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Let \(A=\left[\begin{array}{lll}{1} & {1} & {1} \\ {1} & {2} & {3} \\ {1} & {4} & {5}\end{array}\right]\) and \(D=\left[\begin{array}{lll}{2} & {0} & {0} \\\ {0} & {3} & {0} \\ {0} & {0} & {5}\end{array}\right] .\) Compute \(A D\) and \(D A .\) Explain how the columns or rows of \(A\) change when \(A\) is multiplied by \(D\) on the right or on the left. Find a \(3 \times 3\) matrix \(B,\) not the identity matrix or the zero matrix, such that \(A B=B A\)

Short Answer

Expert verified
AD and DA scale the columns and rows of A. B is a diagonal matrix like \( \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \).

Step by step solution

01

Define Matrices A and D

First, let's clearly define the matrices involved. Matrix \( A \) is a \( 3 \times 3 \) matrix given by:\[A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 4 & 5 \end{bmatrix}\]Matrix \( D \) is a diagonal \( 3 \times 3 \) matrix given by:\[D = \begin{bmatrix} 2 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 5 \end{bmatrix}\]
02

Compute AD

To compute \( AD \), multiply each row of matrix \( A \) by the corresponding column of matrix \( D \). Since \( D \) is diagonal, it scales the columns of \( A \):\[AD = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 4 & 5 \end{bmatrix} \begin{bmatrix} 2 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 5 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5 \ 2 & 6 & 15 \ 2 & 12 & 25 \end{bmatrix}\]
03

Compute DA

Now, compute \( DA \) by multiplying each row of \( D \) by the corresponding column of \( A \). The rows of \( A \) are scaled by the diagonal elements of \( D \):\[DA = \begin{bmatrix} 2 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 4 & 5 \end{bmatrix} = \begin{bmatrix} 2 & 2 & 2 \ 3 & 6 & 9 \ 5 & 20 & 25 \end{bmatrix}\]
04

Analyze the Changes

When \( A \) is multiplied by \( D \) on the right, each column of \( A \) is stretched by the corresponding diagonal element in \( D \). When \( D \) multiplies \( A \) on the left, each row of \( A \) is stretched by the corresponding diagonal element in \( D \).
05

Find Matrix B such that AB = BA

We need to find a matrix \( B \) that commutes with \( A \). Since matrices \( A \) and \( D \) are both \( 3 \times 3 \), a diagonal matrix different from the zero matrix or identity matrix could commute with \( A \). Consider:\[B = \begin{bmatrix} 3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3 \end{bmatrix}\]Matrix \( B \) commutes with \( A \) because each element of \( B \) scales corresponding elements in \( A \), thus \( AB = BA \).
06

Verify AB = BA

Verify by calculating:\[AB = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 4 & 5 \end{bmatrix} \begin{bmatrix} 3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 3 & 3 \ 3 & 6 & 9 \ 3 & 12 & 15 \end{bmatrix}\]\[BA = \begin{bmatrix} 3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 4 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 3 & 3 \ 3 & 6 & 9 \ 3 & 12 & 15 \end{bmatrix}\]This confirms \( AB = BA \).

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Key Concepts

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

Diagonal Matrix
A diagonal matrix is a special type of matrix where all elements outside the main diagonal are zero. In other words, it has non-zero numbers only along the diagonal from the top-left to the bottom-right. Diagonal matrices are easy to work with, especially when performing matrix multiplication or other operations, since they simplify computations.

For example, consider matrix \( D \), which is a diagonal matrix:
  • The entry at the first row and first column is 2
  • The entry at the second row and second column is 3
  • The entry at the third row and third column is 5
All other entries are zero. The simplicity of diagonal matrices lies in their ability to scale another matrix, like stretching or shrinking its rows or columns, when they're used in multiplication.

Diagonal matrices are used frequently in linear algebra because they easily encode transformations like scaling. They are also efficient in computations involving eigenvalues and eigenvectors, simplifying more complex matrices into a manageable form by having zero entries off the main diagonal.
Commutativity in Matrices
In algebra, commutativity is a fundamental property, but it gets a bit tricky when it comes to matrices. Unlike regular numbers, matrices generally do not commute. This means that for matrices \( A \) and \( B \), it is not usually true that \( AB = BA \).

However, there are some exceptions to this rule, particularly when one deals with diagonal matrices or when the matrices are of certain special types. For instance, when both matrices are diagonal, they are more likely to commute. This special property is crucial when solving problems related to matrix equations and transformations.

In the context of the exercise, we found a diagonal matrix \( B \) different from both the identity matrix and the zero matrix that commutes with matrix \( A \). Specifically, \(B = \begin{bmatrix}3 & 0 & 0 \0 & 3 & 0 \0 & 0 & 3 \end{bmatrix}\).

This particular commutativity happens because each element in \( B \) scales corresponding elements in \( A \), making the order of multiplication irrelevant. Commuting matrices often imply potential simplifications in computations, which is valuable in both theoretical and practical applications of linear algebra.
Matrix Scaling
Matrix scaling refers to the process of multiplying a matrix by a scalar value, often represented by a diagonal matrix. This type of operation comes into play heavily in transformations and graphical manipulations.

When a matrix is multiplied by a diagonal matrix, it results in each row or column of the original matrix being multiplied by the corresponding diagonal element. This is effectively a scaling of each row or column by a different factor. For instance, when matrix \( A \) is multiplied by matrix \( D \), the diagonal elements in \( D \) act as scaling factors:
  • The first column of \( A \) is multiplied by 2.
  • The second column of \( A \) is multiplied by 3.
  • The third column of \( A \) is multiplied by 5.
This results in a new matrix that is transformed via these scaling operations. Similarly, when matrix \( D \) multiplies matrix \( A \) on the left, the rows of \( A \) are scaled, resulting in a different pattern of transformation.

Matrix scaling is invaluable in fields like computer graphics, where transformations of space through scaling are routinely applied to models and animations. The usage of diagonal matrices for these operations provides a clear and simplified mechanism to achieve such scaling effortlessly.

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Most popular questions from this chapter

In Exercises \(3-8,\) find the \(3 \times 3\) matrices that produce the described composite 2 \(\mathrm{D}\) transformations, using homogeneous coordinates. Rotate points \(30^{\circ},\) and then reflect through the \(x\) -axis.

[M] The consumption matrix \(C\) below is based on input- output data for the U.S. economy in 1958 , with data for 81 sectors grouped into 7 larger sectors: \((1)\) nonmetal household and personal products, \((2)\) final metal products (such as motor vehicles), (3) basic metal products and mining, (4) basic nonmetal products and agriculture, (5) energy, (6) services, and (7) entertainment and miscellaneous products. Find the production levels needed to satisfy the final demand d. Units are in millions of dollars.) $$ \left[\begin{array}{ccccccc}{.1588} & {.0064} & {.0025} & {.0304} & {.0014} & {.0083} & {.1594} \\ {.0057} & {.2645} & {.0436} & {.0099} & {.0083} & {.0201} & {.3413} \\ {.0264} & {1506} & {.3557} & {.0139} & {.0142} & {.0070} & {.0236} \\ {.3299} & {.0565} & {.0495} & {.3636} & {.0204} & {.0483} & {.0649} \\ {.0089} & {.0081} & {.0333} & {.0295} & {.3412} & {.04837} & {.0020} \\ {.1190} & {.0901} & {.0396} & {.1260} & {.1722} & {.2368} & {.3369} \\ {.0063} & {.0126} & {.0196} & {.0098} & {.0064} & {.0132} & {.0012}\end{array}\right] $$ $$ \mathbf{d}=\left[\begin{array}{c}{74,000} \\ {56,000} \\ {10,500} \\\ {25,000} \\ {17,500} \\ {196,000} \\ {5,000}\end{array}\right] $$

Let \(C\) be a consumption matrix such that \(C^{m} \rightarrow 0\) as \(m \rightarrow \infty,\) and for \(m=1,2, \ldots,\) let \(D_{m}=I+C+\cdots+\) \(C^{m} .\) Find a difference equation that relates \(D_{m}\) and \(D_{m+1}\) and thereby obtain an iterative procedure for computing formula (8) for \((I-C)^{-1} .\)

In Exercises \(3-6,\) the vector \(\mathbf{x}\) is in a subspace \(H\) with a basis \(\mathcal{B}=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\\} .\) Find the \(\mathcal{B}\) -coordinate vector of \(\mathbf{x} .\) $$ \mathbf{b}_{1}=\left[\begin{array}{r}{1} \\ {-3}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{r}{-3} \\ {5}\end{array}\right], \mathbf{x}=\left[\begin{array}{r}{-7} \\ {5}\end{array}\right] $$

In Exercises 37 and 38, construct bases for the column space and the null space of the given matrix A. Justify your work. $$ A=\left[\begin{array}{rrrrr}{3} & {-5} & {0} & {-1} & {3} \\ {-7} & {9} & {-4} & {9} & {-11} \\ {-5} & {7} & {-2} & {5} & {-7} \\ {3} & {-7} & {-3} & {4} & {0}\end{array}\right] $$

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