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Suppose the first two columns, by and \(\mathbf{b}_{2},\) of \(B\) are equal. What can you say about the columns of \(A B\) (if \(A B\) is defined)? Why?

Short Answer

Expert verified
The first two columns of \(AB\) are identical because \(b_1 = b_2\).

Step by step solution

01

Understanding Matrix Multiplication

Matrix multiplication involves taking the dot product of the rows of the first matrix (in this case, matrix \(A\)) with the columns of the second matrix (\(B\)). If \(B\) has repeated columns, like \(b_1 = b_2\), this will have implications for the result of \(AB\).
02

Analyzing Repeated Columns in B

Since the columns \(b_1\) and \(b_2\) of matrix \(B\) are equal, when you multiply \(A\) by \(B\), taking the dot product of each row of \(A\) with these repeated columns will yield the same result for the first two columns of \(AB\).
03

Resulting Columns in AB

The product matrix \(AB\) consists of columns that are combinations of the columns of \(A\). Due to the repetition of \(b_1\) and \(b_2\), the corresponding columns in \(AB\), namely \((AB)_{1}\) and \((AB)_{2}\), will thus be equal. Both will be formed by the same linear combinations determined by \(A\).
04

Conclusion on the Columns of AB

Therefore, the first two columns of \(AB\) will be identical. This is because the multiplication of any row of \(A\) by \(b_1\) or \(b_2\) results in the identical combinations being added to each column.

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Key Concepts

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

Repeated Columns in Matrix
When you encounter repeated columns in a matrix, it means that at least two columns are exactly the same. In the context of matrix multiplication, this has interesting implications. Imagine a matrix \( B \) where the first two columns, \( \mathbf{b}_{1} \) and \( \mathbf{b}_{2} \), are repeated.

When you multiply another matrix, say \( A \), by \( B \), you're effectively performing a series of operations involving these columns. Specifically, the repeated nature of columns will lead to identical outcomes in the resulting product matrix for these specific columns.
  • The operation is reliant on the dot products between rows of \( A \) and columns of \( B \).
  • Identical columns in \( B \) will lead to identical column results in the product matrix \( AB \).
This happens because the exact same linear transformations are applied to identical data (those repeated columns), resulting in identical outcomes.
Dot Product in Linear Algebra
The dot product is a fundamental operation in linear algebra, especially when dealing with matrices. It defines how the multiplication between matrices is carried out.

In essence, to obtain an element in the product matrix \( AB \), you take a row from matrix \( A \) and a column from matrix \( B \), then compute the sum of the products of their corresponding elements. This is the dot product.
  • It simplifies multidimensional interactions into a single scalar value.
  • This operation is repeated for each combination of rows from \( A \) and columns from \( B \) to build the entire product matrix.
Understanding this operation is crucial as it underpins the entire concept of matrix multiplication, determining how matrices interact and combine their information.
Matrix Products and Columns
In matrix multiplication, the product matrix columns consist of special linear combinations of columns from the original matrices. When two matrices \( A \) and \( B \) are multiplied, the resulting matrix's columns are derived from \( A \) interacting with each column in \( B \). This relationship showcases the interplay between the structure of \( A \) and the content of \( B \).

Particularly, when \( B \) has repeated columns, like \( \mathbf{b}_{1} = \mathbf{b}_{2} \), it simplifies our expectations for the columns in the resulting product matrix:
  • Each column in the product matrix is influenced by each row from \( A \) and each column from \( B \).
  • In the case of repeated columns in \( B \), the columns of the resulting matrix \( AB \) will be repeated as well.
This naturally follows because every linear combination derived from \( A \) interacting with these repeated columns will be identical, repeating across those columns in the product matrix.

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Most popular questions from this chapter

[MI The band matrix \(A\) shown below can be used to estimate the unsteady conduction of heat in a rod when the temperatures at points \(p_{1}, \ldots, p_{5}\) on the rod change with time. The constant \(C\) in the matrix depends on the physical nature of the rod, the distance \(\Delta x\) between the points on the rod, and the length of time \(\Delta t\) between successive temperature measurements. Suppose that for \(k=0,1,2, \ldots,\) a vector \(\mathbf{t}_{k}\) in \(\mathbb{R}^{5}\) lists the temperatures at time \(k \Delta t\) . If the two ends of the rod are maintained at \(0^{\circ}\) , then the temperature vectors satisfy the equation \(A \mathbf{t}_{k+1}=\mathbf{t}_{k}(k=0,1, \ldots),\) where \(A=\left[\begin{array}{ccccc}{(1+2 C)} & {-C} & {} & {} \\ {-C} & {(1+2 C)} & {-C} & {} \\ {} & {-C} & {(1+2 C)} & {-C} \\ {} & {} & {-C} & {(1+2 C)} & {-C} \\ {} & {} & {} & {-C} & {(1+2 C)}\end{array}\right]\) a. Find the LU factorization of \(A\) when \(C=1 .\) A matrix such as \(A\) with three nonzero diagonals is called a tridiagonal matrix. The \(L\) and \(U\) factors are bidiagonal matrices. b. Suppose \(C=1\) and \(t_{0}=(10,12,12,12,10) .\) Use the LU factorization of \(A\) to find the temperature distributions \(\mathbf{t}_{1}, \mathbf{t}_{2}, \mathbf{t}_{3},\) and \(\mathbf{t}_{4} .\)

Exercises \(9-12\) display a matrix \(A\) and an echelon form of \(A\) . Find bases for \(\operatorname{Col} A\) and \(\mathrm{Nul} A,\) and then state the dimensions of these subspaces. $$ \begin{aligned} A &=\left[\begin{array}{rrrrr}{1} & {-2} & {9} & {5} & {4} \\\ {1} & {-1} & {6} & {5} & {-3} \\ {-2} & {0} & {-6} & {1} & {-2} \\ {4} & {1} & {9} & {1} & {-9}\end{array}\right] \\ & \sim\left[\begin{array}{rrrrr}{1} & {-2} & {9} & {5} & {4} \\ {0} & {1} & {-3} & {0} & {-7} \\ {0} & {0} & {0} & {1} & {-2} \\ {0} & {0} & {0} & {0} & {0}\end{array}\right] \end{aligned} $$

Construct a nonzero \(3 \times 3\) matrix \(A\) and a nonzero vector \(\mathbf{b}\) such that \(\mathbf{b}\) is in Nul \(A .\)

In Exercises 1 and \(2,\) find the vector \(\mathbf{x}\) determined by the given coordinate vector \([\mathbf{x}]_{\mathcal{B}}\) and the given basis \(\mathcal{B}\) . Illustrate your answer with a figure, as in the solution of Practice Problem \(2 .\) $$\mathcal{B}=\left\\{\left[\begin{array}{l}{1} \\\ {1}\end{array}\right],\left[\begin{array}{r}{2} \\\ {-1}\end{array}\right]\right\\},[\mathbf{x}]_{\mathcal{B}}=\left[\begin{array}{l}{3} \\\ {2}\end{array}\right]$$

Exercises 1–4 refer to an economy that is divided into three sectors—manufacturing, agriculture, and services. For each unit of output, manufacturing requires .10 unit from other companies in that sector, .30 unit from agriculture, and .30 unit from services. For each unit of output, agriculture uses .20 unit of its own output, .60 unit from manufacturing, and .10 unit from services. For each unit of output, the services sector consumes .10 unit from services, .60 unit from manufacturing, but no agricultural products. Determine the production levels needed to satisfy a final demand of 18 units for manufacturing, 18 units for agriculture, and 0 units for services.

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