right)\) and \(a_{2} \cdot B = \left(\begin{array}{ll} 3 & -1 \end{array}\right)\) (c) The product \(A \cdot B = \left(\begin{array}{ll} 3 & 4 \\ 3 & -1 \end{array}\right)\), where its column vectors are the vectors from part (a) and its row vectors are the vectors from part (b).

Step by step solution

01

Part (a): Calculating A*b1 and A*b2

First, we need to find 饾憦1 and 饾憦2. These are the column vectors of matrix B: \( b_{1} = \left(\begin{array}{rr} 2 \\ 1 \end{array}\right) \) and \( b_{2} = \left(\begin{array}{rr} 1 \\ 3 \end{array}\right) \) Now, we can calculate the product of the matrix A and column vectors 饾憦1 and 饾憦2. For A*b1: \[ A \cdot b_{1} = \left(\begin{array}{rr} 1 & 1 \\ 2 & -1 \end{array}\right) \cdot \left(\begin{array}{rr} 2 \\ 1 \end{array}\right) = \left(\begin{array}{rr} 3 \\ 3 \end{array}\right) \] For A*b2: \[ A \cdot b_{2} = \left(\begin{array}{rr} 1 & 1 \\ 2 & -1 \end{array}\right) \cdot \left(\begin{array}{rr} 1 \\ 3 \end{array}\right) = \left(\begin{array}{rr} 4 \\ -1 \end{array}\right) \]
02

Part (b): Calculating a1*B and a2*B

Here, we need to find 饾憥1 and 饾憥2. These are the row vectors of matrix A: \( a_{1} = \left(\begin{array}{ll} 1 &1 \end{array}\right) \) and \( a_{2} = \left(\begin{array}{ll} 2 & -1 \end{array}\right) \) Now, we can calculate the product of the row vectors 饾憥1, 饾憥2 with the matrix B. For 饾憥1 * B: \[ a_{1} \cdot B = \left(\begin{array}{ll} 1 & 1 \end{array}\right) \cdot \left(\begin{array}{ll} 2 & 1 \\ 1 & 3 \end{array}\right) = \left(\begin{array}{ll} 3 & 4 \end{array}\right) \] For 饾憥2 * B: \[ a_{2} \cdot B = \left(\begin{array}{ll} 2 & -1 \end{array}\right) \cdot \left(\begin{array}{ll} 2 & 1 \\ 1 & 3 \end{array}\right) = \left(\begin{array}{ll} 3 &-1 \end{array}\right) \]
03

Part (c): Calculating AB and verifying results

Now, we will calculate AB. \[ A \cdot B = \left(\begin{array}{rr} 1 & 1 \\ 2 & -1 \end{array}\right) \cdot \left(\begin{array}{ll} 2 & 1 \\ 1 & 3 \end{array}\right) = \left(\begin{array}{ll} 3 & 4 \\ 3 & -1 \end{array}\right) \] According to the results in parts (a) and (b), A*b1 and A*b2 should be the column vectors, and a1*B and a2*B should be the row vectors of AB. AB is: \[ \left(\begin{array}{ll} 3 & 4 \\ 3 & -1 \end{array}\right), \] which has column vectors A*b1 and A*b2: \[ \left(\begin{array}{rr} 3 \\ 3 \end{array}\right) \] \[ \left(\begin{array}{rr} 4 \\ -1 \end{array}\right) \] and row vectors a1*B and a2*B: \[ \left(\begin{array}{ll} 3 & 4 \end{array}\right) \] \[ \left(\begin{array}{ll} 3 & -1 \end{array}\right) \] Based on the calculations, our part (c) results are consistent with parts (a) and (b).

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Key Concepts

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

Linear Algebra
Linear algebra is a significant branch of mathematics that deals with vectors, vector spaces, linear mappings, and systems of linear equations. It provides a framework for analyzing patterns and discovering relationships in data and is pivotal in diverse fields such as engineering, physics, computer science, and economics.

In linear algebra, we often use matrices, which are rectangular arrays of numbers, to represent linear transformations and solve systems of linear equations. A matrix can be multiplied by another matrix or by a vector, resulting in another matrix or vector that conveys important information about the transformation or system.
Matrix Product
The matrix product is a way of combining two matrices to form a new matrix. It is essential to note that matrix multiplication is not commutative, meaning that the order of the matrices matters; in general, \(AB eq BA\).

To multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. The resulting matrix's element in the \(i\)-th row and \(j\)-th column is computed by taking the dot product of the \(i\)-th row of the first matrix with the \(j\)-th column of the second matrix.

Matrix multiplication can be thought of as a series of operations involving the row vectors of the first matrix and the column vectors of the second matrix, as we saw in the textbook exercise. These computations reveal how one matrix transforms the dimensions of the other, embodying concepts such as rotation, scaling, and shearing in geometric transformations.
Column Vectors
Column vectors are matrices with a single column, representing points or directions in n-dimensional space, where \(n\) is the number of rows in the vector. In the context of multiplication with another matrix, each column vector from the second matrix contributes to one column of the resulting product matrix.

As seen in the textbook exercise, when matrix \(A\) is multiplied by column vector \(\mathbf{b}_1\), it results in a new column vector. This operation reflects the transformation applied by matrix \(A\) to the vector, which can represent anything from a spatial transformation to a change in state in a system.
Row Vectors
Row vectors are matrices with a single row, and in many ways, they can be considered as the transpose of column vectors. They are useful in matrix multiplication when pre-multiplying a matrix, as with \(\overrightarrow{\mathbf{a}}_{1} B\) in our exercise.

When we multiply a row vector by a matrix, the result is another row vector. This operation combines the elements of the row vector with each row of the matrix in a weighted sum. Row vectors are often used to represent coefficients in linear combinations or constraints in optimization problems.

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Most popular questions from this chapter

Nitric acid is prepared commercially by a series of three chemical reactions. In the first reaction, nitro\(\operatorname{gen}\left(\mathrm{N}_{2}\right)\) is combined with hydrogen \(\left(\mathrm{H}_{2}\right)\) to form ammonia \(\left(\mathrm{NH}_{3}\right) .\) Next, the ammonia is combined with oxygen \(\left(\mathrm{O}_{2}\right)\) to form nitrogen dioxide \(\left(\mathrm{NO}_{2}\right)\) and water. Finally, the \(\mathrm{NO}_{2}\) reacts with some of the water to form nitric acid (HNO \(_{3}\) ) and nitric oxide (NO). The amounts of each of the components of these reactions are measured in moles (a standard unit of measurement for chemical reactions). How many moles of nitrogen, hydrogen, and oxygen are necessary to produce 8 moles of nitric acid?

Let $$A=\left(\begin{array}{ll} 2 & 1 \\ 6 & 4 \end{array}\right)$$ (a) Express \(A\) as a product of elementary matrices. (b) Express \(A^{-1}\) as a product of elementary matrices.

Let \(A=\left(\begin{array}{lll}2 & 1 & 1 \\ 6 & 4 & 5 \\ 4 & 1 & 3\end{array}\right)\) (a) Find elementary matrices \(E_{1}, E_{2}, E_{3}\) such that \\[ E_{3} E_{2} E_{1} A=U \\] where \(U\) is an upper triangular matrix. (b) Determine the inverses of \(E_{1}, E_{2}, E_{3}\) and set \(L=E_{1}^{-1} E_{2}^{-1} E_{3}^{-1} .\) What type of matrix is \(L ?\) Verify that \(A=L U\)

Consider the matrix \\[ A=\left(\begin{array}{lllll} 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 \end{array}\right) \\] (a) Draw a graph that has \(A\) as its adjacency matrix. Be sure to label the vertices of the graph. (b) By inspecting the graph, determine the number of walks of length 2 from \(V_{2}\) to \(V_{3}\) and from \(V_{2}\) to \(V_{5}\) (c) Compute the second row of \(A^{3},\) and use it to determine the number of walks of length 3 from \(V_{2}\) to \(V_{3}\) and from \(V_{2}\) to \(V_{5}\).

In general, matrix multiplication is not commutative (i.e., \(A B \neq B A\) ). However, in certain special cases the commutative property does hold. Show that (a) if \(D_{1}\) and \(D_{2}\) are \(n \times n\) diagonal matrices, then \(D_{1} D_{2}=D_{2} D_{1}\) (b) if \(A\) is an \(n \times n\) matrix and $$B=a_{0} I+a_{1} A+a_{2} A^{2}+\cdots+a_{k} A^{k}$$ where \(a_{0}, a_{1}, \ldots, a_{k}\) are scalars, then \(A B=B A\)
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