Could a set of three vectors in R* span all of R*? Explain. What about n vectors in R" when n is less than m? Could a set of three vectors in R* span all of R*? Explain. Choose the correct answer below. O A. Yes. A set of n vectors in R" can span R" when n

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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Linear Algebra

**Question:**

Could a set of three vectors in \(\mathbb{R}^4\) span all of \(\mathbb{R}^4\)? Explain. What about \(n\) vectors in \(\mathbb{R}^m\) when \(n\) is less than \(m\)?

Could a set of three vectors in \(\mathbb{R}^4\) span all of \(\mathbb{R}^4\)? Explain. Choose the correct answer below.

**Options:**

- **A.** Yes. A set of \(n\) vectors in \(\mathbb{R}^m\) can span \(\mathbb{R}^m\) when \(n < m\). There is a sufficient number of rows in the matrix \(A\) formed by the vectors to have enough pivot points to show that the vectors span \(\mathbb{R}^m\).

- **B.** Yes. Any number of vectors in \(\mathbb{R}^4\) will span all of \(\mathbb{R}^4\).

- **C.** No. There is no way for any number of vectors in \(\mathbb{R}^4\) to span all of \(\mathbb{R}^4\).

- **D.** No. The matrix \(A\) whose columns are the three vectors has four rows. To have a pivot in each row, \(A\) would have to have at least four columns (one for each pivot).
Transcribed Image Text:**Question:** Could a set of three vectors in \(\mathbb{R}^4\) span all of \(\mathbb{R}^4\)? Explain. What about \(n\) vectors in \(\mathbb{R}^m\) when \(n\) is less than \(m\)? Could a set of three vectors in \(\mathbb{R}^4\) span all of \(\mathbb{R}^4\)? Explain. Choose the correct answer below. **Options:** - **A.** Yes. A set of \(n\) vectors in \(\mathbb{R}^m\) can span \(\mathbb{R}^m\) when \(n < m\). There is a sufficient number of rows in the matrix \(A\) formed by the vectors to have enough pivot points to show that the vectors span \(\mathbb{R}^m\). - **B.** Yes. Any number of vectors in \(\mathbb{R}^4\) will span all of \(\mathbb{R}^4\). - **C.** No. There is no way for any number of vectors in \(\mathbb{R}^4\) to span all of \(\mathbb{R}^4\). - **D.** No. The matrix \(A\) whose columns are the three vectors has four rows. To have a pivot in each row, \(A\) would have to have at least four columns (one for each pivot).
Suppose \( A \) is a \( 4 \times 3 \) matrix and \( b \) is a vector in \( \mathbb{R}^4 \) with the property that \( Ax = b \) has a unique solution. What can you say about the reduced echelon form of \( A \)? Justify your answer.

Choose the correct answer below.

- **A.** The first term of the first row will be a 1 and all other terms will be 0. There is only one variable \( x_m \), so there is only one possible solution.

- **B.** There will be a pivot position in each row. If a row did not have a pivot position then the equation \( Ax = b \) would be inconsistent.

- **C.** The first 3 rows will have a pivot position and the last row will be all zeros. If a row had more than one 1, then there would be an infinite number of solutions for \( a_m x_m = b_m \).

- **D.** The first row will have a pivot position and all other rows will be all zeros. There is only one equation to solve, so there is only one solution.
Transcribed Image Text:Suppose \( A \) is a \( 4 \times 3 \) matrix and \( b \) is a vector in \( \mathbb{R}^4 \) with the property that \( Ax = b \) has a unique solution. What can you say about the reduced echelon form of \( A \)? Justify your answer. Choose the correct answer below. - **A.** The first term of the first row will be a 1 and all other terms will be 0. There is only one variable \( x_m \), so there is only one possible solution. - **B.** There will be a pivot position in each row. If a row did not have a pivot position then the equation \( Ax = b \) would be inconsistent. - **C.** The first 3 rows will have a pivot position and the last row will be all zeros. If a row had more than one 1, then there would be an infinite number of solutions for \( a_m x_m = b_m \). - **D.** The first row will have a pivot position and all other rows will be all zeros. There is only one equation to solve, so there is only one solution.
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