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Linear Algebra and its Applications
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Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Suppose the coefficient matrix of a linear system of three equations in three variables has a pivot position in each column. Explain why the system has a unique solution.

The system has a unique solution because it is row reduced to an identity matrix of the order \(3 \times 3\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Identify the condition for a unique solution

The system of three equations is fulfilled if the three planes cross at a single location.

As a result, the echelon form that can be produced by solving this system of equations is an identity matrix of the order \(3 \times 3\).

Thus, the echelon form of the consistent linear system is an identity matrix of the order \(3 \times 3\).

Step 2: Prove that the system has a unique solution

It is given that the coefficient matrix of a linear system of three equations in three variables has a pivot position in each column. So, the system in span \({\mathbb{R}^3}\) is row reduced to identity matrix echelon form.

Thus, the system has a unique solution.

Most popular questions for Math Textbooks

In Exercises 15 and 16, list five vectors in Span \(\left\{ {{v_1},{v_2}} \right\}\). For each vector, show the weights on \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) used to generate the vector and list the three entries of the vector. Do not make a sketch.

15. \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\)

Write the reduced echelon form of a \(3 \times 3\) matrix A such that the first two columns of A are pivot columns and

\(A = \left( {\begin{aligned}{*{20}{c}}3\\{ - 2}\\1\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\\0\end{aligned}} \right)\).

Let \(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&3&{ - 2}\\{ - 2}&6&3\end{array}} \right]\) and \(b = \left[ {\begin{array}{*{20}{c}}4\\1\\{ - 4}\end{array}} \right]\). Denote the columns of \(A\) by \({{\mathop{\rm a}\nolimits} _1},{a_2},{a_3}\) and let \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).

  1. Is \(b\) in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)? How many vectors are in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)?
  2. Is \(b\) in \(W\)? How many vectors are in W.
  3. Show that \({a_1}\) is in W. [Hint: Row operations are unnecessary.]

Suppose A is an \(n \times n\) matrix with the property that the equation \(A{\mathop{\rm x}\nolimits} = 0\) has at least one solution for each b in \({\mathbb{R}^n}\). Without using Theorem 5 or 8, explain why each equation Ax = b has in fact exactly one solution.

In Exercises 6, write a system of equations that is equivalent to the given vector equation.

6. \({x_1}\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}8\\5\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}1\\{ - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\)
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