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Chapter 2: Q2.3-31Q (page 93)

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

Let \({{\bf{r}}_1} \ldots ,{{\bf{r}}_p}\) be vectors in \({\mathbb{R}^{\bf{n}}}\), and let Qbe an\(m \times n\)matrix. Write the matrix\(\left( {\begin{aligned}{*{20}{c}}{Q{{\bf{r}}_1}}& \cdots &{Q{{\bf{r}}_p}}\end{aligned}} \right)\)as a productof two matrices (neither of which is an identity matrix).

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)

Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).

Suppose P is invertible and \(A = PB{P^{ - 1}}\). Solve for Bin terms of A.
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