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Suppose Ais 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.

Short Answer

Expert verified

The equation Ax = b has a unique solution.

Step by step solution

01

State that the equation Ax = b has at least one solution

Theorem 4states that if \(A\) is an \({\rm{m}} \times n\) matrix, then the following statements are equivalent.

  1. For each \({\mathop{\rm b}\nolimits} \) in \({\mathbb{R}^m}\), the equation \(Ax = b\) has a solution.
  2. Each \({\mathop{\rm b}\nolimits} \) in \({\mathbb{R}^m}\) is a linear combination of the columns of A.
  3. The columns of \(A\) span .
  4. \(A\)has a pivot position in every row.

By theorem 4, matrix A has a pivot position in each row because the equation Ax = b has a solution for each b.

02

Explain that equation Ax = b has exactly one solution

Since Ais a square matrix, ithas a pivot in each column. The equation Ax = b has no free variables, which demonstrates that the solution is unique.

Thus, the equation Ax = b has a unique solution.

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Most popular questions from this chapter

Question: Determine whether the statements that follow are true or false, and justify your answer.

16: There exists a 2x2 matrix such that

A[11]=[12]andA[22]=[21].

In (a) and (b), suppose the vectors are linearly independent. What can you say about the numbers \(a,....,f\) ? Justify your answers. (Hint: Use a theorem for (b).)

  1. \(\left( {\begin{aligned}{*{20}{c}}a\\0\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}b\\c\\d\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}d\\e\\f\end{aligned}} \right)\)
  2. \(\left( {\begin{aligned}{*{20}{c}}a\\1\\0\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}b\\c\\1\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}d\\e\\f\\1\end{aligned}} \right)\)

Consider the problem of determining whether the following system of equations is consistent for all \({b_1},{b_2},{b_3}\):

\(\begin{aligned}{c}{\bf{2}}{x_1} - {\bf{4}}{x_2} - {\bf{2}}{x_3} = {b_1}\\ - {\bf{5}}{x_1} + {x_2} + {x_3} = {b_2}\\{\bf{7}}{x_1} - {\bf{5}}{x_2} - {\bf{3}}{x_3} = {b_3}\end{aligned}\)

  1. Define appropriate vectors, and restate the problem in terms of Span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\). Then solve that problem.
  1. Define an appropriate matrix, and restate the problem using the phrase “columns of A.”
  1. Define an appropriate linear transformation T using the matrix in (b), and restate the problem in terms of T.

In Exercises 9, write a vector equation that is equivalent to

the given system of equations.

9. \({x_2} + 5{x_3} = 0\)

\(\begin{array}{c}4{x_1} + 6{x_2} - {x_3} = 0\\ - {x_1} + 3{x_2} - 8{x_3} = 0\end{array}\)

Determine the value(s) of \(a\) such that \(\left\{ {\left( {\begin{aligned}{*{20}{c}}1\\a\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}a\\{a + 2}\end{aligned}} \right)} \right\}\) is linearly independent.

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