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Chapter 6: Problem 14

Use elementary row operations to transform the following matrices to the form: $$ \left[\begin{array}{lll} 1 & 0 & x \\ 0 & 1 & x \end{array}\right] $$ where \(x\) represents some arbitrary number. (a) \(\left[\begin{array}{lll}2 & 1 & 1 \\ 1 & 2 & 0\end{array}\right]\) (b) \(\left[\begin{array}{rrr}1 & 4 & 1 \\ 9 & 35 & 7\end{array}\right]\)

Short Answer

Expert verified
For (a): \(\begin{bmatrix}1 & 0 & \frac{2}{3} \ 0 & 1 & -\frac{1}{3} \ \end{bmatrix}\). For (b): \(\begin{bmatrix}1 & 0 & -7 \ 0 & 1 & 2 \ \end{bmatrix}\).

Step by step solution

01

(a) - Obtain a Leading 1 in the First Row, First Column

To make the element in the first row, first column of matrix (a) equal to 1, divide the entire first row by 2. The matrix becomes: \[ \begin{bmatrix}1 & \frac{1}{2} & \frac{1}{2} \ 1 & 2 & 0 \ d{bmatrix} \]
02

(a) - Eliminate the First Column, Second Row

To make the element in the second row, first column zero, subtract row 1 from row 2. The matrix becomes: \[ \begin{bmatrix}1 & \frac{1}{2} & \frac{1}{2} \ 0 & \frac{3}{2} & -\frac{1}{2} \ d{bmatrix} \]
03

(a) - Obtain a Leading 1 in the Second Row, Second Column

To make the element in the second row, second column equal to 1, divide the entire second row by \(\frac{3}{2}\). The matrix becomes: \[ \begin{bmatrix}1 & \frac{1}{2} & \frac{1}{2} \ 0 & 1 & -\frac{1}{3} \ d{bmatrix} \]
04

(a) - Eliminate the Second Column, First Row

To make the element in the first row, second column zero, subtract \(\frac{1}{2}\) times the second row from the first row. The matrix becomes: \[ \begin{bmatrix}1 & 0 & \frac{2}{3} \ 0 & 1 & -\frac{1}{3} \ d{bmatrix} \]
05

(b) - Obtain a Leading 1 in the First Row, First Column

The first row already has a 1 in the first column, so no row operation is needed for it. The matrix remains: \[ \begin{bmatrix}1 & 4 & 1 \ 9 & 35 & 7 \ d{bmatrix} \]
06

(b) - Eliminate the First Column, Second Row

To make the element in the second row, first column zero, subtract 9 times the first row from the second row. The matrix becomes: \[ \begin{bmatrix}1 & 4 & 1 \ 0 & -1 & -2 \ d{bmatrix} \]
07

(b) - Obtain a Leading 1 in the Second Row, Second Column

To make the element in the second row, second column equal to 1, multiply the entire second row by -1. The matrix becomes: \[ \begin{bmatrix}1 & 4 & 1 \ 0 & 1 & 2 \ d{bmatrix} \]
08

(b) - Eliminate the Second Column, First Row

To make the element in the first row, second column zero, subtract 4 times the second row from the first row. The matrix becomes: \[ \begin{bmatrix}1 & 0 & -7 \ 0 & 1 & 2 \ d{bmatrix} \]

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Key Concepts

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

matrix transformation
Matrix transformation involves applying a series of operations to change the form of a matrix. In elementary row operations, we perform three types of operations on the rows of a matrix:
  • Swapping two rows
  • Multiplying a row by a non-zero scalar
  • Adding or subtracting multiples of rows
The goal is to reach the desired matrix form by systematically applying these operations.
Gauss-Jordan elimination
Gauss-Jordan elimination is a method to transform a matrix into reduced row echelon form (RREF). This method simplifies solving systems of linear equations and finding matrix inverses. The goal is to make every leading entry in a row 1, and all other entries in that column 0. The steps involved are:
  • Obtain a leading 1 in each row
  • Make zeros below and above each leading 1
  • Ensure each leading 1 is the only non-zero entry in its column
In this exercise, Gauss-Jordan elimination helps us transform the given matrices to the desired form.
linear algebra
Linear algebra is the branch of mathematics that deals with vectors, vector spaces, and linear transformations. Key concepts include:
  • Vectors and vector spaces
  • Matrices and matrix operations
  • Systems of linear equations
  • Determinants and eigenvalues
This exercise demonstrates matrix transformations and operations that are fundamental in linear algebra.

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Most popular questions from this chapter

In order that $$ A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] $$ will commute with $$ B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] $$ what conditions must be satisfied by \(a, b, c\), and \(d ?\)

Let \(A\) be a square matrix. Show that $$ \left(I+2 A+3 A^{2}\right)(2 I-A)=(2 I-A)\left(I+2 A+3 A^{2}\right) $$ thus showing that the matrices \(I+2 A+3 A^{2}\) and \(2 I-A\) commute. Remark: Let $$ \begin{aligned} &p(t)=a_{0}+a_{1} t+\cdots+a_{k} t^{k} \\ &q(t)=b_{0}+b_{1} t+\cdots+b_{\ell} t^{2} \end{aligned} $$ be polynomials of degrees \(k\) and \(\ell\). Define matrices $$ p(A)=a_{0} I+a_{1} A+\cdots+a_{k} A^{k} $$ $$ q(A)=b_{0} I+b_{1} A+\cdots+b_{\ell} A^{\ell} $$ It can be shown that \(p(A) q(A)=q(A) p(A)\).

Consider the matrix $$ A=\left[\begin{array}{ccccc} 3 & 1 & 0 & \cdots & 0 \\ 1 & 3 & 1 & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & 1 & 3 & 1 \\ 0 & \cdots & 0 & 1 & 3 \end{array}\right] $$ and the vectors $$ b=\left[\begin{array}{c} 4 \\ 5 \\ 5 \\ \vdots \\ 5 \\ 4 \end{array}\right], \quad x=\left[\begin{array}{c} 1 \\ 1 \\ 1 \\ \vdots \\ 1 \\ 1 \end{array}\right] $$ Set up the linear system (6.2) of order \(n\) associated with \(A\) and \(b\). Verify that the given \(x\) solves this linear system.

Define \(B=w w^{T}\), with \(w\) a column vector of length \(n\). (a) Give an operations count for forming \(B .\) Be as efficient as possible. (b) Let \(A\) be an arbitrary matrix of order \(n \times n\). Give the additional number of operations needed to form the product \(A\) and \(B\), using the matrix \(B\) formed in (a). (c) Give an alternative and less costly way, in operations, to form the product \(A B=A\left(w w^{T}\right)\)

Let \(A\) be a square matrix of order \(n\), and let it satisfy $$ a_{i j}=0 \quad \text { for } \quad i>j $$ Such a matrix is called upper triangular. Show that the sum and products of such matrices are also upper triangular. Is the inverse upper triangular?
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